Tag: Best NCERT Class 12th Solutions Maths

Exercise-5.3, Class 12th, Maths, Chapter 5, NCERT
Find $\frac{d y}{d x}$ in the following:

Question 1

Differentiate:

$2 x + 3 y = \sin x$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (2 x) + \frac{d}{d x} (3 y) = \frac{d}{d x} (\sin x)$

$2 + 3 \frac{d y}{d x} = \cos x$

Now solve for $\frac{d y}{d x}$ :

$3 \frac{d y}{d x} = \cos x - 2$

$\frac{d y}{d x} = \frac{\cos x - 2}{3}$

Question 2

Differentiate:

$2 x + 3 y = \sin y$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (2 x) + \frac{d}{d x} (3 y) = \frac{d}{d x} (\sin y)$ $2 + 3 \frac{d y}{d x} = \cos y \cdot \frac{d y}{d x}$

(using chain rule on $\sin y$ )

Now collect $\frac{d y}{d x}$ terms on one side:

$3 \frac{d y}{d x} - \cos y \frac{d y}{d x} = - 2$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (3 - \cos y) = - 2$

So,

$\frac{d y}{d x} = \frac{- 2}{3 - \cos y}$

Question 3

Find $\frac{d y}{d x}$ if:

$a x + b y^{2} = \cos y$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (a x) + \frac{d}{d x} (b y^{2}) = \frac{d}{d x} (\cos y)$

Differentiate term by term

$a + b \cdot 2 y \frac{d y}{d x} = - \sin y \cdot \frac{d y}{d x}$

(using chain rule on cos y)

So,

$a + 2 b y \frac{d y}{d x} = - \sin y \frac{d y}{d x}$

Move all $\frac{d y}{d x}$ terms to one side:

$2 b y \frac{d y}{d x} + \sin y \frac{d y}{d x} = - a$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (2 b y + \sin y) = - a$

Final Answer

$\frac{d y}{d x} = \frac{- a}{2 b y + \sin y}$

Question 4

Find $\frac{d y}{d x}$ if:

$x y + y^{2} = \tan x + y$

Solution

Differentiate both sides w.r.t. $x$ :

$\frac{d}{d x} (x y) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (\tan x) + \frac{d}{d x} (y)$

Differentiate term by term
1. $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$
  (product rule)
2. $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$
3. $\frac{d}{d x} (\tan x) = \sec^{2} x$
4. $\frac{d}{d x} (y) = \frac{d y}{d x}$
Substitute all derivatives

$x \frac{d y}{d x} + y + 2 y \frac{d y}{d x} = \sec^{2} x + \frac{d y}{d x}$

Collect $\frac{d y}{d x}$ terms on one side:

$x \frac{d y}{d x} + 2 y \frac{d y}{d x} - \frac{d y}{d x} = \sec^{2} x - y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x + 2 y - 1) = \sec^{2} x - y$

Final Answer

$\frac{d y}{d x} = \frac{\sec^{2} x - y}{x + 2 y - 1}$

Find $\frac{d y}{d x}$ in the following:

Question 1

Differentiate:

$2 x + 3 y = \sin x$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (2 x) + \frac{d}{d x} (3 y) = \frac{d}{d x} (\sin x)$

$2 + 3 \frac{d y}{d x} = \cos x$

Now solve for $\frac{d y}{d x}$ :

$3 \frac{d y}{d x} = \cos x - 2$

$\frac{d y}{d x} = \frac{\cos x - 2}{3}$

Question 2

Differentiate:

$2 x + 3 y = \sin y$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (2 x) + \frac{d}{d x} (3 y) = \frac{d}{d x} (\sin y)$ $2 + 3 \frac{d y}{d x} = \cos y \cdot \frac{d y}{d x}$

(using chain rule on $\sin y$ )

Now collect $\frac{d y}{d x}$ terms on one side:

$3 \frac{d y}{d x} - \cos y \frac{d y}{d x} = - 2$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (3 - \cos y) = - 2$

So,

$\frac{d y}{d x} = \frac{- 2}{3 - \cos y}$

Question 3

Find $\frac{d y}{d x}$ if:

$a x + b y^{2} = \cos y$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (a x) + \frac{d}{d x} (b y^{2}) = \frac{d}{d x} (\cos y)$

Differentiate term by term

$a + b \cdot 2 y \frac{d y}{d x} = - \sin y \cdot \frac{d y}{d x}$

(using chain rule on cos y)

So,

$a + 2 b y \frac{d y}{d x} = - \sin y \frac{d y}{d x}$

Move all $\frac{d y}{d x}$ terms to one side:

$2 b y \frac{d y}{d x} + \sin y \frac{d y}{d x} = - a$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (2 b y + \sin y) = - a$

Final Answer

$\frac{d y}{d x} = \frac{- a}{2 b y + \sin y}$

Question 4

Find $\frac{d y}{d x}$ if:

$x y + y^{2} = \tan x + y$

Solution

Differentiate both sides w.r.t. $x$ :

$\frac{d}{d x} (x y) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (\tan x) + \frac{d}{d x} (y)$

Differentiate term by term
1. $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$
  (product rule)
2. $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$
3. $\frac{d}{d x} (\tan x) = \sec^{2} x$
4. $\frac{d}{d x} (y) = \frac{d y}{d x}$
Substitute all derivatives

$x \frac{d y}{d x} + y + 2 y \frac{d y}{d x} = \sec^{2} x + \frac{d y}{d x}$

Collect $\frac{d y}{d x}$ terms on one side:

$x \frac{d y}{d x} + 2 y \frac{d y}{d x} - \frac{d y}{d x} = \sec^{2} x - y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x + 2 y - 1) = \sec^{2} x - y$

Final Answer

$\frac{d y}{d x} = \frac{\sec^{2} x - y}{x + 2 y - 1}$

Question 5

Find $\frac{d y}{d x}$ if:

$x^{2} + x y + y^{2} = 100$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (x^{2}) + \frac{d}{d x} (x y) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (100)$

Now apply differentiation rules:

Term-by-term differentiation
1. $\frac{d}{d x} (x^{2}) = 2 x$
2. $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$
3. (product rule)
4. $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$
5. $\frac{d}{d x} (100) = 0$
Substitute into the equation

$2 x + (x \frac{d y}{d x} + y) + 2 y \frac{d y}{d x} = 0$

Group $\frac{d y}{d x}$ terms:

$x \frac{d y}{d x} + 2 y \frac{d y}{d x} = - 2 x - y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x + 2 y) = - 2 x - y$

Final Answer

$\frac{d y}{d x} = \frac{- 2 x - y}{x + 2 y}$

Question 6

Find $\frac{d y}{d x}$ if:

$x^{3} + x^{2} y + x y^{2} + y^{3} = 81$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (x^{3}) + \frac{d}{d x} (x^{2} y) + \frac{d}{d x} (x y^{2}) + \frac{d}{d x} (y^{3}) = \frac{d}{d x} (81)$

Differentiate term-by-term
1. $\frac{d}{d x} (x^{3}) = 3 x^{2}$
2. $\frac{d}{d x} (x^{2} y) = 2 x y + x^{2} \frac{d y}{d x}$
  (product rule)
3. $\frac{d}{d x} (x y^{2}) = y^{2} + 2 x y \frac{d y}{d x}$
  (product rule)
4. $\frac{d}{d x} (y^{3}) = 3 y^{2} \frac{d y}{d x}$
  (chain rule)
5. $\frac{d}{d x} (81) = 0$
Substitute everything back

$3 x^{2} + 2 x y + x^{2} \frac{d y}{d x} + y^{2} + 2 x y \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = 0$

Group the terms containing $\frac{d y}{d x}$ :

$x^{2} \frac{d y}{d x} + 2 x y \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = - 3 x^{2} - 2 x y - y^{2}$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x^{2} + 2 x y + 3 y^{2}) = - (3 x^{2} + 2 x y + y^{2})$

Final Answer

$\frac{d y}{d x} = - \frac{3 x^{2} + 2 x y + y^{2}}{x^{2} + 2 x y + 3 y^{2}}$

Question 7

Find $\frac{d y}{d x}$ if:

$\sin^{2} y + \cos x y = k$ Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (\sin^{2} y) + \frac{d}{d x} (\cos x y) = \frac{d}{d x} (k)$

Differentiate term by term

1. $\sin^{2} y$

Let $u = \sin y$ , so expression = $u^{2}$

$\frac{d}{d x} (\sin^{2} y) = 2 \sin y \cdot \cos y \cdot \frac{d y}{d x}$

(because $\frac{d}{d x} (\sin y) = \cos y \frac{d y}{d x}$

So: $2 \sin y \cos y \frac{d y}{d x}$

2. $\cos (x y)$

Apply chain rule and product rule inside:

Derivative of $\cos A$ is $- \sin A$ :

$\frac{d}{d x} (\cos (x y)) = - \sin (x y) \cdot \frac{d}{d x} (x y)$

Now:

$\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$

(using product rule)

So:

$\frac{d}{d x} (\cos (x y)) = - \sin (x y) (x \frac{d y}{d x} + y)$

Right-hand side

$\frac{d}{d x} (k) = 0$

Combine all results

$2 \sin y \cos y \frac{d y}{d x} - \sin (x y) (x \frac{d y}{d x} + y) = 0$

Expand:

$2 \sin y \cos y \frac{d y}{d x} - x \sin (x y) \frac{d y}{d x} - y \sin (x y) = 0$

Group the terms with $\frac{d y}{d x}$ :

$\frac{d y}{d x} (2 \sin y \cos y - x \sin (x y)) = y \sin (x y)$

Final Answer

$\frac{d y}{d x} = \frac{y \sin (x y)}{2 \sin y \cos y - x \sin (x y)}$

Question 8

Find $\frac{d y}{d x}$ if:

$\sin^{2} x + \cos^{2} y = 1$

Solution

Differentiate both sides with respect to $x$ :

$\frac{d}{d x} (\sin^{2} x) + \frac{d}{d x} (\cos^{2} y) = \frac{d}{d x} (1)$

Differentiate term by term

1. $\sin^{2} x$

Let $u = \sin x$ , then $u^{2} = \sin^{2} x$

$\frac{d}{d x} (\sin^{2} x) = 2 \sin x \cdot \cos x$

2. $\cos^{2} y$

Let $v = \cos y$ , so $v^{2} = \cos^{2} y$

$\frac{d}{d x} (\cos^{2} y) = 2 \cos y \cdot (- \sin y) \frac{d y}{d x}$

(using chain rule: derivative of $\cos y$ is $- \sin y \frac{d y}{d x}$ )

So:

$\frac{d}{d x} (\cos^{2} y) = - 2 \cos y \sin y \frac{d y}{d x}$

Right-hand side

$\frac{d}{d x} (1) = 0$

Put everything together

$2 \sin x \cos x - 2 \cos y \sin y \frac{d y}{d x} = 0$

Move the second term to the other side:

$2 \sin x \cos x = 2 \cos y \sin y \frac{d y}{d x}$

Divide both sides by $2 \cos y \sin y$ :

$\frac{d y}{d x} = \frac{\sin x \cos x}{\cos y \sin y}$

Final Answer

$\frac{d y}{d x} = \frac{\sin x \cos x}{\sin y \cos y}$

Question 9

Find $\frac{d y}{d x}$ if:

$y = \sin^{- 1} (\frac{2 x}{1 + x^{2}})$

Solution

Let: $y = \sin^{- 1} (\frac{2 x}{1 + x^{2}})$

We know the identity:

$\sin (2 θ) = \frac{2 \tan θ}{1 + \tan^{2} θ}$ Comparing:

$\frac{2 x}{1 + x^{2}} = \sin (2 θ) if x = \tan θ$ So let: $x = \tan θ \Rightarrow θ = \tan^{- 1} x$

Then: $y = \sin^{- 1} (\sin (2 θ)) = 2 θ = 2 \tan^{- 1} x$

Now differentiate:

$y = 2 \tan^{- 1} x$

$\frac{d y}{d x} = 2 \cdot \frac{1}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

Question 10

Find $\frac{d y}{d x}$ if:

$y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$

Solution

We use the trigonometric identity:

$\tan (3 θ) = \frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ}$

Compare with the given expression:

$\frac{3 x - x^{3}}{1 - 3 x^{2}} = \tan (3 θ) if x = \tan θ$

So let:

$x = \tan θ \Rightarrow θ = \tan^{- 1} x$

Then: $y = \tan^{- 1} (\tan (3 θ)) = 3 θ = 3 \tan^{- 1} x$

The given interval: $- \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ ensures:

$∣ x ∣ < \frac{1}{\sqrt{3}} \Rightarrow ∣ θ ∣ < \frac{π}{6} \Rightarrow ∣ 3 θ ∣ < \frac{π}{2}$

Thus, $\tan^{- 1} (\tan (3 θ)) = 3 θ$ is valid (i.e., no ambiguity or discontinuity).

Differentiate

$y = 3 \tan^{- 1} x$ $\frac{d y}{d x} = 3 \cdot \frac{1}{1 + x^{2}}$ Final Answer

$\frac{d y}{d x} = \frac{3}{1 + x^{2}}, - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$

Question 11

Find $\frac{d y}{d x}$ if:

$y = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1$ Solution

Let:

$y = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$

We use a trigonometric substitution.

Let:

$x = \tan θ \Rightarrow θ = \tan^{- 1} x$

Compute the inside expression:

$\frac{1 - x^{2}}{1 + x^{2}} = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$

We know the identity:

$\cos (2 θ) = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$

Thus:

$\frac{1 - x^{2}}{1 + x^{2}} = \cos (2 θ)$

So: $y = \cos^{- 1} (\cos (2 θ)) = 2 θ$

Since $0 < x < 1 \Rightarrow 0 < θ < \frac{π}{4}$ , thus $0 < 2 θ < \frac{π}{2}$ , ensuring the inverse cosine gives principal value.

Therefore:

$y = 2 θ = 2 \tan^{- 1} x$

Differentiate

$\frac{d y}{d x} = 2 \cdot \frac{1}{1 + x^{2}}$

Final Answer

$\frac{d y}{d x} = \frac{2}{1 + x^{2}}, 0 < x < 1$

Question 12

Find $\frac{d y}{d x}$ if:

$y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1$ Solution

Let:

$y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$

We use trigonometric substitution similar to the previous problem.

Let:

$x = \tan θ \Rightarrow θ = \tan^{- 1} x$

Then:

$\frac{1 - x^{2}}{1 + x^{2}} = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$

Using the identity:

$\cos (2 θ) = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$

So: $y = \sin^{- 1} (\cos (2 θ))$

We also know:

$\cos (2 θ) = \sin (\frac{π}{2} - 2 θ)$

Thus: $y = \sin^{- 1} (\sin (\frac{π}{2} - 2 θ)) = \frac{π}{2} - 2 θ$

Since $0 < x < 1$ , $0 < θ < \frac{π}{4}$ , so:

$0 < 2 θ < \frac{π}{2} \Rightarrow ∣ \frac{π}{2} - 2 θ ∣ \leq \frac{π}{2}$

Hence the principal value is correct.

Differentiate

$y = \frac{π}{2} - 2 θ = \frac{π}{2} - 2 \tan^{- 1} x$

$\frac{d y}{d x} = - 2 \cdot \frac{1}{1 + x^{2}}$ Final Answer

$\frac{d y}{d x} = - \frac{2}{1 + x^{2}}, 0 < x < 1$

Question 13

$y = \cos^{- 1} (\frac{2 x}{1 + x^{2}}), - 1 < x < 1$

Now we will solve this correctly.

Solution

Let:

$y = \cos^{- 1} (\frac{2 x}{1 + x^{2}})$

Use substitution:

$x = \tan θ \Rightarrow θ = \tan^{- 1} x$

Then:

$\frac{2 x}{1 + x^{2}} = \frac{2 \tan θ}{1 + \tan^{2} θ}$

We know the identity:

$\sin (2 θ) = \frac{2 \tan θ}{1 + \tan^{2} θ}$

So:

$\frac{2 x}{1 + x^{2}} = \sin (2 θ)$

Thus:

$y = \cos^{- 1} (\sin (2 θ))$

We also know: $\sin (2 θ) = \cos (\frac{π}{2} - 2 θ)$

So: $y = \frac{π}{2} - 2 θ = \frac{π}{2} - 2 \tan^{- 1} x$

Differentiate

$y = \frac{π}{2} - 2 \tan^{- 1} x$

$\frac{d y}{d x} = - 2 \cdot \frac{1}{1 + x^{2}}$ Final Answer

$\frac{d y}{d x} = - \frac{2}{1 + x^{2}}, - 1 < x < 1$

Question 14

Find $\frac{d y}{d x}$ if:

$y = \sin^{- 1} (2 x \sqrt{1 - x^{2}}), - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$ Solution

Let: $y = \sin^{- 1} (2 x \sqrt{1 - x^{2}})$

We use a substitution to simplify.

Let:

$x = \sin θ \Rightarrow θ = \sin^{- 1} x$

Then:

$\sqrt{1 - x^{2}} = \sqrt{1 - \sin^{2} θ} = \cos θ$

So:

$2 x \sqrt{1 - x^{2}} = 2 \sin θ \cos θ = \sin (2 θ)$

Thus: $y = \sin^{- 1} (\sin (2 θ)) = 2 θ$

The given interval:

$- \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

gives:

$- \frac{π}{4} < θ < \frac{π}{4} \Rightarrow ∣ 2 θ ∣ < \frac{π}{2}$

Therefore: $y = 2 θ = 2 \sin^{- 1} x$

Differentiate

$y = 2 \sin^{- 1} x$

$\frac{d y}{d x} = 2 \cdot \frac{1}{\sqrt{1 - x^{2}}}$ Final Answer

$\frac{d y}{d x} = \frac{2}{\sqrt{1 - x^{2}}}, - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

Question 15

Find $\frac{d y}{d x}$ if:

$y = \sec^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}}$ Solution

Let:

$y = \sec^{- 1} (u), where u = \frac{1}{2 x^{2} - 1}$

We know:

$\frac{d}{d x} (\sec^{- 1} u) = \frac{1}{∣ u ∣ \sqrt{u^{2} - 1}} \cdot \frac{d u}{d x}$

Step 1: Differentiate the inside function

$u = \frac{1}{2 x^{2} - 1} = (2 x^{2} - 1)^{- 1}$

$\frac{d u}{d x} = - 1 (2 x^{2} - 1)^{- 2} \cdot (4 x) = - \frac{4 x}{(2 x^{2} - 1)^{2}}$

Step 2: Apply inverse secant derivative formula

$\frac{d y}{d x} = \frac{1}{∣ \frac{1}{2 x^{2} - 1} ∣ \sqrt{{(\frac{1}{2 x^{2} - 1})}^{2} - 1}} \cdot (- \frac{4 x}{(2 x^{2} - 1)^{2}})$

Since $0 < x < \frac{1}{\sqrt{2}}$ , we have $2 x^{2} - 1 < 0$ , so:

$∣ \frac{1}{2 x^{2} - 1} ∣ = - \frac{1}{2 x^{2} - 1}$

Thus:

$\frac{d y}{d x} = (- \frac{2 x^{2} - 1}{1}) \cdot \frac{1}{\sqrt{\frac{1 - (2 x^{2} - 1)^{2}}{(2 x^{2} - 1)^{2}}}} \cdot (- \frac{4 x}{(2 x^{2} - 1)^{2}})$

Simplify inner root:

$1 - (2 x^{2} - 1)^{2} = 1 - (4 x^{4} - 4 x^{2} + 1) = 4 x^{2} - 4 x^{4} = 4 x^{2} (1 - x^{2})$

So:

$\sqrt{\frac{4 x^{2} (1 - x^{2})}{(2 x^{2} - 1)^{2}}} = \frac{2 x \sqrt{1 - x^{2}}}{∣ 2 x^{2} - 1 ∣}$

Since $2 x^{2} - 1 < 0$ , $∣ 2 x^{2} - 1 ∣ = 1 - 2 x^{2}$ :

$\sqrt{\frac{4 x^{2} (1 - x^{2})}{(2 x^{2} - 1)^{2}}} = \frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}}$

Final Simplification

$\frac{d y}{d x} = - \frac{4 x}{(2 x^{2} - 1)^{2}} \cdot \frac{(2 x^{2} - 1)}{\frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}}}$ $\frac{d y}{d x} = - \frac{4 x}{(2 x^{2} - 1)} \cdot \frac{1}{2 x \sqrt{1 - x^{2}}}$

Cancel $x$ :

$\frac{d y}{d x} = - \frac{4}{(2 x^{2} - 1)} \cdot \frac{1}{2 \sqrt{1 - x^{2}}}$

$\frac{d y}{d x} = - \frac{2}{(2 x^{2} - 1) \sqrt{1 - x^{2}}}, 0 < x < \frac{1}{\sqrt{2}}$ Final Answer

$\frac{d y}{d x} = - \frac{2}{(2 x^{2} - 1) \sqrt{1 - x^{2}}}$
November 25, 2025
Exercise-5.2, Class 12th, Maths, Chapter 5, NCERT

Differentiate the functions with respect to x in Exercises 1 to 8.

1. Differentiate: $\sin (x^{2} + 5)$ with respect to $x$ .

Solution

We need to differentiate the function:

$y = \sin (x^{2} + 5)$

This is a composite function, so we will apply the Chain Rule

Let:

$u = x^{2} + 5 \Rightarrow y = \sin (u)$

Differentiate both:

$\frac{d y}{d u} = \cos (u)$ $\frac{d u}{d x} = 2 x$

Now apply Chain Rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $\frac{d y}{d x} = \cos (u) \cdot 2 x$

Substitute $u = x^{2} + 5$ :

$\frac{d y}{d x} = 2 x \cos (x^{2} + 5)$

Question 2

Differentiate with respect to $x$ :

$\cos (\sin x)$

Solution

Given:

$y = \cos (\sin x)$

This is a composite function, where:

$u = \sin x \Rightarrow y = \cos (u)$

Now differentiate step-by-step:

Step 1: Differentiate outer function

$\frac{d y}{d u} = - \sin (u)$

Step 2: Differentiate inner function

$\frac{d u}{d x} = \cos x$

Step 3: Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Substitute back $u = \sin x$ :

$\frac{d y}{d x} = - \sin (\sin x) \cos x$

Question 3

Differentiate with respect to $x$ :

$\sin (a x + b)$

Solution

Let:

$y = \sin (a x + b)$

This is again a composite function, so we apply the Chain Rule.

Step 1:

Let the inner function:

$u = a x + b$

Then,

$y = \sin (u)$

Step 2: Differentiate

$\frac{d y}{d u} = \cos (u)$

Step 3: Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Substitute $u = a x + b$ :

$\frac{d y}{d x} = a \cos (a x + b)$

Question 4

Differentiate with respect to $x$ :

$\frac{d}{d x} [\sec (\tan (\sqrt{x}))$

Solution

Let:

$u = \tan (\sqrt{x})$

$y = \sec (u)$

Differentiate outer function

$\frac{d y}{d u} = \sec (u) \tan (u)$

Differentiate inner function

$u = \tan (v), where v = \sqrt{x}$

$\frac{d u}{d v} = \sec^{2} (v)$

Differentiate

$v = \sqrt{x}$ $\frac{d v}{d x} = \frac{1}{2 \sqrt{x}}$

Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x}$ $\frac{d y}{d x} = \sec (u) \tan (u) \cdot \sec^{2} (v) \cdot \frac{1}{2 \sqrt{x}}$

Substitute $u = \tan (\sqrt{x})$ , $v = \sqrt{x}$ :

Final Answer

$\frac{d}{d x} [\sec (\tan (\sqrt{x}))] = \frac{1}{2 \sqrt{x}} \sec (\tan (\sqrt{x})) \tan (\tan (\sqrt{x})) \sec^{2} (\sqrt{x})$

Question 5

Differentiate with respect to $x$ :

$\frac{\sin (a x + b)}{\cos (c x + d)}$ Solution

Let: $y = \frac{\sin (a x + b)}{\cos (c x + d)}$

This is a quotient, so we use the Quotient Rule:

$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$

where

$u = \sin (a x + b), v = \cos (c x + d)$

Step 1: Differentiate $u$

$\frac{d u}{d x} = \cos (a x + b) \cdot a$

Step 2: Differentiate $v$

$\frac{d v}{d x} = - \sin (c x + d) \cdot c$

Step 3: Apply Quotient Rule

$\frac{d y}{d x} = \frac{\cos (c x + d) \cdot (a \cos (a x + b)) - \sin (a x + b) \cdot (- c \sin (c x + d))}{\cos^{2} (c x + d)}$

Simplify the numerator:

$= \frac{a \cos (a x + b) \cos (c x + d) + c \sin (a x + b) \sin (c x + d)}{\cos^{2} (c x + d)}$

Final Answer

$\frac{d}{d x} (\frac{\sin (a x + b)}{\cos (c x + d)}) = \frac{a \cos (a x + b) \cos (c x + d) + c \sin (a x + b) \sin (c x + d)}{\cos^{2} (c x + d)}$

Question 6

Differentiate with respect to $x$ :

$\cos (x^{3}) \cdot \sin^{2} (x^{5})$

Solution

The given function is a product, so we apply the Product Rule:

$\frac{d}{d x} (u \cdot v) = u^{'} v + u v^{'}$

Let:

$u = \cos (x^{3})$

Step 1: Differentiate $u = \cos (x^{3})$

Using chain rule:

$u^{'} = - \sin (x^{3}) \cdot 3 x^{2}$

$u^{'} = - 3 x^{2} \sin (x^{3})$

Step 2: Differentiate $v = \sin^{2} (x^{5})$

Rewrite:

$v = (\sin (x^{5}))^{2}$

Let $t = \sin (x^{5})$ , then $v = t^{2}$

$\frac{d v}{d t} = 2 t = 2 \sin (x^{5})$

$\frac{d t}{d x} = \cos (x^{5}) \cdot 5 x^{4}$

So,

$v^{'} = 2 \sin (x^{5}) \cdot 5 x^{4} \cos (x^{5})$

Step 3: Apply Product Rule

$\frac{d y}{d x} = u^{'} v + u v^{'}$

Substitute:

$\frac{d y}{d x} = (- 3 x^{2} \sin (x^{3})) \cdot \sin^{2} (x^{5}) + \cos (x^{3}) \cdot 10 x^{4} \sin (x^{5}) \cos (x^{5})$

Final Answer

$\frac{d}{d x} [\cos (x^{3}) \sin^{2} (x^{5})] = - 3 x^{2} \sin (x^{3}) \sin^{2} (x^{5}) + 10 x^{4} \cos (x^{3}) \sin (x^{5}) \cos (x^{5})$

$y = 2 cot (x^{2})$

Let’s differentiate step by step.

Question 7

$y = 2 \sqrt{\cot (x^{2})}$

Rewrite using exponent form:

$y = 2 (\cot (x^{2}))^{1 / 2}$ Solution

Let:

$u = \cot (x^{2})$ $y = 2 u^{1 / 2}$

Step 1: Differentiate outer function

$\frac{d y}{d u} = 2 \cdot \frac{1}{2} u^{- 1 / 2} = u^{- 1 / 2}$

Step 2: Differentiate inner function

$u = \cot (x^{2})$

Derivative of $\cot t$ is $- \csc^{2} t$

So, by chain rule:

$\frac{d u}{d x} = - \csc^{2} (x^{2}) \cdot 2 x$

Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

$\frac{d y}{d x} = u^{- 1 / 2} \cdot (- 2 x \csc^{2} (x^{2}))$

Substitute back $u = \cot (x^{2})$ :

$\frac{d y}{d x} = - 2 x \csc^{2} (x^{2}) \cdot (\cot (x^{2}))^{- 1 / 2}$

Rewrite using square root:

$\frac{d y}{d x} = - \frac{2 x \csc^{2} (x^{2})}{\sqrt{\cot (x^{2})}}$

Final Answer

$\frac{d}{d x} [2 \sqrt{\cot (x^{2})}] = - \frac{2 x \csc^{2} (x^{2})}{\sqrt{\cot (x^{2})}}$

Question 8

Differentiate with respect to $x$ :

$\cos (\sqrt{x})$ Solution

Let: $y = \cos (\sqrt{x})$

This is a composite function, so we will apply the Chain Rule.

Step 1: Identify inner and outer functions

$u = \sqrt{x} = x^{1 / 2}$

Step 2: Differentiate each part

$\frac{d y}{d u} = - \sin (u)$

$\frac{d u}{d x} = \frac{1}{2 \sqrt{x}}$

Step 3: Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Substitute $u = \sqrt{x}$ :

Final Answer

$\frac{d}{d x} [\cos (\sqrt{x})] = - \frac{\sin (\sqrt{x})}{2 \sqrt{x}}$

Question 9

Prove that the function

$f (x) = ∣ x - 1 ∣, x \in R$

is not differentiable at $x = 1$ .

Solution

First, write the function in piecewise form:

$f (x) = {\begin{cases} 1 - x, & if x < 1 \\ x - 1, & if x > 1 \end{cases}$

Also,

$f (1) = ∣ 1 - 1 ∣ = 0$

To check differentiability at $x = 1$ , evaluate the left-hand derivative and the right-hand derivative.

Left-hand derivative (LHD) at $x = 1$

For $x < 1$ , $f (x) = 1 - x$

$\frac{d}{d x} (1 - x) = - 1$ So,

$LHD at x = 1 = \lim_{h \to 0^{-}} \frac{f (1 + h) - f (1)}{h} = - 1$

Right-hand derivative (RHD) at $x = 1$

For $x > 1$ , $f (x) = x - 1$

$\frac{d}{d x} (x - 1) = 1$

So,

$RHD at x = 1 = \lim_{h \to 0^{+}} \frac{f (1 + h) - f (1)}{h} = 1$

Conclusion

$LHD = - 1, RHD = 1$

Since: $LHD \neq RHD$

$∴ f (x) = ∣ x - 1 ∣ is not differentiable at x = 1.$

Reason

The graph of $∣ x - 1 ∣$ has a sharp corner (cusp) at $x = 1$ , which makes the slope undefined there.

Question 10

Prove that the greatest integer function defined by

$f (x) = [x], 0 < x < 3$

is not differentiable at $x = 1$ and $x = 2$ .

Solution:

Understanding the function

The function $[x]$ defines the greatest integer less than or equal to $x$ .

For $0 < x < 3$ , the function behaves as follows:

$f (x) = {\begin{cases} 0, & 0 < x < 1 \\ 1, & 1 < x < 2 \\ 2, & 2 < x < 3 \end{cases}$

Continuity & Differentiability at $x = 1$

Left-hand limit approaching 1:

For $x < 1$ , $f (x) = 0$

$\lim_{x \to 1^{-}} f (x) = 0$

Right-hand limit approaching 1:

For $x > 1$ , $f (x) = 1$

$\lim_{x \to 1^{+}} f (x) = 1$

Since

$\lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x),$ $f (x) is discontinuous at x = 1.$ Differentiability result

A function must be continuous at a point to be differentiable there.

Since $f (x)$ is not continuous at $x = 1$ , it cannot be differentiable at $x = 1$ .

$f (x) is not differentiable at x = 1.$

Similarly at $x = 2$

Left-hand limit approaching 2

For $x < 2$ , $f (x) = 1$

$\lim_{x \to 2^{-}} f (x) = 1$

Right-hand limit approaching 2

For $x > 2$ , $f (x) = 2$

$\lim_{x \to 2^{+}} f (x) = 2$

Since

$\lim_{x \to 2^{-}} f (x) \neq \lim_{x \to 2^{+}} f (x),$ $f (x) is discontinuous at x = 2.$ Thus,

$f (x) is not differentiable at x = 2.$

November 25, 2025
Exercise-5.1, Class 12th, Maths, Chapter 5, NCERT
Question 25

Examine the continuity of the function

$f (x) = {\begin{cases} \sin x - \cos x, & if x \neq 0 \\ - 1, & if x = 0 \end{cases}$

Answer

We are given the function:

$f (x) = {\begin{cases} \sin x - \cos x, & x \neq 0 \\ - 1, & x = 0 \end{cases}$

We need to examine the continuity at $x = 0$ .

Step 1: Value of the function at $x = 0$

$f (0) = - 1$

Step 2: Find the limit of $f (x)$ as $x \to 0$

$\lim_{x \to 0} (\sin x - \cos x)$

We know standard limits:

$\sin 0 = 0, \cos 0 = 1$

So,

$\lim_{x \to 0} (\sin x - \cos x) = 0 - 1 = - 1$

Step 3: Compare Limit and Function Value

Expression Value

$\lim_{x \to 0} f (x)$ $- 1$

$f (0)$ $- 1$

Both values are equal.

Conclusion

Since:

$\lim_{x \to 0} f (x) = f (0)$ The function $f (x)$ is continuous at $x = 0$ .

Question 26

Find the value of $k$ so that the function $f$ is continuous at $x = \frac{π}{2}$ :

$f (x) = {\begin{cases} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{cases}$

Solution

For continuity at $x = \frac{π}{2}$ , we need:

$\lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$

Given:

$f (\frac{π}{2}) = 3$

Now compute the limit:

$\lim_{x \to \frac{π}{2}} \frac{k \cos x}{π - 2 x}$

Substitute $x = \frac{π}{2}$ :

$\frac{k \cos (\frac{π}{2})}{π - 2 (\frac{π}{2})} = \frac{k (0)}{0}$

This is the indeterminate form $\frac{0}{0}$ , so apply L’Hôpital’s Rule.

Differentiate numerator and denominator:

$\lim_{x \to \frac{π}{2}} \frac{- k \sin x}{- 2} = \lim_{x \to \frac{π}{2}} \frac{k \sin x}{2}$

Now substitute $x = \frac{π}{2}$ :

$\frac{k \sin (\frac{π}{2})}{2} = \frac{k (1)}{2} = \frac{k}{2}$

For continuity: $\frac{k}{2} = 3$ $k = 6$

Final Answer

$k = 6$

Question 27

Find the value of $k$ so that the function $f$ is continuous at $x = 2$ :

$f (x) = {\begin{cases} k x^{2}, & x \leq 2 \\ 3, & x > 2 \end{cases}$

Solution

For continuity at $x = 2$ :

$\lim_{x \to 2} f (x) = f (2)$

Step 1: Value at the point

$f (2) = k (2^{2}) = 4 k$

Step 2: Limit as $x \to 2$

Since the function changes definition at $x = 2$ , we use left-hand limit (LHL) and right-hand limit (RHL):

Left-hand limit (LHL)

$\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} k x^{2} = k (2^{2}) = 4 k$

Right-hand limit (RHL)

$\lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} 3 = 3$

Condition for continuity

$LHL = RHL = f (2)$

So, $4 k = 3$

$k = \frac{3}{4}$

Final Answer

$k = \frac{3}{4}$

Thus, the function is continuous at $x = 2$ when $k = \frac{3}{4}$ .

Question 28

Find the value of $k$ so that the function $f$ is continuous at $x = π$ :

$f (x) = {\begin{cases} k x + 1, & x \leq π \\ \cos x, & x > π \end{cases}$

Solution

For continuity at $x = π$ , we require:

$\lim_{x \to π} f (x) = f (π)$

Step 1: Value of the function at $x = π$

Since $x = π$ falls in the first part $(x \leq π)$ :

$f (π) = k π + 1$ Step 2: Left-hand limit (LHL)

$\lim_{x \to π^{-}} f (x) = \lim_{x \to π^{-}} (k x + 1) = k π + 1$

Step 3: Right-hand limit (RHL)

$\lim_{x \to π^{+}} f (x) = \lim_{x \to π^{+}} \cos x = \cos π$ $\cos π = - 1$

Condition for Continuity

$LHL = RHL = f (π)$ $k π + 1 = - 1$

Solve for $k$

$k π = - 1 - 1 = - 2$

$k = \frac{- 2}{π}$

Question 29

Find the value of $k$ so that the function $f$ is continuous at $x = 5$ :

$f (x) = {\begin{cases} k x + 1, & x \leq 5 \\ 3 x - 5, & x > 5 \end{cases}$

Solution

For continuity at $x = 5$ , we need:

$\lim_{x \to 5} f (x) = f (5)$

Step 1: Function value at the point

Since $x \leq 5$ :

$f (5) = k (5) + 1 = 5 k + 1$

Step 2: Left-hand limit (LHL) as $x \to 5^{-}$

$\lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{-}} (k x + 1) = 5 k + 1$

Step 3: Right-hand limit (RHL) as $x \to 5^{+}$

$\lim_{x \to 5^{+}} f (x) = \lim_{x \to 5^{+}} (3 x - 5) = 3 (5) - 5 = 15 - 5 = 10$

Step 4: Apply continuity condition

$LHL = RHL = f (5)$ $5 k + 1 = 10$

$5 k = 9$

$k = \frac{9}{5}$

Question 30

Find the values of $a$ and $b$ such that the function

$f (x) = {\begin{cases} 5, & x \leq 2 \\ a x + b, & 2 < x < 10 \\ 21, & x \geq 10 \end{cases}$

is continuous.

Solution

For continuity, the function must be continuous at:
- $x = 2$ (where left part meets the middle part)
- $x = 10$ (where middle part meets the right part)
Continuity at $x = 2$

$\lim_{x \to 2^{-}} f (x) = f (2) = 5$ $\lim_{x \to 2^{+}} f (x) = a (2) + b = 2 a + b$

For continuity:

$2 a + b = 5 (Equation 1)$

Continuity at $x = 10$

$\lim_{x \to 10^{-}} f (x) = a (10) + b = 10 a + b$ $f (10) = 21$

For continuity:

$10 a + b = 21 (Equation 2)$

Solve Equations (1) and (2)

$2 a + b = 5$ $10 a + b = 21$

Subtract (1) from (2):

$(10 a + b) - (2 a + b) = 21 - 5$ $8 a = 16$

$a = 2$

Substitute $a = 2$ into Equation (1):

$2 (2) + b = 5$

$4 + b = 5$

$b = 1$

Question 31

Show that the function

$f (x) = \cos (x^{2})$

is a continuous function.

Solution

We know that:
- $x^{2}$ is a polynomial → continuous everywhere
- $\cos x$ is a continuous function for all real numbers
A composition of two continuous functions is also continuous.

Here,

$f (x) = \cos (x^{2}) = \cos [g (x)] where g (x) = x^{2}$

Since both $g (x)$ and $\cos x$ are continuous, therefore:

$f (x) = \cos (x^{2}) is continuous for all x \in R$

Answer:

$\cos (x^{2}) is continuous everywhere.$

Question 32

Show that the function

$f (x) = ∣ \cos x ∣$

is a continuous function.

Solution
- $\cos x$ is continuous for all real numbers
- The absolute value function $∣ x ∣$ is also continuous
- The composition of continuous functions is continuous
$f (x) = ∣ \cos x ∣ = ∣ g (x) ∣, where g (x) = \cos x$

Since both $g (x)$ and $∣ x ∣$ are continuous,

$f (x) = ∣ \cos x ∣ is continuous for all x$

Answer:

$∣ \cos x ∣ is continuous for all real x .$

Question 33

Examine whether

$f (x) = \sin ∣ x ∣$

is a continuous function.

Solution
- $∣ x ∣$ is continuous for all real numbers
- $\sin x$ is continuous for all real numbers
- Composition of continuous functions is continuous
Thus:

$f (x) = \sin (∣ x ∣) = \sin (g (x)), g (x) = ∣ x ∣$

Since both are continuous,

$\sin ∣ x ∣ is continuous for all x$

Answer:

$\sin ∣ x ∣ is continuous everywhere.$

Question 34

Find all the points of discontinuity of

$f (x) = ∣ x ∣ - ∣ x + 1 ∣$

Solution

Check where each absolute expression changes form.

Break points occur where inside values become zero:

$x = 0 and x = - 1$

So evaluate function in intervals:

Case 1: $x < - 1$

$∣ x ∣ = - x, ∣ x + 1 ∣ = - (x + 1)$ $f (x) = - x - (- (x + 1)) = - x + x + 1 = 1$

Case 2: $- 1 \leq x < 0$

$∣ x ∣ = - x, ∣ x + 1 ∣ = x + 1$ $f (x) = - x - (x + 1) = - 2 x - 1$

Case 3: $x \geq 0$

$∣ x ∣ = x, ∣ x + 1 ∣ = x + 1$ $f (x) = x - (x + 1) = - 1$

Check continuity at critical points

At $x = - 1$

$\lim_{x \to - 1^{-}} f (x) = 1$ $\lim_{x \to - 1^{+}} f (x) = - 2 (- 1) - 1 = 2 - 1 = 1$ $f (- 1) = - 2 (- 1) - 1 = 1$

So continuous at $x = - 1$

At $x = 0$

$\lim_{x \to 0^{-}} f (x) = - 2 (0) - 1 = - 1$ $\lim_{x \to 0^{+}} f (x) = - 1$ $f (0) = - 1$

So continuous at $x = 0$

Final Answer

$The function f (x) = ∣ x ∣ - ∣ x + 1 ∣ is continuous everywhere.$
November 24, 2025
Exercise-5.1, Class 12th, Maths, Chapter 5, NCERT
Question 13.

Is the function defined by

$f (x) = {\begin{cases} x + 5, & if x \leq 1 \\ x - 5, & if x > 1 \end{cases}$

a continuous function?

Answer

To check the continuity of the function at $x = 1$ , evaluate the following:

Value of the function at $x = 1$

Since $x \leq 1$ :

$f (1) = 1 + 5 = 6$

Left-hand limit (LHL) as $x \to 1^{-}$

$\lim_{x \to 1^{-}} (x + 5) = 6$

Right-hand limit (RHL) as $x \to 1^{+}$

$\lim_{x \to 1^{+}} (x - 5) = - 4$

Comparison

$LHL = 6, RHL = - 4, f (1) = 6$

Since

$LHL \neq RHL$

Final Answer

The function is not continuous at $x = 1$ because the left-hand limit and right-hand limit are not equal.

Discuss the continuity of the function f, where f is defined by

Question 14.

Discuss the continuity of the function

$f (x) = {\begin{cases} 3, & if 0 \leq x \leq 1 \\ 4, & if 1 < x < 3 \\ 5, & if 3 \leq x \leq 10 \end{cases}$

Answer

To check continuity, we examine the points where the definition of the function changes:
at $x = 1$ and $x = 3$ .

Continuity on the intervals
- On $[0, 1]$ : $f (x) = 3$ (a constant function → continuous).
- On $(1, 3)$ : $f (x) = 4$ (constant function → continuous).
- On $[3, 10]$ : $f (x) = 5$ (constant function → continuous).
So the only possible discontinuities are at the endpoints where the pieces join.

Check continuity at $x = 1$

Left-hand limit (as $x \to 1^{-}$ )

$\lim_{x \to 1^{-}} f (x) = 3$

Right-hand limit (as $x \to 1^{+}$ )

$\lim_{x \to 1^{+}} f (x) = 4$

Value of the function

$f (1) = 3$

Since

$\lim_{x \to 1^{-}} f (x) = 3 and \lim_{x \to 1^{+}} f (x) = 4$

Left-hand limit ≠ Right-hand limit
→ function is discontinuous at $x = 1$

Check continuity at $x = 3$

Left-hand limit (as $x \to 3^{-}$ )

$\lim_{x \to 3^{-}} f (x) = 4$

Right-hand limit (as $x \to 3^{+}$ )

$\lim_{x \to 3^{+}} f (x) = 5$

Value of the function

$f (3) = 5$

Since

$4 \neq 5$

→ function is discontinuous at $x = 3$ .

Final Conclusion

The function $f (x)$ is:
- Continuous on each open interval $(0, 1)$ , $(1, 3)$ , and $(3, 10)$
- Discontinuous at the points $x = 1$ and $x = 3$
Question 15

Discuss the continuity of the function $f$ defined by:

$f (x) = {\begin{cases} 2 x, & if x < 0 \\ 0, & if 0 < x < 1 \\ 4 x, & if x > 1 \end{cases}$

Answer

This is a piecewise function, and we need to check continuity at the points where the definition changes, i.e., at $x = 0$ and $x = 1$ .

Continuity for $x < 0$ , $0 < x < 1$ , and $x > 1$

In each interval, the function is a polynomial (linear function), and polynomials are continuous everywhere.
So, $f (x)$ is continuous within each interval.

Check continuity at $x = 0$

Left-hand limit (LHL) as $x \to 0^{-}$

Using $2 x$ :

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} 2 x = 0$

Right-hand limit (RHL) as $x \to 0^{+}$

Using 0:

$\lim_{x \to 0^{+}} f (x) = 0$

Value of the function at $x = 0$

Notice: The function definition does not include $x = 0$ in any case, so:

$f (0) does not exist$

Conclusion at $x = 0$

Even though

$LHL = RHL = 0,$

the value of the function at $x = 0$ is not defined.
So, the function is not continuous at $x = 0$ .

Check continuity at $x = 1$

Left-hand limit (LHL) as $x \to 1^{-}$

Using 0:

$\lim_{x \to 1^{-}} f (x) = 0$

Right-hand limit (RHL) as $x \to 1^{+}$

Using $4 x$ :

$\lim_{x \to 1^{+}} f (x) = 4 (1) = 4$

Value of the function at $x = 1$

Not defined in any case, so:

$f (1) does not exist$

Conclusion at $x = 1$

Since

$LHL = 0 \neq 4 = RHL$

The limit does not exist. Therefore, the function is not continuous at $x = 1$ .

Final Conclusion

The function $f (x)$ is:
- Continuous within each open interval $(- \infty, 0)$ , $(0, 1)$ , and $(1, \infty)$
- Not continuous at $x = 0$ (because $f (0)$ is not defined)
- Not continuous at $x = 1$ (because left-hand and right-hand limits are not equal and $f (1)$ is not defined)
Question 16

Discuss the continuity of the function $f (x)$ , where

$f (x) = {\begin{cases} - 2, & if x \leq - 1 \\ 2 x, & if - 1 < x \leq 1 \\ 2, & if x > 1 \end{cases}$

Answer:

To check continuity, we must examine the possible points of discontinuity—here the function changes its definition at $x = - 1$ and $x = 1$ .
So, we check continuity at these points.

Continuity at $x = - 1$

Value of the function at $x = - 1$

Since $x \leq - 1$ :

$f (- 1) = - 2$

Left-hand limit (LHL) as $x \to - 1^{-}$

Using $- 2$ :

$\lim_{x \to - 1^{-}} f (x) = - 2$

Right-hand limit (RHL) as $x \to - 1^{+}$

Using $2 x$ :

$\lim_{x \to - 1^{+}} 2 x = 2 (- 1) = - 2$

Conclusion

$L H L = R H L = f (- 1) = - 2$

So, the function is continuous at $x = - 1$ .

Continuity at $x = 1$

Value of the function at $x = 1$

Using $2 x$ :

$f (1) = 2 (1) = 2$

Left-hand limit (LHL) as $x \to 1^{-}$

Using $2 x$ :

$\lim_{x \to 1^{-}} 2 x = 2$

Right-hand limit (RHL) as $x \to 1^{+}$

Using constant $2$ :

$\lim_{x \to 1^{+}} f (x) = 2$

Conclusion

$L H L = R H L = f (1) = 2$

Thus, the function is continuous at $x = 1$ .

Final Conclusion

Since the function is continuous at both points where it changes its definition ( $x = - 1$ and $x = 1$ ), and there are no other breaks, gaps, or jumps:

$The function f (x) is continuous for all real values of x .$

Question 17

Find the relationship between $a$ and $b$ so that the function $f (x)$ defined by

$f (x) = {\begin{cases} a x + 1, & if x \leq 3 \\ b x + 3, & if x > 3 \end{cases}$

is continuous at $x = 3$ .

Answer

To ensure continuity at $x = 3$ , we require:

$Left-hand limit = Right-hand limit = f (3)$

Value of the function at $x = 3$

Using the first expression since $x \leq 3$ :

$f (3) = a (3) + 1 = 3 a + 1$

Left-hand Limit (LHL) as $x \to 3^{-}$

$\lim_{x \to 3^{-}} f (x) = 3 a + 1$

Right-hand Limit (RHL) as $x \to 3^{+}$

Using $b x + 3$ :

$\lim_{x \to 3^{+}} f (x) = b (3) + 3 = 3 b + 3$

Condition for continuity

$3 a + 1 = 3 b + 3$

Solving

$3 a - 3 b = 2$

Final Answer

$a - b = \frac{2}{3}$

This is the required relationship between $a$ and $b$ for $f (x)$ to be continuous at $x = 3$ .

Question 18

For what value of $λ$ is the function defined by

$f (x) = {\begin{cases} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{cases}$

continuous at $x = 0$ ? What about at $x = 1$ ?

Answer

Checking continuity at $x = 0$

Value of the function at $x = 0$

Since $x \leq 0$ :

$f (0) = λ (0^{2} - 2 \cdot 0) = λ (0) = 0$

Left-hand limit (LHL)

$\lim_{x \to 0^{-}} λ (x^{2} - 2 x) = λ (0 - 0) = 0$

Right-hand limit (RHL)

$\lim_{x \to 0^{+}} (4 x + 1) = 4 (0) + 1 = 1$

Condition for continuity

$L H L = R H L = f (0)$

So,

$0 = 1$

This statement is never true, so no real value of $λ$ can make the function continuous at $x = 0$ .

Checking continuity at $x = 1$

At $x = 1$ , the value is taken from the second piece $4 x + 1$ , and there is no matching left-hand expression at 1, since the first part ends at $x = 0$ .
Thus, the function is not defined in a neighborhood around 1 from the left side.

Therefore, the function cannot be continuous at $x = 1$ .

Final Conclusion

$There is no value of λ that makes the function continuous at x = 0.$ $The function is not continuous at x = 1 because continuity cannot be checked.$

Question 19

Show that the function $g (x) = x - [x]$ is discontinuous at all integral points.
Here $[x]$ denotes the greatest integer less than or equal to $x$ .

Answer

Given Function

$g (x) = x - [x]$

The expression $x - [x]$ represents the fractional part of $x$ , denoted as ${x}$ .
Thus,

$g (x) = {x}$

The fractional part function always satisfies:

$0 \leq {x} < 1$

To show discontinuity at integer points

Let $x = n$ , where $n$ is any integer.

Left-hand limit (LHL) as $x \to n^{-}$

When $x$ approaches $n$ from the left, $x = n - h$ where $h \to 0^{+}$ .
Then the greatest integer function gives:

$[x] = n - 1$

So,

$g (x) = x - [x] = (n - h) - (n - 1) = 1 - h$

Taking limit,

$\lim_{x \to n^{-}} g (x) = \lim_{h \to 0^{+}} (1 - h) = 1$

Right-hand limit (RHL) as $x \to n^{+}$

When approaching from the right, $x = n + h$ , $h \to 0^{+}$ , then

$[x] = n$

So,

$g (x) = (n + h) - n = h$

Taking limit,

$\lim_{x \to n^{+}} g (x) = \lim_{h \to 0^{+}} h = 0$

Value of the function at $x = n$

$g (n) = n - [n] = n - n = 0$

Comparison

$\lim_{x \to n^{-}} g (x) = 1, \lim_{x \to n^{+}} g (x) = 0, g (n) = 0$

Since,

$\lim_{x \to n^{-}} g (x) \neq \lim_{x \to n^{+}} g (x)$

The limit does not exist at $x = n$ , and therefore, the function is discontinuous at every integer $n$ .

Final Conclusion

$g (x) = x - [x] is discontinuous at all integral points.$

Question 20

Is the function defined by

$f (x) = x^{2} - \sin x + 5$

continuous at $x = π$ ?

Answer

The function $f (x) = x^{2} - \sin x + 5$ is composed of the following functions:
- $x^{2}$ → a polynomial function (continuous for all real $x$ )
- $\sin x$ → a trigonometric function (continuous for all real $x$ )
- Constant $5$ → continuous everywhere
The sum or difference of continuous functions is also continuous for all real numbers.

Therefore, $f (x)$ is continuous everywhere, including at $x = π$ .

Check using limits

Value at $x = π$ :

$f (π) = π^{2} - \sin π + 5 = π^{2} - 0 + 5 = π^{2} + 5$

Limit as $x \to π$ :

$\lim_{x \to π} f (x) = \lim_{x \to π} (x^{2} - \sin x + 5) = π^{2} - 0 + 5 = π^{2} + 5$

Comparison

$\lim_{x \to π} f (x) = f (π)$

So the function is continuous at $x = π$ .

Final Answer

$Yes, the function is continuous at x = π .$

Question 21

Discuss the continuity of the following functions:

(a) $f (x) = \sin x + \cos x$
(b) $f (x) = \sin x - \cos x$
(c) $f (x) = \sin x \cdot \cos x$

Answer

To discuss continuity, recall that:
- $\sin x$ and $\cos x$ are continuous functions for all real values of $x$ .
- Sum, difference, and product of continuous functions are also continuous.
So, we analyze each function:

(a) $f (x) = \sin x + \cos x$
- $\sin x$ is continuous for all $x \in R$
- $\cos x$ is continuous for all $x \in R$
The sum of two continuous functions is continuous.

Conclusion

$f (x) = \sin x + \cos x is continuous for all real x .$

(b) $f (x) = \sin x - \cos x$
- Difference of continuous functions remains continuous.
Conclusion

$f (x) = \sin x - \cos x is continuous for all real x .$

(c) $f (x) = \sin x \cdot \cos x$
- Product of two continuous functions is continuous.
Conclusion

$f (x) = \sin x \cdot \cos x is continuous for all real x .$

Question 22

Discuss the continuity of the cosine, cosecant, secant, and cotangent functions.

Answer

To discuss the continuity of these trigonometric functions, recall:
- A function is continuous at a point if the limit exists and equals the value of the function at that point.
- Discontinuity occurs where the function is not defined.
Cosine Function

$f (x) = \cos x$
- $\cos x$ is defined for all real values of $x$ .
- It is smooth and has no breaks or gaps.
Conclusion

$\cos x is continuous for all x \in R .$

Cosecant Function

$f (x) = \csc x = \frac{1}{\sin x}$
- $\csc x$ is not defined where $\sin x = 0$ .
- $\sin x = 0$ at $x = n π$ , where $n$ is any integer.
Conclusion

$\csc x is discontinuous at x = n π .$ $It is continuous for all other real values of x .$

Secant Function

$f (x) = \sec x = \frac{1}{\cos x}$
- $\sec x$ is not defined where $\cos x = 0$ .
- $\cos x = 0$ at $x = \frac{(2 n + 1) π}{2}$ , where $n$ is any integer.
Conclusion

$\sec x is discontinuous at x = \frac{(2 n + 1) π}{2} .$ $It is continuous everywhere else.$

Cotangent Function

$f (x) = \cot x = \frac{\cos x}{\sin x}$
- $\cot x$ is not defined where $\sin x = 0$ .
- Similar to cosecant, discontinuity occurs at $x = n π$ .
Conclusion

$\cot x is discontinuous at x = n π .$ $It is continuous for all other real values of x .$

Question 23

Find all points of discontinuity of

$f (x) = {\begin{cases} \frac{\sin x}{x}, & if x < 0 \\ x + 1, & if x \geq 0 \end{cases}$

Answer

The function changes its definition at $x = 0$ , so discontinuity (if any) must be checked at $x = 0$ .

Step 1: Compute limit from the left side as $x \to 0^{-}$

Consider

$\lim_{x \to 0^{-}} \frac{\sin x}{x}$

We know the standard limit identity:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

So,

$\lim_{x \to 0^{-}} f (x) = 1$

Step 2: Compute right-hand limit as $x \to 0^{+}$

From the second part $f (x) = x + 1$ when $x \geq 0$ :

$\lim_{x \to 0^{+}} (x + 1) = 0 + 1 = 1$

Step 3: Value of the function at $x = 0$

Since $x \geq 0$ ,

$f (0) = 0 + 1 = 1$

Comparison

$\lim_{x \to 0^{-}} f (x) = 1, \lim_{x \to 0^{+}} f (x) = 1, f (0) = 1$

Since all three are equal:

$\lim_{x \to 0} f (x) = f (0)$

Final Conclusion

$The function is continuous at x = 0.$

There are no other points where the definition changes, and both components of the function are continuous in their respective intervals.

$Therefore, f (x) is continuous for all x .$

Question 24

Determine if the function defined by

$f (x) = {\begin{cases} x^{2} \sin (\frac{1}{x}), & if x \neq 0 \\ 0, & if x = 0 \end{cases}$

is continuous at $x = 0$ .

Answer

To check continuity at $x = 0$ , we need to verify:

$\lim_{x \to 0} f (x) = f (0)$

Given:

$f (0) = 0$

Find the limit of $f (x)$ as $x \to 0$

$f (x) = x^{2} \sin (\frac{1}{x})$

We know that:

$- 1 \leq \sin (\frac{1}{x}) \leq 1$

Multiplying the entire inequality by $x^{2} \geq 0$ :

$- x^{2} \leq x^{2} \sin (\frac{1}{x}) \leq x^{2}$

Now take the limit as $x \to 0$ :

$\lim_{x \to 0} - x^{2} = 0 and \lim_{x \to 0} x^{2} = 0$

By Squeeze (Sandwich) Theorem:

$\lim_{x \to 0} x^{2} \sin (\frac{1}{x}) = 0$

Therefore:

$\lim_{x \to 0} f (x) = 0 = f (0)$

Final Conclusion

$Yes, the function f (x) is continuous at x = 0.$
November 23, 2025
Miscellaneous Exercises on Chapter 4, Class 12th, Maths
Question 1.
Prove that the determinant

$∣ \begin{array}{ccc} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{array} ∣$

is independent of $θ$ .

Solution.

Compute the determinant by expanding along the first row:

$\begin{aligned} D & = x ∣ \begin{array}{cc} - x & 1 \\ 1 & x \end{array} ∣ - \sin θ ∣ \begin{array}{cc} - \sin θ & 1 \\ \cos θ & x \end{array} ∣ + \cos θ ∣ \begin{array}{cc} - \sin θ & - x \\ \cos θ & 1 \end{array} ∣ \end{aligned}$

Evaluate each $2 \times 2$ determinant:

$\begin{aligned} x ∣ \begin{array}{cc} - x & 1 \\ 1 & x \end{array} ∣ & = x ((- x) (x) - 1 \cdot 1) = x (- x^{2} - 1) = - x^{3} - x, \\ - \sin θ ∣ \begin{array}{cc} - \sin θ & 1 \\ \cos θ & x \end{array} ∣ & = - \sin θ ((- \sin θ) x - 1 \cdot \cos θ) = - \sin θ (- x \sin θ - \cos θ) \\ = x \sin^{2} θ + \sin θ \cos θ, \\ \cos θ ∣ \begin{array}{cc} - \sin θ & - x \\ \cos θ & 1 \end{array} ∣ & = \cos θ ((- \sin θ) \cdot 1 - (- x) \cos θ) = \cos θ (- \sin θ + x \cos θ) \\ = - \sin θ \cos θ + x \cos^{2} θ \end{aligned}$ Add the three parts:

$\begin{aligned} D & = (- x^{3} - x) + (x \sin^{2} θ + \sin θ \cos θ) + (- \sin θ \cos θ + x \cos^{2} θ) \\ = - x^{3} - x + x (\sin^{2} θ + \cos^{2} θ) + (\sin θ \cos θ - \sin θ \cos θ) \\ = - x^{3} - x + x \cdot 1 + 0 = - x^{3} \end{aligned}$ Thus the determinant equals $- x^{3}$ , which does not depend on $θ$ .

Question 2.
Evaluate the determinant

$∣ \begin{array}{ccc} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{array} ∣$

Solution:

Let

$D = ∣ \begin{array}{ccc} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{array} ∣$

We will expand along the second row since it has a zero term.

Step 1: Expansion along second row

$D = (- \sin β) (- 1)^{2 + 1} ∣ \begin{array}{cc} \cos α \sin β & - \sin α \\ \sin α \sin β & \cos α \end{array} ∣$

$+ (\cos β) (- 1)^{2 + 2} ∣ \begin{array}{cc} \cos α \cos β & - \sin α \\ \sin α \cos β & \cos α \end{array} ∣$

Simplify the cofactors:

$D = \sin β ∣ \begin{array}{cc} \cos α \sin β & - \sin α \\ \sin α \sin β & \cos α \end{array} ∣ + \cos β ∣ \begin{array}{cc} \cos α \cos β & - \sin α \\ \sin α \cos β & \cos α \end{array} ∣$

Step 2: Evaluate the 2×2 determinants

For the first:

$∣ \begin{array}{cc} \cos α \sin β & - \sin α \\ \sin α \sin β & \cos α \end{array} ∣ = (\cos α \sin β) (\cos α) - (- \sin α) (\sin α \sin β)$

$= \sin β (\cos^{2} α + \sin^{2} α) = \sin β .$

For the second:

$∣ \begin{array}{cc} \cos α \cos β & - \sin α \\ \sin α \cos β & \cos α \end{array} ∣ = (\cos α \cos β) (\cos α) - (- \sin α) (\sin α \cos β)$

$= \cos β (\cos^{2} α + \sin^{2} α) = \cos β .$

Step 3: Substitute back

$D = \sin β (\sin β) + \cos β (\cos β) = \sin^{2} β + \cos^{2} β = 1$

Final Answer:

$D = 1$

Question 3.
If

$A^{- 1} = (\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}) and B = (\begin{array}{ccc} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{array}),$

find $(A B)^{- 1}$

Solution:

We know the property of inverse of a product:

$(A B)^{- 1} = B^{- 1} A^{- 1}$

Step 1: Find $B^{- 1}$

Let

$B = (\begin{array}{ccc} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{array})$

To find $B^{- 1}$ , compute by the adjoint method (or via row operations).
After simplification, we get:

$B^{- 1} = (\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array})$

Step 2: Compute $(A B)^{- 1} = B^{- 1} A^{- 1}$

$A^{- 1} = (\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array})$

Now multiply $B^{- 1}$ and $A^{- 1}$ :

$(A B)^{- 1} = (\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}) (\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array})$

Step 3: Matrix multiplication

$\begin{aligned} (A B)_{11}^{- 1} & = 3 (3) + 2 (- 15) + 6 (5) = 9 - 30 + 30 = 9, \\ (A B)_{12}^{- 1} & = 3 (- 1) + 2 (6) + 6 (- 2) = - 3 + 12 - 12 = - 3, \\ (A B)_{13}^{- 1} & = 3 (1) + 2 (- 5) + 6 (2) = 3 - 10 + 12 = 5, \\ (A B)_{21}^{- 1} & = 1 (3) + 1 (- 15) + 2 (5) = 3 - 15 + 10 = - 2, \\ (A B)_{22}^{- 1} & = 1 (- 1) + 1 (6) + 2 (- 2) = - 1 + 6 - 4 = 1, \\ (A B)_{23}^{- 1} & = 1 (1) + 1 (- 5) + 2 (2) = 1 - 5 + 4 = 0, \\ (A B)_{31}^{- 1} & = 2 (3) + 2 (- 15) + 5 (5) = 6 - 30 + 25 = 1, \\ (A B)_{32}^{- 1} & = 2 (- 1) + 2 (6) + 5 (- 2) = - 2 + 12 - 10 = 0, \\ (A B)_{33}^{- 1} & = 2 (1) + 2 (- 5) + 5 (2) = 2 - 10 + 10 = 2. \end{aligned}$

Step 4: Write the result

$(A B)^{- 1} = (\begin{array}{ccc} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{array})$

Final Answer:

$(A B)^{- 1} = (\begin{array}{ccc} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{array})$

Question 4.
Let

$A = (\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array})$

Verify that
(i) $[adj A]^{- 1} = adj (A^{- 1})$
(ii) $(A^{- 1})^{- 1} = A$

Answer & proof

(i) General identity and verification.

For any invertible square matrix $A$ we have the identity

$adj A = (\det A) A^{- 1}$

Taking inverses on both sides (and using $(c M)^{- 1} = c^{- 1} M^{- 1}$ for scalar $c \neq 0$ ) gives

$(adj A)^{- 1} = ((\det A) A^{- 1})^{- 1} = (\det A)^{- 1} (A^{- 1})^{- 1} = (\det A)^{- 1} A$

Also for the inverse matrix $A^{- 1}$ we have

$adj (A^{- 1}) = (\det (A^{- 1})) (A^{- 1})^{- 1}$

But $\det (A^{- 1}) = (\det A)^{- 1}$ and $(A^{- 1})^{- 1} = A$ , so

$adj (A^{- 1}) = (\det A)^{- 1} A$

Comparing the two expressions we obtain the required equality for any invertible $A$ :

$(adj A)^{- 1} = adj (A^{- 1})$

Verification for the given matrix $A$ .

Compute $\det A$ , $adj A$ and the two sides explicitly:

$\det A = - 5$

One finds

$adj A = (\begin{array}{ccc} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{array}), A^{- 1} = \frac{1}{- 5} (\begin{array}{ccc} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{array}) = (\begin{array}{ccc} - 14 / 5 & 9 / 5 & 1 / 5 \\ 9 / 5 & - 4 / 5 & - 1 / 5 \\ 1 / 5 & - 1 / 5 & 1 / 5 \end{array})$

Then

$(adj A)^{- 1} = (1 / \det A) A = \frac{1}{- 5} (\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}) = (\begin{array}{ccc} - 1 / 5 & - 2 / 5 & - 1 / 5 \\ - 2 / 5 & - 3 / 5 & - 1 / 5 \\ - 1 / 5 & - 1 / 5 & - 1 \end{array})$

And using $adj (A^{- 1}) = (\det A^{- 1}) (A^{- 1})^{- 1} = (\det A)^{- 1} A$ gives exactly the same matrix. So the identity holds for this $A$ .

(ii):

Given

$A = (\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}),$

verify that

$(A^{- 1})^{- 1} = A$

Step 1. Find $A^{- 1}$

We already know from earlier computation that

$\det (A) = - 5$

Now, let’s find the adjoint of $A$ :

Compute cofactors of $A$ :

$Cofactor matrix of A = (\begin{array}{ccc} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{array})$

So the adjoint of $A$ is its transpose:

$adj (A) = (\begin{array}{ccc} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{array})$

Hence,

$A^{- 1} = \frac{1}{\det (A)} adj (A) = \frac{1}{- 5} (\begin{array}{ccc} 14 & - 9 & - 1 \\ - 9 & 4 & 1 \\ - 1 & 1 & - 1 \end{array}) = (\begin{array}{ccc} - \frac{14}{5} & \frac{9}{5} & \frac{1}{5} \\ \frac{9}{5} & - \frac{4}{5} & - \frac{1}{5} \\ \frac{1}{5} & - \frac{1}{5} & \frac{1}{5} \end{array})$

Step 2. Find $(A^{- 1})^{- 1}$

By the definition of matrix inverse:

$A^{- 1} A = I$

Multiplying both sides on the left by $(A^{- 1})^{- 1}$ , we get:

$(A^{- 1})^{- 1} (A^{- 1}) A = (A^{- 1})^{- 1} I,$

which simplifies to:

$A = (A^{- 1})^{- 1}$

Step 3. Verify numerically

If you actually take the inverse of $A^{- 1}$ (by determinant and adjoint, or by direct computation),
you’ll get back the original matrix $A$ :

$(A^{- 1})^{- 1} = (\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}) = A$

Final Answer:

$(A^{- 1})^{- 1} = A .$

Question 5.
Evaluate

$∣ \begin{array}{ccc} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{array} ∣$

Solution:

Let

$D = ∣ \begin{array}{ccc} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{array} ∣$

We will simplify this determinant using column operations.

Step 1: Simplify columns

Let’s apply the operation

$C_{3} \to C_{3} - C_{1} - C_{2}$

(i.e., replace the third column with $C_{3} - C_{1} - C_{2}$ )

Compute the new third column entries:
- For first row: $(x + y) - x - y = 0$
- For second row: $x - y - (x + y) = x - y - x - y = - 2 y$
- For third row: $y - (x + y) - x = y - x - y - x = - 2 x$
So the new determinant becomes:

$D = ∣ \begin{array}{ccc} x & y & 0 \\ y & x + y & - 2 y \\ x + y & x & - 2 x \end{array} ∣$

Step 2: Expand along the first row (since it has a zero)

$D = x ∣ \begin{array}{cc} x + y & - 2 y \\ x & - 2 x \end{array} ∣ - y ∣ \begin{array}{cc} y & - 2 y \\ x + y & - 2 x \end{array} ∣$

Step 3: Compute each $2 \times 2$ determinant
1. For the first one:
$∣ \begin{array}{cc} x + y & - 2 y \\ x & - 2 x \end{array} ∣ = (x + y) (- 2 x) - (- 2 y) (x)$

$= - 2 x (x + y) + 2 x y = - 2 x^{2} - 2 x y + 2 x y = - 2 x^{2} .$
1. For the second one:
$∣ \begin{array}{cc} y & - 2 y \\ x + y & - 2 x \end{array} ∣ = y (- 2 x) - (- 2 y) (x + y)$

$= - 2 x y + 2 y (x + y) = - 2 x y + 2 x y + 2 y^{2} = 2 y^{2}$

Step 4: Substitute back

$D = x (- 2 x^{2}) - y (2 y^{2}) = - 2 x^{3} - 2 y^{3} = - 2 (x^{3} + y^{3})$

Final Answer:

$∣ \begin{array}{ccc} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{array} ∣ = - 2 (x^{3} + y^{3}) .$

Question 6.
Evaluate

$∣ \begin{array}{ccc} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{array} ∣$

Solution:

Let

$D = ∣ \begin{array}{ccc} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{array} ∣$

We’ll simplify this determinant using row operations.

Step 1: Simplify rows

Perform the following operations:

$R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$

Compute the new rows:
- $R_{2} = (1, x + y, y) - (1, x, y) = (0, y, 0)$
- $R_{3} = (1, x, x + y) - (1, x, y) = (0, 0, y)$
So the new determinant becomes:

$D = ∣ \begin{array}{ccc} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & y \end{array} ∣$

Step 2: Expand along the first column

$D = 1 \cdot ∣ \begin{array}{cc} y & 0 \\ 0 & y \end{array} ∣ = 1 \cdot (y \cdot y - 0 \cdot 0) = y^{2}$

Final Answer:

$∣ \begin{array}{ccc} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{array} ∣ = y^{2} .$

Question 7.
Solve the system

${\begin{cases} \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \\ \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \\ \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2 \end{cases}$

Solution (matrix method)

Put $u = \frac{1}{x}, v = \frac{1}{y}, w = \frac{1}{z}$ . Then the system becomes linear in $u, v, w$ :

$\begin{aligned} 2 u + 3 v + 10 w & = 4, \\ 4 u - 6 v + 5 w & = 1, \\ 6 u + 9 v - 20 w & = 2 \end{aligned}$

In matrix form $A t = b$ where

$A = (\begin{array}{ccc} 2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20 \end{array}), t = (\begin{array}{c} u \\ v \\ w \end{array}), b = (\begin{array}{c} 4 \\ 1 \\ 2 \end{array})$

Solve $t = A^{- 1} b$ . Doing this (or by Cramer’s rule / row reduction) gives

$t = (\begin{array}{c} u \\ v \\ w \end{array}) = (\begin{array}{c} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{array})$

So

$u = \frac{1}{2}, v = \frac{1}{3}, w = \frac{1}{5}$

Convert back to $x, y, z$ :

$x = \frac{1}{u} = 2, y = \frac{1}{v} = 3, z = \frac{1}{w} = 5$

Check

Substitute $(x, y, z) = (2, 3, 5)$ into the original equations:

$\frac{2}{2} + \frac{3}{3} + \frac{10}{5} = 1 + 1 + 2 = 4, \frac{4}{2} - \frac{6}{3} + \frac{5}{5} = 2 - 2 + 1 = 1$

$, \frac{6}{2} + \frac{9}{3} - \frac{20}{5} = 3 + 3 - 4 = 2$

all hold.

Answer: $x = 2, y = 3, z = 5.$

Question 8.
If $x, y, z$ are nonzero real numbers, then the inverse of the matrix

$A = (\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array})$

is which of the following?

(A) $(\begin{array}{ccc} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{array})$

(B) $x y z (\begin{array}{ccc} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{array})$

(C) $\frac{1}{x y z} (\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array})$

(D) $\frac{1}{x y z} (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array})$

Answer: (A).

Reason: For a diagonal matrix, the inverse (when all diagonal entries are nonzero) is the diagonal matrix of reciprocals. Check $A \cdot A^{- 1} = I$ :

$(\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}) (\begin{array}{ccc} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{array}) = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array})$

Question 9.
Let

$A = (\begin{array}{ccc} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{array}), 0 \leq θ \leq 2 π$

Which of the following is true?
(A) $\det (A) = 0$
(B) $\det (A) \in (2, \infty)$
(C) $\det (A) \in (2, 4)$
(D) $\det (A) \in [2, 4]$

Solution

Compute the determinant by expansion along the first row:

$\begin{aligned} \det A & = 1 \cdot ∣ \begin{array}{cc} 1 & \sin θ \\ - \sin θ & 1 \end{array} ∣ - \sin θ \cdot ∣ \begin{array}{cc} - \sin θ & \sin θ \\ - 1 & 1 \end{array} ∣ + 1 \cdot ∣ \begin{array}{cc} - \sin θ & 1 \\ - 1 & - \sin θ \end{array} ∣ \\ = 1 \cdot (1 \cdot 1 - \sin θ \cdot (- \sin θ)) - \sin θ \cdot ((- \sin θ) \cdot 1 - \sin θ \cdot (- 1)) + \end{aligned}$

$\begin{aligned} 1 \cdot ((- \sin θ) (- \sin θ) - 1 \cdot (- 1)) \\ = 1 + \sin^{2} θ - \sin θ (- \sin θ + \sin θ) + \sin^{2} θ + 1 \\ = 1 + \sin^{2} θ + 0 + \sin^{2} θ + 1 = 2 + 2 \sin^{2} θ \end{aligned}$

So $\det A = 2 (1 + \sin^{2} θ)$

Since $0 \leq \sin^{2} θ \leq 1$ , we have

$2 (1 + 0) \leq \det A \leq 2 (1 + 1) ⟹ 2 \leq \det A \leq 4$

Thus $\det (A) \in [2, 4]$ . The correct choice is (D).
October 29, 2025
Exercise-4.5-Part-2, Class 12th, Maths, NCERT

Question 15.
If

$A = (\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}),$

find $A^{- 1}$ . Using $A^{- 1}$ , solve the following system of equations:

${\begin{cases} 2 x - 3 y + 5 z = 11, \\ 3 x + 2 y - 4 z = - 5, \\ x + y - 2 z = - 3. \end{cases}$

Solution:

We have

$A = (\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array})$

Let $b = (\begin{array}{c} 11 \\ - 5 \\ - 3 \end{array})$

We know that

$A x = b \Rightarrow x = A^{- 1} b$

Step 1: Find $A^{- 1}$

Using the adjoint and determinant method,

$\det (A) = - 1$

The adjugate (adjoint) of $A$ is:

$adj (A) = (\begin{array}{ccc} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array})$

Hence,

$A^{- 1} = \frac{1}{\det (A)} \cdot adj (A) = - 1 \times (\begin{array}{ccc} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array}) = (\begin{array}{ccc} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}) .$

Step 2: Use $A^{- 1}$ to find $x, y, z$

$(\begin{array}{c} x \\ y \\ z \end{array}) = A^{- 1} b = (\begin{array}{ccc} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}) (\begin{array}{c} 11 \\ - 5 \\ - 3 \end{array})$

Now multiply:

$\begin{aligned} x & = 0 (11) + 1 (- 5) + (- 2) (- 3) = - 5 + 6 = 1, \\ y & = - 2 (11) + 9 (- 5) + (- 23) (- 3) = - 22 - 45 + 69 = 2, \\ z & = - 1 (11) + 5 (- 5) + (- 13) (- 3) = - 11 - 25 + 39 = 3. \end{aligned}$

Final Answers:

$A^{- 1} = (\begin{array}{ccc} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}), x = 1, y = 2, z = 3$

Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60.
The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90.
The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70.
Find the cost of each item per kilogram using the matrix method.

Solution:

Let the cost per kilogram of onion, wheat and rice be ₹ $o$ , ₹ $w$ and ₹ $r$ respectively.

Then, according to the question, we have:

$\begin{aligned} 4 o + 3 w + 2 r & = 60, \\ 2 o + 4 w + 6 r & = 90, \\ 6 o + 2 w + 3 r & = 70 \end{aligned}$

Step 1: Write in matrix form

$(\begin{array}{ccc} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{array}) (\begin{array}{c} o \\ w \\ r \end{array}) = (\begin{array}{c} 60 \\ 90 \\ 70 \end{array})$

That is,

$A x = b,$

where

$A = (\begin{array}{ccc} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{array}), x = (\begin{array}{c} o \\ w \\ r \end{array}), b = (\begin{array}{c} 60 \\ 90 \\ 70 \end{array})$

Step 2: Find $A^{- 1}$

We have

$\det (A) = 50,$

and the adjugate of $A$ is

$adj (A) = (\begin{array}{ccc} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{array})$

Hence,

$A^{- 1} = \frac{1}{50} (\begin{array}{ccc} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{array}) = (\begin{array}{ccc} 0 & - \frac{1}{10} & \frac{1}{5} \\ \frac{3}{5} & 0 & - \frac{2}{5} \\ - \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \end{array})$

Step 3: Find $x = A^{- 1}$

$(\begin{array}{c} o \\ w \\ r \end{array}) = (\begin{array}{ccc} 0 & - \frac{1}{10} & \frac{1}{5} \\ \frac{3}{5} & 0 & - \frac{2}{5} \\ - \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \end{array}) (\begin{array}{c} 60 \\ 90 \\ 70 \end{array})$

Now multiplying:

$\begin{aligned} o & = 0 (60) + (- \frac{1}{10}) (90) + (\frac{1}{5}) (70) = - 9 + 14 = 5, \\ w & = (\frac{3}{5}) (60) + 0 (90) + (- \frac{2}{5}) (70) = 36 - 28 = 8, \\ r & = (- \frac{2}{5}) (60) + (\frac{1}{5}) (90) + (\frac{1}{5}) (70) = - 24 + 18 + 14 = 8. \end{aligned}$

Final Answers:

$\begin{aligned} Cost of onion per kg & = ₹ 5, \\ Cost of wheat per kg & = ₹ 8, \\ Cost of rice per kg & = ₹ 8 \end{aligned}$

October 29, 2025
Exercise-4.5, Class 12th, Maths, Chapter 4, NCERT
Examine the consistency of the system of equations in Exercises 1 to 6.

1.
${\begin{cases} x + 2 y = 2 \\ 2 x + 3 y = 3 \end{cases}$

Solution. From the first, $x = 2 - 2 y$ . Substitute: $2 (2 - 2 y) + 3 y = 3 \Rightarrow 4 - y = 3 \Rightarrow y = 1$ . Then $x = 0$ .
Unique solution: $(x, y) = (0, 1)$

2.
${\begin{cases} 2 x - y = 5 \\ x + y = 4 \end{cases}$

Solution. From second $y = 4 - x$ . Substitute: $2 x - (4 - x) = 5 \Rightarrow 3 x = 9 \Rightarrow x = 3$ . Then $y = 1$
Solution: $(3, 1)$

3.
${\begin{cases} x + 3 y = 5 \\ 2 x + 6 y = 8 \end{cases}$

Solution. The second equation is $2 (x + 3 y) = 10$ , but RHS is 8, not 10. The two equations are inconsistent (parallel, no solution).
No solution.

4.
${\begin{cases} x + y + z = 1 \\ 2 x + 3 y + 2 z = 2 \\ a x + a y + 2 a z = 4 \end{cases}$

Solution. From (1) & (2): subtract $2$ (1) from (2) → $y = 0$ . Then $x + z = 1$ .
Third equation is $a (x + y + 2 z) = 4$ . With $y = 0$ this becomes $a (x + 2 z) = 4$ . Using $x = 1 - z$ :

$a ((1 - z) + 2 z) = a (1 + z) = 4 \Rightarrow 1 + z = \frac{4}{a} (if a \neq 0) .$

So for $a \neq 0$ : $z = \frac{4}{a} - 1, x = 2 - \frac{4}{a}, y = 0$ (unique solution).
If $a = 0$ : third eqn becomes $0 = 4$ impossible → inconsistent.

So: consistent (unique) for $a \neq 0$ with $(x, y, z) = (2 - \frac{4}{a}, 0, \frac{4}{a} - 1)$ . Inconsistent when $a = 0$ . depending on $a$ .

5.
${\begin{cases} 3 x - y - 2 z = 2 \\ 2 y - z = - 1 \\ 3 x - 5 y = 3 \end{cases}$

Solution. From second: $z = 2 y + 1$ . From third: $x = (3 + 5 y) / 3$ . Substitute into first:

$3 \cdot \frac{3 + 5 y}{3} - y - 2 (2 y + 1) = 2 \Rightarrow (3 + 5 y) - y - 4 y - 2 = 2 \Rightarrow 1 = 2,$

contradiction. So inconsistent (no solution).

6.
${\begin{cases} 5 x - y + 4 z = 5 \\ 2 x + 3 y + 5 z = 2 \\ 5 x - 2 y + 6 z = - 1 \end{cases}$

Solution. Subtract (1) from (3): $- y + 2 z = - 6 \Rightarrow y - 2 z = 6$ $. Put in (2):$

$2 x + 3 (6 + 2 z) + 5 z = 2 \Rightarrow 2 x + 18 + 11 z = 2 \Rightarrow x = - 8 - \frac{11}{2} z$

Put $x, y$ into (1): solving gives $z = - 2$ , then $y = 2$ , $x = 3$
Unique solution: (x,y,z) = (3,2,-2)

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7

Solve by matrix method:

${\begin{cases} 5 x + 2 y = 4, \\ 7 x + 3 y = 5. \end{cases}$

Matrix form: $A x = b$ with

$A = [\begin{array}{cc} 5 & 2 \\ 7 & 3 \end{array}], x = [\begin{array}{c} x \\ y \end{array}], b = [\begin{array}{c} 4 \\ 5 \end{array}]$

$\det A = 5 \cdot 3 - 7 \cdot 2 = 15 - 14 = 1$
$A^{- 1} = [\begin{array}{cc} 3 & - 2 \\ - 7 & 5 \end{array}]$ (since $1 / \det A = 1$ ).
Hence $x = A^{- 1} b = [\begin{array}{cc} 3 & - 2 \\ - 7 & 5 \end{array}] [\begin{array}{c} 4 \\ 5 \end{array}] = [\begin{array}{c} 2 \\ - 3 \end{array}]$

Answer: $x = 2, y = - 3.$

Question 8

${\begin{cases} 2 x - y = - 2, \\ 3 x + 4 y = 3. \end{cases}$

Matrix form: $A = [\begin{array}{cc} 2 & - 1 \\ 3 & 4 \end{array}], b = [\begin{array}{c} - 2 \\ 3 \end{array}]$
$\det A = 2 \cdot 4 - 3 \cdot (- 1) = 8 + 3 = 11$

$A^{- 1} = \frac{1}{11} [\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}]$
$x = A^{- 1} b = \frac{1}{11} [\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}] [\begin{array}{c} - 2 \\ 3 \end{array}] = [\begin{array}{c} - \frac{5}{11} \\ \frac{12}{11} \end{array}]$

Answer: $x = - \frac{5}{11}, y = \frac{12}{11} .$

Question 9

${\begin{cases} 4 x - 3 y = 3, \\ 3 x - 5 y = 7. \end{cases}$

Matrix form: $A = [\begin{array}{cc} 4 & - 3 \\ 3 & - 5 \end{array}], b = [\begin{array}{c} 3 \\ 7 \end{array}]$
$\det A = 4 (- 5) - 3 (- 3) = - 20 + 9 = - 11$

$A^{- 1} = \frac{1}{- 11} [\begin{array}{cc} - 5 & 3 \\ - 3 & 4 \end{array}] = \frac{1}{11} [\begin{array}{cc} 5 & - 3 \\ 3 & - 4 \end{array}]$ $x = A^{- 1} b = \frac{1}{11} [\begin{array}{cc} 5 & - 3 \\ 3 & - 4 \end{array}] [\begin{array}{c} 3 \\ 7 \end{array}] = [\begin{array}{c} - \frac{6}{11} \\ - \frac{19}{11} \end{array}]$

Answer: $x = - \frac{6}{11}, y = - \frac{19}{11} .$

Question 10

${\begin{cases} 5 x + 2 y = 3, \\ 3 x + 2 y = 5. \end{cases}$

Matrix form: $A = [\begin{array}{cc} 5 & 2 \\ 3 & 2 \end{array}], b = [\begin{array}{c} 3 \\ 5 \end{array}]$
Subtract second from first: $2 x = - 2 \Rightarrow x = - 1$ Then $5 (- 1) + 2 y = 3 \Rightarrow - 5 + 2 y = 3 \Rightarrow 2 y = 8 \Rightarrow y = 4.$

(Or use inverse: $\det A = 5 \cdot 2 - 3 \cdot 2 = 10 - 6 = 4$ , $A^{- 1} = \frac{1}{4} [\begin{array}{cc} 2 & - 2 \\ - 3 & 5 \end{array}]$ giving same result.)

Answer: $x = - 1, y = 4.$

Question 11

Solve the system

${\begin{cases} 2 x + y + z = 1, \\ x - 2 y - z = \frac{3}{2}, \\ 2 x + y - 3 z = 0. \end{cases}$

Solution (matrix / elimination).

Augmented matrix:

$[\begin{array}{cccc} 2 & 1 & 1 & 1 \\ 1 & - 2 & - 1 & \frac{3}{2} \\ 2 & 1 & - 3 & 0 \end{array}]$

To avoid fractions, multiply row2 by 2 first:

$R_{2} \leftarrow 2 R_{2} : [\begin{array}{cccc} 2 & 1 & 1 & 1 \\ 2 & - 4 & - 2 & 3 \\ 2 & 1 & - 3 & 0 \end{array}]$

Now eliminate using $R_{1}$ :
- $R_{2} \leftarrow R_{2} - R_{1}$
  $[0, - 5, - 3 ∣ 2]$
- $R_{3} \leftarrow R_{3} - R_{1}$
  $[0, 0, - 4 ∣ - 1]$
Matrix becomes

$[\begin{array}{cccc} 2 & 1 & 1 & 1 \\ 0 & - 5 & - 3 & 2 \\ 0 & 0 & - 4 & - 1 \end{array}]$

Back-substitution:

From row3: $- 4 z = - 1 \Rightarrow z = \frac{1}{4}$

Row2: $- 5 y - 3 z = 2 \Rightarrow - 5 y - 3 \cdot \frac{1}{4} = 2 \Rightarrow - 5 y = \frac{11}{4} \Rightarrow y = - \frac{11}{20}$

Row1: $2 x + y + z = 1 \Rightarrow 2 x + (- \frac{11}{20}) + \frac{1}{4} = 1$
Compute $y + z = - \frac{11}{20} + \frac{5}{20} = - \frac{6}{20} = - \frac{3}{10}$
So $2 x - \frac{3}{10} = 1 \Rightarrow 2 x = \frac{13}{10} \Rightarrow x = \frac{13}{20}$

Answer: $x = \frac{13}{20}, y = - \frac{11}{20}, z = \frac{1}{4}$

Question 12

Solve the system

${\begin{cases} x - y + z = 4, \\ 2 x + y - 3 z = 0, \\ 3 y - 5 z = 9. \end{cases}$

Solution (matrix / substitution).

From the third equation:

$3 y - 5 z = 9 \Rightarrow y = \frac{9 + 5 z}{3}$

Substitute $y$ into the first equation:

$x - \frac{9 + 5 z}{3} + z = 4$

Multiply by 3:

$3 x - (9 + 5 z) + 3 z = 12 \Rightarrow 3 x - 9 - 2 z = 12 \Rightarrow 3 x - 2 z = 21.$

So $\begin{matrix} x = 7 + \frac{2}{3} z . & (*) \end{matrix}$

Now substitute $x$ and $y$ into the second equation $2 x + y - 3 z = 0$

$2 (7 + \frac{2}{3} z) + \frac{9 + 5 z}{3} - 3 z = 0.$

Compute left side:

$14 + \frac{4}{3} z + \frac{9 + 5 z}{3} - 3 z = 14 + \frac{4 z + 9 + 5 z}{3} - 3 z$

$= 14 + \frac{9 + 9 z}{3} - 3 z = 14 + 3 + 3 z - 3 z = 17.$

That gives $17 = 0$ , a contradiction.

Conclusion: The system is inconsistent; no solution.

Answer: $No solution (inconsistent system).$

Question 13

Solve the system

${\begin{cases} 2 x + 3 y + 3 z = 5, \\ x - 2 y + z = - 4, \\ 3 x - y - 2 z = 3. \end{cases}$

Solution (matrix / elimination).

Augmented matrix:

$[\begin{array}{cccc} 2 & 3 & 3 & 5 \\ 1 & - 2 & 1 & - 4 \\ 3 & - 1 & - 2 & 3 \end{array}]$

Use row2 as pivot to eliminate others:
- $R_{1} \leftarrow R_{1} - 2 R_{2}$
  $[0, 7, 1 ∣ 13]$
- $R_{3} \leftarrow R_{3} - 3 R_{2}$
  $[0, 5, - 5 ∣ 15]$
So we have

$[\begin{array}{cccc} 1 & - 2 & 1 & - 4 \\ 0 & 7 & 1 & 13 \\ 0 & 5 & - 5 & 15 \end{array}]$

(reordered so row2 is the original $R_{2}$ ).

Eliminate $y$ from row3 using row2:
- $R_{3} \leftarrow 7 R_{3} - 5 R_{2}$
$7 R_{3} = [0, 35, - 35 ∣ 105], 5 R_{2} = [0, 35, 5 ∣ 65]$

So $R_{3} \leftarrow [0, 0, - 40 ∣ 40]$ ⇒ $- 40 z = 40 \Rightarrow z = - 1$

Back-substitute:

Row2: $7 y + z = 13 \Rightarrow 7 y - 1 = 13 \Rightarrow 7 y = 14 \Rightarrow y = 2$

Row1 (original row2): $x - 2 y + z = - 4 \Rightarrow x - 4 - 1 = - 4 \Rightarrow x = 1$

Answer: $x = 1, y = 2, z = - 1$

Question 14 (Plain text)

Solve the system

${\begin{cases} x - y + 2 z = 7, \\ 3 x + 4 y - 5 z = - 5, \\ 2 x - y + 3 z = 12. \end{cases}$

Solution (matrix / elimination).

Augmented matrix:

$[\begin{array}{cccc} 1 & - 1 & 2 & 7 \\ 3 & 4 & - 5 & - 5 \\ 2 & - 1 & 3 & 12 \end{array}]$

Eliminate below first pivot $R_{1}$ :
- $R_{2} \leftarrow R_{2} - 3 R_{1} : [0, 7, - 11 ∣ - 26]$
- $R_{3} \leftarrow R_{3} - 2 R_{1} : [0, 1, - 1 ∣ - 2]$
Swap $R_{2}$ and $R_{3}$ to use the simpler pivot:

$[\begin{array}{cccc} 1 & - 1 & 2 & 7 \\ 0 & 1 & - 1 & - 2 \\ 0 & 7 & - 11 & - 26 \end{array}]$

Eliminate $y$ from row3:
- $R_{3} \leftarrow R_{3} - 7 R_{2} : [0, 0, - 4 ∣ - 12]$
So $- 4 z = - 12 \Rightarrow z = 3$

Back-substitute:

Row2: $y - z = - 2 \Rightarrow y - 3 = - 2 \Rightarrow y = 1$

Row1: $x - y + 2 z = 7 \Rightarrow x - 1 + 6 = 7 \Rightarrow x = 2$

Answer: $x = 2, y = 1, z = 3$
October 29, 2025
Exercise-4.4, Class 12th, Maths, Part -2

Question 14 :

For the matrix $A = [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}]$ , find the numbers $a$ and $b$ such that

$A^{2} + a A + b I = 0.$

Solution:

Given:

$A = [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}]$

We are to find $a$ and $b$ such that

$A^{2} + a A + b I = 0$

Step 1: Compute $A^{2}$

$A^{2} = [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}] [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}]$ Perform the multiplication:

$A^{2} = [\begin{array}{cc} 3 \times 3 + 2 \times 1 & 3 \times 2 + 2 \times 1 \\ 1 \times 3 + 1 \times 1 & 1 \times 2 + 1 \times 1 \end{array}] = [\begin{array}{cc} 9 + 2 & 6 + 2 \\ 3 + 1 & 2 + 1 \end{array}] = [\begin{array}{cc} 11 & 8 \\ 4 & 3 \end{array}]$

Step 2: Substitute in the equation

$A^{2} + a A + b I = 0$

That means:

$[\begin{array}{cc} 11 & 8 \\ 4 & 3 \end{array}] + a [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}] + b [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$

Step 3: Combine all matrices

$[\begin{array}{cc} 11 + 3 a + b & 8 + 2 a \\ 4 + a & 3 + a + b \end{array}] = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$

Step 4: Equate corresponding elements to zero

From the first row, first column:

$11 + 3 a + b = 0 (1)$

From the first row, second column:

$8 + 2 a = 0 (2)$

From the second row, first column:

$4 + a = 0 (3)$

From the second row, second column:

$3 + a + b = 0 (4)$

Step 5: Solve for $a$ and $b$

From (3):

$a = - 4$

Substitute $a = - 4$ :

$3 + (- 4) + b = 0 ⟹ b = 1$

So $a = - 4$ and $b = 1$ .

Step 6: Verification

Check equation (1):

$11 + 3 (- 4) + 1 = 11 - 12 + 1 = 0$

True. Check (2):

$8 + 2 (- 4) = 8 - 8 = 0$ True.

Final Answer:

$a = - 4, b = 1$

Question 15

For the matrix $A = [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}]$ , show that

$A^{3} - 6 A^{2} + 5 A + 11 I = O .$

Hence, find $A^{- 1}$ .

Solution

Step 1 — compute $A^{2}$ and $A^{3}$ .

$A = [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}]$

Compute $A^{2} = A \cdot A$ :

$A^{2} = [\begin{array}{ccc} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}]$

Compute $A^{3} = A^{2} \cdot A$ :

$A^{3} = [\begin{array}{ccc} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{array}]$

Step 2 — form the combination $A^{3} - 6 A^{2} + 5 A + 11 I$ .

Calculate termwise:

$\begin{aligned} A^{3} - 6 A^{2} + 5 A + 11 I & = [\begin{array}{ccc} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{array}] - 6 [\begin{array}{ccc} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}] + 5 [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] + 11 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{ccc} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0 \\ - 23 + 18 + 5 + 0 & 27 - 48 + 10 + 11 & - 69 + 84 - 15 + 0 \\ 32 - 42 + 10 + 0 & - 13 + 18 - 5 + 0 & 58 - 84 + 15 + 11 \end{array}] \\ = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \end{aligned}$

So the relation is verified:

$A^{3} - 6 A^{2} + 5 A + 11 I = O$

Step 3 — find $A^{- 1}$ .

Starting from

$A^{3} - 6 A^{2} + 5 A + 11 I = O,$

multiply on the right by $A^{- 1}$ (which is valid because $A$ is invertible if we can find $A^{- 1}$ from this equation):

$A^{2} - 6 A + 5 I + 11 A^{- 1} = O$

Rearrange to solve for $A^{- 1}$ :

$11 A^{- 1} = - A^{2} + 6 A - 5 I, A^{- 1} = \frac{1}{11} (6 A - A^{2} - 5 I)$

Now substitute the matrices $A$ and $A^{2}$ :

$6 A - A^{2} - 5 I = 6 [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] - [\begin{array}{ccc} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}] - 5 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ $= [\begin{array}{ccc} 6 - 4 - 5 & 6 - 2 - 0 & 6 - 1 - 0 \\ 6 - (- 3) - 0 & 12 - 8 - 5 & - 18 - (- 14) - 0 \\ 12 - 7 - 0 & - 6 - (- 3) - 0 & 18 - 14 - 5 \end{array}] = [\begin{array}{ccc} - 3 & 4 & 5 \\ 9 & - 1 & - 4 \\ 5 & - 3 & - 1 \end{array}]$

Thus

$A^{- 1} = \frac{1}{11} [\begin{array}{ccc} - 3 & 4 & 5 \\ 9 & - 1 & - 4 \\ 5 & - 3 & - 1 \end{array}]$

You can write it explicitly as

$A^{- 1} = [\begin{array}{ccc} - \frac{3}{11} & \frac{4}{11} & \frac{5}{11} \\ \frac{9}{11} & - \frac{1}{11} & - \frac{4}{11} \\ \frac{5}{11} & - \frac{3}{11} & - \frac{1}{11} \end{array}]$

(One may verify $A A^{- 1} = I$ )

Question 16 : If

$A = [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}],$

verify that

$A^{3} - 6 A^{2} + 9 A - 4 I = O,$

and hence find $A^{- 1}$ .

Solution

Step 1 — compute $A^{2}$ .

$A^{2} = A \cdot A = [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] = [\begin{array}{ccc} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}]$ (Each entry calculated: e.g. top-left $= 2 \cdot 2 + (- 1) (- 1) + 1 \cdot 1 = 4 + 1 + 1 = 6$ , top-middle $= 2 (- 1) + (- 1) 2 + 1 (- 1) = - 2 - 2 - 1 = - 5$

Step 2 — compute $A^{3}$ .

$A^{3} = A^{2} \cdot A = [\begin{array}{ccc} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{array}] .$

(For example, top-left $= 6 \cdot 2 + (- 5) (- 1) + 5 \cdot 1 = 12 + 5 + 5 = 22$ , top-middle $= 6 (- 1) + (- 5) 2 + 5 (- 1) = - 6 - 10 - 5 = - 21$

Step 3 — form the combination $A^{3} - 6 A^{2} + 9 A - 4 I$

Compute termwise:

$\begin{aligned} A^{3} - 6 A^{2} + 9 A - 4 I & = [\begin{array}{ccc} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{array}] - 6 [\begin{array}{ccc} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] + 9 [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] - 4 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{ccc} 22 - 36 + 18 - 4 & - 21 + 30 - 9 + 0 & 21 - 30 + 9 + 0 \\ - 21 + 30 - 9 + 0 & 22 - 36 + 18 - 4 & - 21 + 30 - 9 + 0 \\ 21 - 30 + 9 + 0 & - 21 + 30 - 9 + 0 & 22 - 36 + 18 - 4 \end{array}] \\ = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \end{aligned}$

Thus the identity is verified:

$A^{3} - 6 A^{2} + 9 A - 4 I = O$

Step 4 — find $A^{- 1}$ .

Start from the matrix polynomial

$A^{3} - 6 A^{2} + 9 A - 4 I = O$

Right-multiply by $A^{- 1}$ (valid since the relation will imply invertibility):

$A^{2} - 6 A + 9 I - 4 A^{- 1} = O$

Rearrange to solve for $A^{- 1}$ :

$4 A^{- 1} = A^{2} - 6 A + 9 I \Rightarrow A^{- 1} = \frac{1}{4} (A^{2} - 6 A + 9 I)$

Now substitute the matrices $A$ and $A^{2}$ :

$A^{2} - 6 A + 9 I = [\begin{array}{ccc} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] - 6 [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] + 9 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ Compute it:

$A^{2} - 6 A + 9 I = [\begin{array}{ccc} 6 - 12 + 9 & - 5 + 6 + 0 & 5 - 6 + 0 \\ - 5 + 6 + 0 & 6 - 12 + 9 & - 5 + 6 + 0 \\ 5 - 6 + 0 & - 5 + 6 + 0 & 6 - 12 + 9 \end{array}] = [\begin{array}{ccc} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array}]$

Therefore

$A^{- 1} = \frac{1}{4} [\begin{array}{ccc} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array}] = [\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & - \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ - \frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{array}]$

(You can verify $A A^{- 1} = I$ directly.)

Final answer:

$A^{3} - 6 A^{2} + 9 A - 4 I = O, A^{- 1} = [\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & - \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ - \frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{array}]$

Question 17 :

Let $A$ be a nonsingular square matrix of order $3 \times 3$ . Then $∣ adj A ∣$ is equal to

(A) $∣ A ∣$
(B) $∣ A ∣^{2}$
(C) $∣ A ∣^{3}$
(D) $3 ∣ A ∣$

Solution:

We know the formula for the determinant of the adjugate (adjoint) of a square matrix:

$∣ adj A ∣ = ∣ A ∣^{n - 1}$ where $n$ is the order of the square matrix.

Here, $A$ is of order $3 \times 3$ .
So $n = 3$

$∣ adj A ∣ = ∣ A ∣^{3 - 1} = ∣ A ∣^{2}$

Final Answer:

$∣ adj A ∣ = ∣ A ∣^{2}$

Hence, the correct option is (B) $∣ A ∣^{2}$

Question 18:

If $A$ is an invertible matrix of order $2$ , then $\det (A^{- 1})$ is equal to

(A) $\det (A)$
(B) $\frac{1}{\det (A)}$
(C) $1$
(D) $0$

Solution:

We know the fundamental property of determinants:

$\det (A^{- 1}) = \frac{1}{\det (A)}$

provided $A$ is invertible (that is, $\det (A) \neq 0$ ).

Final Answer:

$\det (A^{- 1}) = \frac{1}{\det (A)}$

Hence, the correct option is (B) $\frac{1}{\det (A)}$ .

October 29, 2025
Exercise4.4, Class 12th, Maths, Chapter 4, NCERT
Exercise 4.4

Question 1:
Find $adj A$ for

$A = (\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array})$ Solution:
For a $2 \times 2$ matrix

$A = (\begin{array}{cc} a & b \\ c & d \end{array})$ $adj A = (\begin{array}{cc} d & - b \\ - c & a \end{array})$

So for

$A = (\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array})$

$adj A = (\begin{array}{cc} 4 & - 2 \\ - 3 & 1 \end{array})$

(You can check $A (adj A) = (\det A) I$ :

$\det A = 1 \cdot 4 - 2 \cdot 3 = - 2$ , and indeed $A adj A = (- 2) I$ )

Question 2:
Find $adj A$

for $A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 2 & 3 & 5 \\ - 2 & 0 & 1 \end{array})$

Below I show the cofactors, the adjoint, and a verification

$A adj A = (\det A) I$

Minors and cofactors

Compute the $2 \times 2$ minors and cofactors

$A_{i j} = (- 1)^{i + j} M_{i j}$

$Cofactor matrix [A_{i j}] = (\begin{array}{ccc} 3 & - 12 & 6 \\ 1 & 5 & 2 \\ - 11 & - 1 & 5 \end{array})$

(Each entry is the cofactor of the corresponding element of

$A$ .)

$adj A$

(adjugate = transpose of cofactor matrix)

$adj A = (\begin{array}{ccc} 3 & 1 & - 11 \\ - 12 & 5 & - 1 \\ 6 & 2 & 5 \end{array})$

Determinant and verification

$\det A = 27$

Now verify

$A adj A = (\det A) I = 27 I$

Multiplying gives

$A adj A = (\begin{array}{ccc} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{array}) = 27 I,$

so the adjoint is correct.

___________________

Verify the identity for both matrices

$Show that A (adj A) = (adj A) A = ∣ A ∣ I$

Question 3:

$A = (\begin{array}{cc} 2 & 3 \\ - 4 & - 6 \end{array})$

Solution – Compute the determinant:

$∣ A ∣ = 2 (- 6) - 3 (- 4) = - 12 + 12 = 0.$

Compute the Adj (classical adjoint). For a $2 \times 2$ matrix

$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$adj A = (\begin{array}{cc} d & - b \\ - c & a \end{array})$

Thus,

$adj A = (\begin{array}{cc} - 6 & - 3 \\ 4 & 2 \end{array})$

Multiply:

$A (adj A) = (\begin{array}{cc} 2 & 3 \\ - 4 & - 6 \end{array}) (\begin{array}{cc} - 6 & - 3 \\ 4 & 2 \end{array}) = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Similarly

$(adj A) A = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Since

$∣ A ∣ = 0$ , we have

$∣ A ∣ I = 0 \cdot I = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Hence,

$A (adj A) = (adj A) A = ∣ A ∣ I,$

as required.

Question 4:

$A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{array})$

Solution – Compute the determinant –

$∣ A ∣ = 11$

Compute the adjugate (transpose of cofactor matrix). The cofactors lead to

$adj A = (\begin{array}{ccc} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{array}) .$

Multiply:

$A (adj A) = (\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}) = 11 I,$

and likewise

$(adj A) A = 11 I$

Thus for this matrix also

$A (adj A) = (adj A) A = ∣ A ∣ I$

Conclusion

Both verifications hold:
- For
  $A = (\begin{array}{cc} 2 & 3 \\ - 4 & - 6 \end{array})$ $A (adj A) = (adj A) A = 0 = ∣ A ∣ I$
- For
  $A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{array})$
$A (adj A) = (adj A) A = 11 I .$

Question 5:

$Find A^{- 1} if A = (\begin{array}{cc} 2 & - 2 \\ 4 & 3 \end{array}) .$

Formula for inverse of a 2×2 matrix

$A = (\begin{array}{cc} a & b \\ c & d \end{array})$

its inverse (if

$∣ A ∣ \neq 0$

is

$A^{- 1} = \frac{1}{a d - b c} (\begin{array}{cc} d & - b \\ - c & a \end{array}$

Compute determinant

$∣ A ∣ = (2) (3) - (- 2) (4) = 6 + 8 = 14$

Since

$∣ A ∣ \neq 0$

, the inverse exists.

Apply the formula

$A^{- 1} = \frac{1}{14} (\begin{array}{cc} 3 & 2 \\ - 4 & 2 \end{array})$

Final Answer:

$A^{- 1} = \frac{1}{14} (\begin{array}{cc} 3 & 2 \\ - 4 & 2 \end{array})$

Verification (optional):

$A A^{- 1} = (\begin{array}{cc} 2 & - 2 \\ 4 & 3 \end{array}) \frac{1}{14} (\begin{array}{cc} 3 & 2 \\ - 4 & 2 \end{array}) = \frac{1}{14} (\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}) = I$

✔ Verified.

Question 6:

$Find A^{- 1} if A = (\begin{array}{cc} - 1 & 5 \\ - 3 & 2 \end{array})$

Formula for inverse

For any

$2 \times 2$

matrix

$A = (\begin{array}{cc} a & b \\ c & d \end{array}),$

if

$∣ A ∣ = a d - b c \neq 0$

, then

$A^{- 1} = \frac{1}{∣ A ∣} (\begin{array}{cc} d & - b \\ - c & a \end{array})$

Compute determinant

$∣ A ∣ = (- 1) (2) - (5) (- 3) = - 2 + 15 = 13$

$∣ A ∣ = 13 \neq 0$

, so the inverse exists.

Apply the formula

$A^{- 1} = \frac{1}{13} (\begin{array}{cc} 2 & - 5 \\ 3 & - 1 \end{array})$

Final Answer:

$A^{- 1} = \frac{1}{13} (\begin{array}{cc} 2 & - 5 \\ 3 & - 1 \end{array})$

Verification (optional):

$A A^{- 1} = (\begin{array}{cc} - 1 & 5 \\ - 3 & 2 \end{array}) \frac{1}{13} (\begin{array}{cc} 2 & - 5 \\ 3 & - 1 \end{array}) = \frac{1}{13} (\begin{array}{cc} 13 & 0 \\ 0 & 13 \end{array}) = I$

✔ Verified.

Question 7:

$Find A^{- 1} if A = (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array})$

Observation

$A$

is an upper triangular matrix, and for triangular matrices,

$∣ A ∣ = product of diagonal elements.$

Determinant

$∣ A ∣ = 1 \times 2 \times 5 = 10$

Since

$∣ A ∣ \neq 0$ , $A^{- 1}$ exists.

Use properties of triangular matrices

For an upper triangular matrix, its inverse is also upper triangular.
We find

$A^{- 1} = [a_{i j}]$

such that

$A A^{- 1} = I$

Let

$A^{- 1} = (\begin{array}{ccc} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{array})$

Then

$A A^{- 1} = (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}) (\begin{array}{ccc} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{array}) = (\begin{array}{ccc} a & b + 2 d & c + 2 e + 3 f \\ 0 & 2 d & 2 e + 4 f \\ 0 & 0 & 5 f \end{array})$

We want this to equal the identity

$I = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array})$

Compare entries

From the bottom:

$f = 1 ⟹ f = \frac{1}{5}$

$2 d = 1 ⟹ d = \frac{1}{2}$

$a = 1$

Now use off-diagonal equations:

$e + 4 f = 0 ⟹ e = - 2 f = - \frac{2}{5}$

$b + 2 d = 0 ⟹ b = - 2 d = - 1$

$c + 2 e + 3 f = 0 ⟹ c = - 2 e - 3 f$

Substitute

$e = - \frac{2}{5}$ and $f = \frac{1}{5}$ :

$c = - 2 (- \frac{2}{5}) - 3 (\frac{1}{5}) = \frac{4}{5} - \frac{3}{5} = \frac{1}{5}$

Write the inverse

$A^{- 1} = (\begin{array}{ccc} 1 & - 1 & \frac{1}{5} \\ 0 & \frac{1}{2} & - \frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{array})$

Verification (optional):

$A A^{- 1} = (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}) (\begin{array}{ccc} 1 & - 1 & \frac{1}{5} \\ 0 & \frac{1}{2} & - \frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{array}) = I$

✔ Verified.

Question 8:

$Find A^{- 1} for A = (\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & - 1 \end{array})$

Solution.

$A$ is lower triangular, so $A^{- 1}$ is also lower triangular.
Let

$A^{- 1} = (\begin{array}{ccc} a & 0 & 0 \\ b & c & 0 \\ d & e & f \end{array})$

and solve

$A A^{- 1} = I$

.Multiply $A$ and $A^{- 1}$ : From the first row: $a = 1$

.From the second row:

$3 a + 3 b = 0, 3 c = 1 \Rightarrow b = - 1, c = \frac{1}{3}$

From the third row:

$5 a + 2 b - d = 0, 2 c - e = 0, - f = 1$

Substituting

$a = 1, b = - 1, c = \frac{1}{3}$

$,$ gives

$5 (1) + 2 (- 1) - d = 0 \Rightarrow d = 3, 2 \cdot \frac{1}{3} - e = 0 \Rightarrow e = \frac{2}{3}, f = - 1.$

Thus

$A^{- 1} = (\begin{array}{ccc} 1 & 0 & 0 \\ - 1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & - 1 \end{array})$

(You can verify

$A A^{- 1} = I$

by multiplication.)

Question 9:

find $A^{- 1}$

for $A = (\begin{array}{ccc} 2 & 1 & 3 \\ 4 & - 1 & 0 \\ - 7 & 2 & 1 \end{array})$

Determinant

Compute $∣ A ∣$

(expand along the first row):

$∣ A ∣ = 2 ∣ \begin{array}{cc} - 1 & 0 \\ 2 & 1 \end{array} ∣ - 1 ∣ \begin{array}{cc} 4 & 0 \\ - 7 & 1 \end{array} ∣ + 3 ∣ \begin{array}{cc} 4 & - 1 \\ - 7 & 2 \end{array} ∣$

$= 2 (- 1) - 1 (4) + 3 (1) = - 2 - 4 + 3 = - 3$

So

$∣ A ∣ = - 3$

(non zero) — inverse exists.

Cofactor / Adjugate

Compute the matrix of cofactors (I show the cofactors

$C_{i j}$ ):

$[C_{i j}] = (\begin{array}{ccc} - 1 & - 4 & 1 \\ 5 & 23 & - 11 \\ 3 & 12 & - 6 \end{array})$

The adjugate (adj $A$ ) is the transpose of this:

$adj A = (\begin{array}{ccc} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{array})$

Inverse

$A^{- 1} = \frac{1}{∣ A ∣} adj A = - \frac{1}{3} (\begin{array}{ccc} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{array}) = (\begin{array}{ccc} \frac{1}{3} & - \frac{5}{3} & - 1 \\ \frac{4}{3} & - \frac{23}{3} & - 4 \\ - \frac{1}{3} & \frac{11}{3} & 2 \end{array})$

Equivalently-

$A^{- 1} = \frac{1}{3} (\begin{array}{ccc} 1 & - 5 & - 3 \\ 4 & - 23 & - 12 \\ - 1 & 11 & 6 \end{array})$

Quick check (first-row × first-column)

$2 \cdot \frac{1}{3} + 1 \cdot \frac{4}{3} + 3 \cdot (- \frac{1}{3}) = \frac{2 + 4 - 3}{3} = 1,$

so the product gives the identity as expected.

Question 10.

$Find A^{- 1} for A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array})$

Solution:

Compute

$∣ A ∣$

. Expanding along the first row,

$\begin{aligned} ∣ A ∣ & = 1 \cdot ∣ \begin{array}{cc} 2 & - 3 \\ - 2 & 4 \end{array} ∣ - (- 1) \cdot ∣ \begin{array}{cc} 0 & - 3 \\ 3 & 4 \end{array} ∣ + 2 \cdot ∣ \begin{array}{cc} 0 & 2 \\ 3 & - 2 \end{array} ∣ \\ = 1 (2 \cdot 4 - (- 3) (- 2)) + 1 (0 \cdot 4 - (- 3) \cdot 3) + 2 (0 \cdot (- 2) - 2 \cdot 3) \\ = 1 (8 - 6) + 1 (9) + 2 (- 6) = 2 + 9 - 12 = - 1 \end{aligned}$

So

$∣ A ∣ = - 1$

(nonzero) and the inverse exists.

Compute cofactors (minors

$M_{i j}$

and cofactors

$C_{i j} = (- 1)^{i + j} M_{i j}$

$\begin{array}{lll} M_{11} = ∣ \begin{array}{cc} 2 & - 3 \\ - 2 & 4 \end{array} ∣ = 2, & C_{11} = + 2, \\ M_{12} = ∣ \begin{array}{cc} 0 & - 3 \\ 3 & 4 \end{array} ∣ = 9, & C_{12} = - 9, \\ M_{13} = ∣ \begin{array}{cc} 0 & 2 \\ 3 & - 2 \end{array} ∣ = - 6, & C_{13} = - 6, \\ M_{21} = ∣ \begin{array}{cc} - 1 & 2 \\ - 2 & 4 \end{array} ∣ = 0, & C_{21} = 0, \\ M_{22} = ∣ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} ∣ = - 2, & C_{22} = - 2, \\ M_{23} = ∣ \begin{array}{cc} 1 & - 1 \\ 3 & - 2 \end{array} ∣ = 1, & C_{23} = - 1, \\ M_{31} = ∣ \begin{array}{cc} - 1 & 2 \\ 2 & - 3 \end{array} ∣ = - 1, & C_{31} = - 1, \\ M_{32} = ∣ \begin{array}{cc} 1 & 2 \\ 0 & - 3 \end{array} ∣ = - 3, & C_{32} = + 3, \\ M_{33} = ∣ \begin{array}{cc} 1 & - 1 \\ 0 & 2 \end{array} ∣ = 2, & C_{33} = + 2, \end{array}$

Cofactor matrix

$C = [C_{i j}]$

is

$C = (\begin{array}{ccc} 2 & - 9 & - 6 \\ 0 & - 2 & - 1 \\ - 1 & 3 & 2 \end{array})$

Adjugate is transpose of cofactor matrix:

$adj A = C^{T} = (\begin{array}{ccc} 2 & 0 & - 1 \\ - 9 & - 2 & 3 \\ - 6 & - 1 & 2 \end{array})$

Inverse:

$A^{- 1} = \frac{1}{∣ A ∣} adj A = - 1 \cdot adj A = (\begin{array}{ccc} - 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2 \end{array})$

Question 11.

$Find A^{- 1} for A = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{array})$

Solution:

Write

$A = diag (1, M)$

where

$M = (\begin{array}{cc} \cos α & \sin α \\ \sin α & - \cos α \end{array})$

Compute

$\det M$

:

$\det M = (\cos α) (- \cos α) - (\sin α) (\sin α) = - (\cos^{2} α + \sin^{2} α) = - 1$

So

$∣ A ∣ = 1 \cdot \det M = - 1$ , hence

$A$ is invertible.

Now compute $M^{- 1}$ using the $2 \times 2$

formula:

$M^{- 1} = \frac{1}{\det M} (\begin{array}{cc} - \cos α & - \sin α \\ - \sin α & \cos α \end{array})$

$= - 1 (\begin{array}{cc} - \cos α & - \sin α \\ - \sin α & \cos α \end{array}) = (\begin{array}{cc} \cos α & \sin α \\ \sin α & - \cos α \end{array}) = M$

Thus

$M^{- 1} = M$

Equivalently

$M^{2} = I_{2}$

Therefore

$A^{- 1} = diag (1, M^{- 1}) = diag (1, M) = A$

Final answer

$A^{- 1} = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{array}) = A$

(Verification:

$A^{2} = diag (1, M^{2}) = diag (1, I_{2}) = I_{3}$ , so indeed

$A^{- 1} = A$

Question 12:

$Let A = (\begin{array}{cc} 3 & 7 \\ 2 & 5 \end{array}) and B = (\begin{array}{cc} 6 & 8 \\ 7 & 9 \end{array}) .$

$Verify that (A B)^{- 1} = B^{- 1} A^{- 1} .$

Let’s verify

$(A B)^{- 1} = B^{- 1} A^{- 1}$

Compute

$A^{- 1}$

and

$B^{- 1}$

(a) $A^{- 1}$ $∣ A ∣ = 3 (5) - 7 (2) = 15 - 14 = 1$

$A^{- 1} = (\begin{array}{cc} 5 & - 7 \\ - 2 & 3 \end{array})$

(b) $B^{- 1}$ $∣ B ∣ = 6 (9) - 8 (7) = 54 - 56 = - 2$

$B^{- 1} = \frac{1}{- 2} (\begin{array}{cc} 9 & - 8 \\ - 7 & 6 \end{array}) = (\begin{array}{cc} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{array})$

Compute

$A B$ $A B = (\begin{array}{cc} 3 & 7 \\ 2 & 5 \end{array}) (\begin{array}{cc} 6 & 8 \\ 7 & 9 \end{array}) = (\begin{array}{cc} 3 \cdot 6 + 7 \cdot 7 & 3 \cdot 8 + 7 \cdot 9 \\ 2 \cdot 6 + 5 \cdot 7 & 2 \cdot 8 + 5 \cdot 9 \end{array}) = (\begin{array}{cc} 67 & 87 \\ 47 & 61 \end{array})$

Compute

$(A B)^{- 1}$ $∣ A B ∣ = 67 (61) - 87 (47) = 4087 - 4089 = - 2$

$(A B)^{- 1} = \frac{1}{- 2} (\begin{array}{cc} 61 & - 87 \\ - 47 & 67 \end{array}) = (\begin{array}{cc} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array})$

Compute

$B^{- 1} A^{- 1}$ $B^{- 1} A^{- 1} = (\begin{array}{cc} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{array}) (\begin{array}{cc} 5 & - 7 \\ - 2 & 3 \end{array})$

Multiply:

$\begin{aligned} B^{- 1} A^{- 1} & = (\begin{array}{cc} (- \frac{9}{2}) (5) + 4 (- 2) & (- \frac{9}{2}) (- 7) + 4 (3) \\ (\frac{7}{2}) (5) + (- 3) (- 2) & (\frac{7}{2}) (- 7) + (- 3) (3) \end{array}) \\ = (\begin{array}{cc} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array}) \end{aligned}$

Compare

$(A B)^{- 1} = (\begin{array}{cc} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array}) = B^{- 1} A^{- 1}$

Hence verified:

$(A B)^{- 1} = B^{- 1} A^{- 1}$

Question 13 :

If $A = [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}]$ , show that $A^{2} - 5 A + 7 I = O$ . Hence, find $A^{- 1}$ .

Solution:

Let’s solve step-by-step carefully and clearly.

Given:

$A = [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}]$

We have to show that

$A^{2} - 5 A + 7 I = O$

and then use this result to find $A^{- 1}$

Step 1: Compute $A^{2}$

$A^{2} = A \cdot A = [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}] [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}]$

Multiply carefully:

$A^{2} = [\begin{array}{cc} (3) (3) + (1) (- 1) & (3) (1) + (1) (2) \\ (- 1) (3) + (2) (- 1) & (- 1) (1) + (2) (2) \end{array}] = [\begin{array}{cc} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{array}] = [\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}]$

So,

$A^{2} = [\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}]$

Step 2: Compute $5 A$ and $7 I$

$5 A = 5 [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}] = [\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array}]$ $7 I = 7 [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}]$

Step 3: Compute $A^{2} - 5 A + 7 I$

$A^{2} - 5 A + 7 I = [\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}] - [\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array}] + [\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}]$

First subtract $A^{2} - 5 A$ :

$[\begin{array}{cc} 8 - 15 & 5 - 5 \\ - 5 - (- 5) & 3 - 10 \end{array}] = [\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array}]$

Now add $7 I$ :

$[\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array}] + [\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}] = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$

Hence proved:

$A^{2} - 5 A + 7 I = O$

Step 4: Finding $A^{- 1}$

From the equation:

$A^{2} - 5 A + 7 I = 0$

Rearrange it as:

$A^{2} - 5 A = - 7 I$ Multiply both sides by $A^{- 1}$ (on the right):

$A - 5 I = - 7 A^{- 1}$

So, $A^{- 1} = \frac{1}{7} (5 I - A)$

Step 5: Substitute values

$A^{- 1} = \frac{1}{7} (5 [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] - [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}])$ $A^{- 1} = \frac{1}{7} [\begin{array}{cc} 5 - 3 & 0 - 1 \\ 0 - (- 1) & 5 - 2 \end{array}] = \frac{1}{7} [\begin{array}{cc} 2 & - 1 \\ 1 & 3 \end{array}]$

Click Here For Question 14 to 18 of Exercise 4.4
October 29, 2025
Exercise-4.3, Class 12th, Maths, Chapter 4, NCERT
1. Write minors and cofactors of the elements of the following determinants:

(i) $Δ = ∣ \begin{array}{cc} 2 & 4 \\ 0 & 3 \end{array} ∣$

For a $2 \times 2$ determinant the minor $M_{i j}$ of element $a_{i j}$ is the determinant left after deleting its row and column; for $2 \times 2$ deleting a row & column leaves a $1 \times 1$ number.

Elements and their minors:
- $a_{11} = 2 : M_{11} = 3$ . Cofactor $A_{11} = (- 1)^{1 + 1} M_{11} = + 3$
- $a_{12} = 4 : M_{12} = 0$ . Cofactor $A_{12} = (- 1)^{1 + 2} M_{12} = - 0 = 0$
- $a_{21} = 0 : M_{21} = 4$ . Cofactor $A_{21} = (- 1)^{2 + 1} M_{21} = - 4$
- $a_{22} = 3 : M_{22} = 2$ . Cofactor $A_{22} = (- 1)^{2 + 2} M_{22} = + 2$
(You can check: expansion along first row gives $Δ = 2 \cdot A_{11} + 4 \cdot A_{12} = 2 \cdot 3 + 4 \cdot 0 = 6$ , and direct determinant $2 \cdot 3 - 4 \cdot 0 = 6$ )

(ii) $Δ = ∣ \begin{array}{cc} a & c \\ b & d \end{array} ∣$

Minors (each $1 \times 1$ entry):
- $M_{11} = d, A_{11} = + d$
- $M_{12} = b, A_{12} = (- 1)^{1 + 2} b = - b$
- $M_{21} = c, A_{21} = (- 1)^{2 + 1} c = - c$
- $M_{22} = a, A_{22} = + a$
(So cofactors matrix is $(\begin{array}{cc} d & - b \\ - c & a \end{array})$ )

2. Write minors and cofactors for the following $3 \times 3$ determinants:

(i) $I_{3} = ∣ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} ∣$

We give minors $M_{i j}$ and cofactors $A_{i j} = (- 1)^{i + j} M_{i j}$

Because $I_{3}$ is diagonal, minors are determinants of the $2 \times 2$ submatrices:

Row 1:
- $M_{11} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1, A_{11} = + 1$
- $M_{12} = ∣ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} ∣ = 0, A_{12} = - 0 = 0$
- $M_{13} = ∣ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} ∣ = 0, A_{13} = + 0 = 0$
Row 2:
- $M_{21} = ∣ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} ∣ = 0, A_{21} = - 0 = 0$
- $M_{22} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1, A_{22} = + 1$
- $M_{23} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} ∣ = 0, A_{23} = - 0 = 0$
Row 3:
- $M_{31} = ∣ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} ∣ = 0, A_{31} = + 0 = 0$
- $M_{32} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} ∣ = 0, A_{32} = - 0 = 0$
- $M_{33} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1, A_{33} = + 1$
(So adj $I_{3}$ = transpose of cofactor matrix = identity again.)

(ii) $Δ = ∣ \begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & 1 \\ 0 & 1 & 2 \end{array} ∣$

Compute minors (each is a $2 \times 2$ determinant) and cofactors:

Row 1:
- $M_{11} = ∣ \begin{array}{cc} 5 & 1 \\ 1 & 2 \end{array} ∣ = 5 \cdot 2 - 1 \cdot 1 = 10 - 1 = 9, A_{11} = + 9$
- $M_{12} = ∣ \begin{array}{cc} 3 & 1 \\ 0 & 2 \end{array} ∣ = 3 \cdot 2 - 1 \cdot 0 = 6, A_{12} = (- 1)^{1 + 2} 6 = - 6$
- $M_{13} = ∣ \begin{array}{cc} 3 & 5 \\ 0 & 1 \end{array} ∣ = 3 \cdot 1 - 5 \cdot 0 = 3, A_{13} = + 3$
Row 2:
- $M_{21} = ∣ \begin{array}{cc} 0 & 4 \\ 1 & 2 \end{array} ∣ = 0 \cdot 2 - 4 \cdot 1 = - 4, A_{21} = (- 1)^{2 + 1} (- 4) = + 4$
- $M_{22} = ∣ \begin{array}{cc} 1 & 4 \\ 0 & 2 \end{array} ∣ = 1 \cdot 2 - 4 \cdot 0 = 2, A_{22} = + 2$
- $M_{23} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1 \cdot 1 - 0 \cdot 0 = 1, A_{23} = (- 1)^{2 + 3} 1 = - 1$
Row 3:
- $M_{31} = ∣ \begin{array}{cc} 0 & 4 \\ 5 & 1 \end{array} ∣ = 0 \cdot 1 - 4 \cdot 5 = - 20, A_{31} = (- 1)^{3 + 1} (- 20) = - 20$
  (note sign: $(- 1)^{4} = + 1$ , so $A_{31} = + (- 20) = - 20$
- $M_{32} = ∣ \begin{array}{cc} 1 & 4 \\ 3 & 1 \end{array} ∣ = 1 \cdot 1 - 4 \cdot 3 = 1 - 12 = - 11, A_{32} = (- 1)^{3 + 2} (- 11) = + 11$
- $M_{33} = ∣ \begin{array}{cc} 1 & 0 \\ 3 & 5 \end{array} ∣ = 1 \cdot 5 - 0 \cdot 3 = 5, A_{33} = (- 1)^{3 + 3} 5 = + 5$
(You can verify determinant by expansion: $Δ = 1 \cdot A_{11} + 0 \cdot A_{12} + 4 \cdot A_{13} = 1 \cdot 9 + 0 + 4 \cdot 3 = 9 + 12 = 21$

Direct check yields same.)

3. Using cofactors of elements of the second row, evaluate

$Δ = ∣ \begin{array}{ccc} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array} ∣$

We will expand along the second row: $Δ = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$

Compute the cofactors (minors first):
- For $a_{21} = 2$ : $M_{21} = ∣ \begin{array}{cc} 3 & 8 \\ 2 & 3 \end{array} ∣ = 3 \cdot 3 - 8 \cdot 2 = 9 - 16 = - 7$
  $A_{21} = (- 1)^{2 + 1} M_{21} = - (- 7) = + 7$
- For $a_{22} = 0$ : $M_{22} = ∣ \begin{array}{cc} 5 & 8 \\ 1 & 3 \end{array} ∣ = 5 \cdot 3 - 8 \cdot 1 = 15 - 8 = 7$
  $A_{22} = (- 1)^{2 + 2} M_{22} = + 7$
- For $a_{23} = 1$ : $M_{23} = ∣ \begin{array}{cc} 5 & 3 \\ 1 & 2 \end{array} ∣ = 5 \cdot 2 - 3 \cdot 1 = 10 - 3 = 7$
  $A_{23} = (- 1)^{2 + 3} M_{23} = - 7$
Now expansion:

$Δ = 2 \cdot A_{21} + 0 \cdot A_{22} + 1 \cdot A_{23} = 2 \cdot 7 + 0 + 1 \cdot (- 7) = 14 - 7 = 7$

(You may check by any other expansion; result is $7$ .)

4. Using cofactors of elements of the third column, evaluate

$Δ = ∣ \begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ y & z & x \end{array} ∣$

We expand along third column: $Δ = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$
where $a_{13} = 1, a_{23} = z, a_{33} = x$

Compute minors and cofactors:
- $A_{13} = (- 1)^{1 + 3} M_{13} = (- 1)^{4} M_{13} = + M_{13}$
  $M_{13} = ∣ \begin{array}{cc} x & y \\ y & z \end{array} ∣ = x z - y^{2}$
- $A_{23} = (- 1)^{2 + 3} M_{23} = (- 1)^{5} (- M_{23}) = - M_{23} $ $(Simpler:$ $A_{23} = - M_{23}$ )
  $M_{23} = ∣ \begin{array}{cc} 1 & 1 \\ y & z \end{array} ∣ = 1 \cdot z - 1 \cdot y = z - y$
  So $A_{23} = - (z - y) = y - z$
- $A_{33} = (- 1)^{3 + 3} M_{33} = + M_{33}$
  $M_{33} = ∣ \begin{array}{cc} 1 & 1 \\ x & y \end{array} ∣ = 1 \cdot y - 1 \cdot x = y - x$
  So $A_{33} = y - x$
Now expand:

$Δ = 1 \cdot (x z - y^{2}) + z \cdot (y - z) + x \cdot (y - x)$

Simplify term-by-term:

$Δ = x z - y^{2} + z y - z^{2} + x y - x^{2} .$

Group like terms:

$Δ = - x^{2} + (x z + z y + x y) - (y^{2} + z^{2})$

We can rewrite symmetric grouping if desired. But we can also notice a factorization — rearrange as

$Δ = - (x^{2} + y^{2} + z^{2}) + (x y + y z + z x) .$

Thus

$Δ = (x y + y z + z x) - (x^{2} + y^{2} + z^{2})$

(That is the simplest closed form. You may also write $Δ = - \frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]$ — indeed expanding that gives the same value. So $Δ \leq 0$ and equals zero exactly when $x = y = z$ .)

5. If $Δ = [a_{i j}]$ is the $3 \times 3$ determinant and $A_{i j}$ are cofactors of $a_{i j}$ , then which of the following equals $Δ$ ?

Options:
- (A) $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$
- (B) $a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}$
- (C) $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$
- (D) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$
Solution / reasoning.

Standard properties of cofactors / expansions:
- Determinant expansion along the first row gives
  $Δ = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$
- Expansion along the first column gives
  $Δ = a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$
Option (D) exactly matches the expansion along the first column, so it equals $Δ$ .
Options (A), (B), (C) are mixed-index combinations that do not in general equal $Δ$ (they give either other identities or zero). For example, the sum of elements of one row multiplied by cofactors of a different row equals $0$ .

Therefore the correct choice is

$(D) a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$
October 28, 2025
Exercise-4.2, Class 12th, Maths, Chapter 4, NCERT

Question 1 (Area of triangles)

$Find area of the triangle whose vertices are:$ $(i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8)$

$(iii) (- 2, - 3), (3, 2), (- 1, - 8) .$

Method. Area $= \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$

(i) $x_{1} = 1, y_{1} = 0; x_{2} = 6, y_{2} = 0; x_{3} = 4, y_{3} = 3$

$S = \frac{1}{2} ∣ 1 (0 - 3) + 6 (3 - 0) + 4 (0 - 0) ∣$

$= \frac{1}{2} ∣ - 3 + 18 + 0 ∣ = \frac{1}{2} (15) = \frac{15}{2} .$

(ii) $(2, 7), (1, 1), (10, 8)$

$S = \frac{1}{2} ∣ 2 (1 - 8) + 1 (8 - 7) + 10 (7 - 1) ∣$

$= \frac{1}{2} ∣ - 14 + 1 + 60 ∣ = \frac{1}{2} (47) = \frac{47}{2}$

(iii) $(- 2, - 3), (3, 2), (- 1, - 8)$

$S = \frac{1}{2} ∣ - 2 (2 - (- 8)) + 3 ((- 8) - (- 3)) + (- 1) ((- 3) - 2) ∣$

$= \frac{1}{2} ∣ - 20 - 15 + 5 ∣ = \frac{1}{2} (30) = 15$

Question 2 (Collinearity)

$Show that A (a, b + c), B (b, c + a), C (c, a + b) are collinear.$

Solution. Points are collinear iff area of triangle $A B C$ is zero. Compute

$Δ = ∣ \begin{array}{ccc} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{array} ∣ = a ((c + a) - (a + b)) + b ((a + b) - (b + c)) + c ((b + c) - (c + a))$

Simplify each bracket:

$a (c - b) + b (a - c) + c (b - a) = a c - a b + a b - b c + b c - a c = 0$

Hence $Δ = 0$ so $A, B, C$ are collinear.

Question 3 (Values of $k$ for area = 4)

$Find k if area = 4 for:$ $(i) (k, 0), (4, 0), (0, 2) (ii) (- 2, 0), (0, 4), (0, k) .$

(i) Using determinant formula:

$Δ = k (0 - 2) + 4 (2 - 0) + 0 (0 - 0) = - 2 k + 8.$

Area $= \frac{1}{2} ∣ Δ ∣ = 4 \Rightarrow ∣ ⁣ - 2 k + 8 ⁣ ∣ = 8$
So $- 2 k + 8 = 8$ or $- 2 k + 8 = - 8$ giving $k = 0$ or $k = 8$
$k = 0 or 8$

(ii)

$Δ = - 2 (4 - k) + 0 + 0 = - 8 + 2 k$

$\frac{1}{2} ∣ - 8 + 2 k ∣ = 4 \Rightarrow ∣ - 8 + 2 k ∣ = 8$ . So $- 8 + 2 k = 8$ or $- 8 + 2 k = - 8$
gives $k = 8$ or $k = 0$
$k = 0 or 8.$

Question 4 (Equation of a line using determinants)

$(i) line joining (1, 2) and (3, 6) . (ii) line joining (3, 1) and (9, 3) .$

Use determinant form for line through $(x_{1}, y_{1}), (x_{2}, y_{2})$

$∣ \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array} ∣ = 0.$

(i) $∣ \begin{array}{ccc} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array} ∣ = 0$

$\Rightarrow x (2 - 6) - y (1 - 3) + 1 (6 - 6) = - 4 x + 2 y = 0 \Rightarrow y = 2 x .$

(ii) $∣ \begin{array}{ccc} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{array} ∣ = 0$

$\Rightarrow x (1 - 3) - y (3 - 9) + 1 (9 - 3) = - 2 x + 6 y = 0 \Rightarrow x - 3 y = 0 .$

Question 5 (Find $k$ from given area = 35)

$If area of triangle is 35 with vertices (2, - 6), (5, 4), (k, 4)$

The points $(5, 4)$ and $(k, 4)$ have same $y$ -coordinate, so base $= ∣ k - 5 ∣$ , height from $(2, - 6)$ to line $y = 4$ is $4 - (- 6) = 10$

Area $= \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot ∣ k - 5 ∣ \cdot 10 = 5 ∣ k - 5 ∣ = 35$
So $∣ k - 5 ∣ = 7 \Rightarrow k - 5 = \pm 7 \Rightarrow k = 12 or k = - 2.$

$k = 12 or k = - 2$ (So the correct choice from the options is the pair $12, - 2$ )

October 28, 2025
Exercise-4.1, Class 12th, Maths, Chapter 4, NCERT

DETERMINANTS

Question 1

Evaluate $∣ \begin{array}{cc} 2 & 4 \\ - 5 & - 1 \end{array} ∣$

Solution.
$\det = 2 (- 1) - 4 (- 5) = - 2 + 20 = 18$

Question 2

(i) $∣ \begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array} ∣$

Solution 2 (i):

$\det = \cos θ \cdot \cos θ - (- \sin θ) \cdot \sin θ = \cos^{2} θ + \sin^{2} θ = 1 .$

Question 2 (ii):

$∣ \begin{array}{cc} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} ∣$ Solution:

We know that for a $2 \times 2$ determinant.

Now compute:

$\det = (x^{2} - x + 1) (x + 1) - (x - 1) (x + 1) .$

Step 1: Expand both terms

$(x^{2} - x + 1) (x + 1) = x^{3} + x^{2} - x^{2} - x + x + 1 = x^{3} + 1$ $(x - 1) (x + 1) = x^{2} - 1$

Step 2: Substitute back

$\det = (x^{3} + 1) - (x^{2} - 1) = x^{3} - x^{2} + 2$

Question 3

If $A = (\begin{array}{cc} 1 & 2 \\ 4 & 2 \end{array})$ , show $∣ 2 A ∣ = 4 ∣ A ∣$

Solution.
For an $n \times n$ matrix scaling by scalar $k$ multiplies the determinant by $k^{n}$ . Here $n = 2$ and $k = 2$ . So $∣ 2 A ∣ = 2^{2} ∣ A ∣ = 4 ∣ A ∣$

(Direct check: $∣ A ∣ = 1 \cdot 2 - 2 \cdot 4 = 2 - 8 = - 6$ . Then $∣ 2 A ∣ = \det (\begin{array}{cc} 2 & 4 \\ 8 & 4 \end{array}) = 2 \cdot 4 - 4 \cdot 8 = 8 - 32 = - 24 = 4 (- 6)$

Question 4

If $A = (\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array})$ , show $∣ 3 A ∣ = 27 ∣ A ∣$

Solution.
$A$ is $3 \times 3$ . Scaling by 3 multiplies determinant by $3^{3} = 27$ . Hence $∣ 3 A ∣ = 27 ∣ A ∣$ .

(One can check: $∣ A ∣$ is product of diagonal entries $1 \cdot 1 \cdot 4 = 4$ .

$∣ 3 A ∣$ has diagonal $3, 3, 12$ so determinant $3 \cdot 3 \cdot 12 = 108 = 27 \cdot 4$

Question 5 — Evaluate the following determinants

(i) $∣ \begin{array}{ccc} 3 & - 1 & - 2 \\ 0 & 0 & - 1 \\ 3 & - 5 & 0 \end{array} ∣$

Solution (5.i).
Expand along second row (only one nonzero entry $a_{23} = - 1$ ):

Minor for $a_{23}$ is $∣ \begin{array}{cc} 3 & - 1 \\ 3 & - 5 \end{array} ∣ = 3 (- 5) - (- 1) 3 = - 15 + 3 = - 12$

Cofactor $C_{23} = (- 1)^{2 + 3} \cdot (- 12) = - 1 \cdot (- 12) = 12$ . Then determinant = $a_{23} C_{23} = (- 1) \cdot 12 = - 12$

(ii) $∣ \begin{array}{ccc} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{array} ∣$

Solution (5.ii).
Compute by expansion / rule of Sarrus:

$\det = 3 ∣ \begin{array}{cc} 1 & - 2 \\ 3 & 1 \end{array} ∣ - (- 4) ∣ \begin{array}{cc} 1 & - 2 \\ 2 & 1 \end{array} ∣ + 5 ∣ \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} ∣ .$

Evaluate minors: $\cdot$
So $\det = 3 \cdot 7 + 4 \cdot 5 + 5 \cdot 1 = 21 + 20 + 5 = 46$

(iii) $∣ \begin{array}{ccc} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{array} ∣$

Solution (5.iii).
Compute (expanding first row):

$\det = 0 \cdot (\dots) - 1 \cdot ∣ \begin{array}{cc} - 1 & - 3 \\ - 2 & 0 \end{array} ∣ + 2 \cdot ∣ \begin{array}{cc} - 1 & 0 \\ - 2 & 3 \end{array} ∣$

First minor: $(- 1) \cdot 0 - (- 3) (- 2) = 0 - 6 = - 6$ ; the corresponding contribution is $- 1 \cdot (- 6) = 6$
Second minor: $(- 1) \cdot 3 - 0 \cdot (- 2) = - 3$ ; contribution $2 \cdot (- 3) = - 6$ . Sum $6 - 6 = 0$

(iv) $∣ \begin{array}{ccc} 2 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{array} ∣$

Solution (5.iv).
Expand along first row:

$\det = 2 ∣ \begin{array}{cc} 2 & - 1 \\ - 5 & 0 \end{array} ∣ - (- 1) ∣ \begin{array}{cc} 0 & - 1 \\ 3 & 0 \end{array} ∣ + (- 2) ∣ \begin{array}{cc} 0 & 2 \\ 3 & - 5 \end{array} ∣$

Compute minors: first $= 2 \cdot 0 - (- 1) (- 5) = 0 - 5 = - 5$ ⇒ $2 (- 5) = - 10$
second minor $= 0 \cdot 0 - (- 1) \cdot 3 = 3$ ⇒ sign gives $+ 1 \cdot 3 = + 3$
third minor $= 0 \cdot (- 5) - 2 \cdot 3 = - 6$ ⇒ multiplied by $- 2$ gives $+ 12$ .
Sum $- 10 + 3 + 12 = 5 .$

Question 6

If $A = (\begin{array}{ccc} 1 & 1 & - 2 \\ 2 & 1 & - 3 \\ 5 & 4 & - 9 \end{array})$ , find $∣ A ∣$

Solution.
Expand along first row:

$∣ A ∣ = 1 ∣ \begin{array}{cc} 1 & - 3 \\ 4 & - 9 \end{array} ∣ - 1 ∣ \begin{array}{cc} 2 & - 3 \\ 5 & - 9 \end{array} ∣ + (- 2) ∣ \begin{array}{cc} 2 & 1 \\ 5 & 4 \end{array} ∣$

Compute minors: first = $1 (- 9) - (- 3) 4 = - 9 + 12 = 3$
second = $2 (- 9) - (- 3) 5 = - 18 + 15 = - 3$ . With the minus sign yields $- 1 \cdot (- 3) = + 3$ .
third minor = $2 \cdot 4 - 1 \cdot 5 = 8 - 5 = 3$ , times $- 2$ gives $- 6$ . Sum $3 + 3 - 6 = 0$

So $∣ A ∣ = 0$

Question 7 (i):

$Find the value of x if ∣ \begin{array}{cc} 2 & 4 \\ 5 & 1 \end{array} ∣ = ∣ \begin{array}{cc} 2 x & 4 \\ 6 & x \end{array} ∣$

Solution:

For any $2 \times 2$ matrix
$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$\det = a d - b c$ Step 1: Compute the determinant on the left-hand side (LHS)

$∣ \begin{array}{cc} 2 & 4 \\ 5 & 1 \end{array} ∣ = (2) (1) - (4) (5) = 2 - 20 = - 18$ Step 2: Compute the determinant on the right-hand side (RHS)

$∣ \begin{array}{cc} 2 x & 4 \\ 6 & x \end{array} ∣ = (2 x) (x) - (4) (6) = 2 x^{2} - 24$

Step 3: Set them equal

$- 18 = 2 x^{2} - 24.$

Simplify:

$2 x^{2} = - 18 + 24 = 6.$ $x^{2} = 3.$ $x = \pm \sqrt{3}$

Question 7 (ii):

$Find the value of x if ∣ \begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array} ∣ = ∣ \begin{array}{cc} x & 3 \\ 2 x & 5 \end{array} ∣$

Solution:

For any $2 \times 2$ matrix
$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$\det = a d - b c .$

Step 1: Compute LHS

$∣ \begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array} ∣ = (2) (5) - (3) (4) = 10 - 12 = - 2$

Step 2: Compute RHS

$∣ \begin{array}{cc} x & 3 \\ 2 x & 5 \end{array} ∣ = (x) (5) - (3) (2 x) = 5 x - 6 x = - x$

Step 3: Equate LHS and RHS

$- 2 = - x \Rightarrow x = 2$

Question 8:

$If ∣ \begin{array}{cc} x & 2 \\ 18 & x \end{array} ∣ = ∣ \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} ∣, then find x$

Options:
(A) $6$ (B) $\pm 6$ (C) $- 6$ (D) $0$

Solution:

For any $2 \times 2$ determinant
$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$\det = a d - b c .$

Step 1: Compute LHS

$∣ \begin{array}{cc} x & 2 \\ 18 & x \end{array} ∣ = x \cdot x - 2 \cdot 18 = x^{2} - 36$ Step 2: Compute RHS

$∣ \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} ∣ = 6 \cdot 6 - 2 \cdot 18 = 36 - 36 = 0$

Step 3: Equate LHS and RHS

$x^{2} - 36 = 0$

$x^{2} = 36$

$x = \pm 6$

Final Answer:

$(B) x = \pm 6$

October 28, 2025
Miscellaneous Exercise on Chapter 3, Class 12th, Maths, NCERT
Question 1:

$If A and B are symmetric matrices, prove that A B - B A is a skew-symmetric matrix.$

Answer (proof):

Recall a matrix $M$ is skew-symmetric if $M^{T} = - M$

Given $A$ and $B$ are symmetric, so

$A^{T} = A, B^{T} = B .$

Consider the transpose of $A B - B A$ :

$(A B - B A)^{T} = (A B)^{T} - (B A)^{T} = B^{T} A^{T} - A^{T} B^{T} .$

Using symmetry of $A$ and $B$ we get

$(A B - B A)^{T} = B A - A B = - (A B - B A) .$

Thus $(A B - B A)^{T} = - (A B - B A)$ , so $A B - B A$ is skew-symmetric.

(As a remark, every skew-symmetric matrix has zeros on its diagonal, so the diagonal entries of $A B - B A$ are all zero.)

Question 2:

$Show that the matrix B^{'} A B is symmetric according as A is symmetric or skew-symmetric.$

Answer (proof):

Let $A$ and $B$ be square matrices of the same order, and let $B^{'}$ denote the transpose of $B$ .
We need to show that:
- If $A$ is symmetric, then $B^{'} A B$ is symmetric.
- If $A$ is skew-symmetric, then $B^{'} A B$ is skew-symmetric.
Case 1: $A$ is symmetric

If $A$ is symmetric, then $A^{'} = A$ .

Consider $(B^{'} A B)^{'}$ :

$(B^{'} A B)^{'} = B^{'} A^{'} B .$

But since $A^{'} = A$ ,

$(B^{'} A B)^{'} = B^{'} A B .$

Hence, $B^{'} A B$ is symmetric.

Case 2: $A$ is skew-symmetric

If $A$ is skew-symmetric, then $A^{'} = - A$ .

Now, take the transpose of $B^{'} A B$ :

$(B^{'} A B)^{'} = B^{'} A^{'} B = B^{'} (- A) B = - B^{'} A B .$

Thus, $(B^{'} A B)^{'} = - B^{'} A B$ ,
which means $B^{'} A B$ is skew-symmetric.

Hence proved:
The matrix $B^{'} A B$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.

Question 3:

$Find the values of x, y, z if A = (\begin{array}{ccc} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{array}) satisfies A^{T} A = I .$

Answer (solution):

Compute $A^{T} A$ . A straightforward multiplication gives

$A^{T} A = (\begin{array}{ccc} 2 x^{2} & 0 & 0 \\ 0 & 6 y^{2} & 0 \\ 0 & 0 & 3 z^{2} \end{array}) .$

For $A^{T} A = I$ we must have the diagonal entries equal to 1, hence

$2 x^{2} = 1, 6 y^{2} = 1, 3 z^{2} = 1$

Therefore

$x^{2} = \frac{1}{2} ⟹ x = \pm \frac{1}{\sqrt{2}}, y^{2} = \frac{1}{6} ⟹ y = \pm \frac{1}{\sqrt{6}}, z^{2} = \frac{1}{3} ⟹ z = \pm \frac{1}{\sqrt{3}} .$

All choices of independent signs are allowed, so the solutions are

$(x, y, z) = (\pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{3}}),$

(8 sign-combinations in total).

Question 4:

$For what values of x does [1 2 1] (\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}) (\begin{array}{c} 0 \\ 2 \\ x \end{array}) = 0 ?$

Answer (solution):

First compute the product of the matrix with the column vector:

$(\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}) (\begin{array}{c} 0 \\ 2 \\ x \end{array}) = (\begin{array}{c} 1 \cdot 0 + 2 \cdot 2 + 0 \cdot x \\ 2 \cdot 0 + 0 \cdot 2 + 1 \cdot x \\ 1 \cdot 0 + 0 \cdot 2 + 2 \cdot x \end{array}) = (\begin{array}{c} 4 \\ x \\ 2 x \end{array})$

Now left-multiply by $[1 2 1]$ :

$[1 2 1] (\begin{array}{c} 4 \\ x \\ 2 x \end{array}) = 1 \cdot 4 + 2 \cdot x + 1 \cdot 2 x = 4 + 4 x$

Set equal to zero:

$4 + 4 x = 0 \Rightarrow x = - 1$

Question 5:

$If A = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}), show that A^{2} - 5 A + 7 I = 0$

Answer (solution):

Given

$A = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}), I = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) .$

Step 1: Compute $A^{2}$

$A^{2} = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) = (\begin{array}{cc} 3 (3) + 1 (- 1) & 3 (1) + 1 (2) \\ (- 1) (3) + 2 (- 1) & (- 1) (1) + 2 (2) \end{array}) = (\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}) .$

Step 2: Compute $5 A$

$5 A = 5 (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) = (\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array})$

Step 3: Compute $7 I$

$7 I = 7 (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) = (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array})$

Step 4: Substitute into $A^{2} - 5 A + 7 I$

$A^{2} - 5 A + 7 I = (\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}) - (\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array}) + (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array})$

Compute step-by-step:

$A^{2} - 5 A = (\begin{array}{cc} 8 - 15 & 5 - 5 \\ - 5 - (- 5) & 3 - 10 \end{array}) = (\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array})$

Now add $7 I$ :

$A^{2} - 5 A + 7 I = (\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array}) + (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}) = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Hence proved:

$A^{2} - 5 A + 7 I = 0$

Question 6:

$Find x if [x - 5 - 1] (\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}) (\begin{array}{c} x \\ 4 \\ 1 \end{array}) = 0.$

Solution:

We will simplify step-by-step.

Step 1: Multiply the matrix with the column vector

$(\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}) (\begin{array}{c} x \\ 4 \\ 1 \end{array}) = (\begin{array}{c} 1 \cdot x + 0 \cdot 4 + 2 \cdot 1 \\ 0 \cdot x + 2 \cdot 4 + 1 \cdot 1 \\ 2 \cdot x + 0 \cdot 4 + 3 \cdot 1 \end{array}) = (\begin{array}{c} x + 2 \\ 9 \\ 2 x + 3 \end{array})$

Step 2: Multiply the row vector $[x - 5 - 1]$ with the result

$[x - 5 - 1] (\begin{array}{c} x + 2 \\ 9 \\ 2 x + 3 \end{array}) = x (x + 2) + (- 5) (9) + (- 1) (2 x + 3)$

Simplify:

$= x^{2} + 2 x - 45 - 2 x - 3 = x^{2} - 48.$

Step 3: Set equal to zero

$x^{2} - 48 = 0 \Rightarrow x^{2} = 48$

$x= \pm 4 \sqrt{3}$

Question 7

$A manufacturer produces three products x, y, z sold in two markets with annual sales$ $S = (\begin{array}{ccc} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{array}) (rows: Market I, Market II; columns: x, y, z).$ $(a) Unit sale prices p = (\begin{array}{c} 2.50 \\ 1.50 \\ 1.00 \end{array}) . (b) Unit costs c = (\begin{array}{c} 2.00 \\ 1.00 \\ 0.50 \end{array}) .$ $Find (a) total revenue in each market, (b) gross profit in each market.$

Solution

Use matrix multiplication. Revenue vector $R$ (marketwise) is

$R = S p .$

Compute component-wise:

Market I revenue

$10000 (2.5) + 2000 (1.5) + 18000 (1) = 25000 + 3000 + 18000 = 46000.$

Market II revenue

$6000 (2.5) + 20000 (1.5) + 8000 (1) = 15000 + 30000 + 8000 = 53000.$

So

$R = (\begin{array}{c} 46000 \\ 53000 \end{array}) rupees.$

(b) Total cost vector $C = S c$ :

Market I cost

$10000 (2) + 2000 (1) + 18000 (0.5) = 20000 + 2000 + 9000 = 31000.$

Market II cost

$6000 (2) + 20000 (1) + 8000 (0.5) = 12000 + 20000 + 4000 = 36000.$

So

$C = (\begin{array}{c} 31000 \\ 36000 \end{array}) rupees.$

Gross profit for each market $G = R - C$ :

$G = (\begin{array}{c} 46000 - 31000 \\ 53000 - 36000 \end{array}) = (\begin{array}{c} 15000 \\ 17000 \end{array}) rupees$ Final answers

(a) Total revenue — Market I: Rs. 46,000; Market II: Rs. 53,000.
(b) Gross profit — Market I: Rs. 15,000; Market II: Rs. 17,000.

(Also note matrix forms: $R = S p, C = S c, G = S (p - c)$

Question 8:

$Find the matrix X such that X (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) .$

Solution:

We are given:

$X (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array})$

Let:

$A = (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}), B = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) .$

We need $X$ such that $X A = B$ .

To isolate $X$ , multiply both sides on the right by $A^{T} (A A^{T})^{- 1}$ :
(since $A$ is $2 \times 3$ , not square, we use this formula)

$X = B A^{T} (A A^{T})^{- 1} .$

Step 1: Compute $A A^{T}$

$A A^{T} = (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) (\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}) = (\begin{array}{cc} 1^{2} + 2^{2} + 3^{2} & 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 \\ 4 \cdot 1 + 5 \cdot 2 + 6 \cdot 3 & 4^{2} + 5^{2} + 6^{2} \end{array}) = (\begin{array}{cc} 14 & 32 \\ 32 & 77 \end{array})$

Step 2: Compute $(A A^{T})^{- 1}$

First find determinant:

$∣ A A^{T} ∣ = (14) (77) - (32) (32) = 1078 - 1024 = 54$

So,

$(A A^{T})^{- 1} = \frac{1}{54} (\begin{array}{cc} 77 & - 32 \\ - 32 & 14 \end{array}) .$

Step 3: Compute $B A^{T}$

$B A^{T} = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) (\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array})$

$= (\begin{array}{cc} (- 7) (1) + (- 8) (2) + (- 9) (3) & (- 7) (4) + (- 8) (5) + (- 9) (6) \\ (2) (1) + (4) (2) + (6) (3) & (2) (4) + (4) (5) + (6) (6) \end{array})$

Compute each entry:

$B A^{T} = (\begin{array}{cc} - 7 - 16 - 27 & - 28 - 40 - 54 \\ 2 + 8 + 18 & 8 + 20 + 36 \end{array}) = (\begin{array}{cc} - 50 & - 122 \\ 28 & 64 \end{array})$

Step 4: Compute $X = B A^{T} (A A^{T})^{- 1}$

$X = \frac{1}{54} (\begin{array}{cc} - 50 & - 122 \\ 28 & 64 \end{array}) (\begin{array}{cc} 77 & - 32 \\ - 32 & 14 \end{array}) .$

Compute the product:

First row:

$(- 50) (77) + (- 122) (- 32) = - 3850 + 3904 = 54,$ $(- 50) (- 32) + (- 122) (14) = 1600 - 1708 = - 108.$

Second row:

$(28) (77) + (64) (- 32) = 2156 - 2048 = 108,$ $(28) (- 32) + (64) (14) = - 896 + 896 = 0$

So:

$X = \frac{1}{54} (\begin{array}{cc} 54 & - 108 \\ 108 & 0 \end{array}) = (\begin{array}{cc} 1 & - 2 \\ 2 & 0 \end{array}) .$

Choose the correct answer in the following questions:

Question 9:

$If A = (\begin{array}{cc} α & β \\ γ & - α \end{array}) is such that A^{2} = I, then find the correct relation among α, β, γ .$

Options:
(A) $1 + α^{2} + β γ = 0$
(B) $1 - α^{2} + β γ = 0$
(C) $1 - α^{2} - β γ = 0$
(D) $1 + α^{2} - β γ = 0$

Solution:

Given

$A = (\begin{array}{cc} α & β \\ γ & - α \end{array})$

Compute $A^{2}$ :

$A^{2} = (\begin{array}{cc} α & β \\ γ & - α \end{array}) (\begin{array}{cc} α & β \\ γ & - α \end{array}) = (\begin{array}{cc} α^{2} + β γ & α β - α β \\ α γ - α γ & γ β + α^{2} \end{array}) = (\begin{array}{cc} α^{2} + β γ & 0 \\ 0 & α^{2} + β γ \end{array})$

Thus,

$A^{2} = (α^{2} + β γ) I .$

We are told $A^{2} = I$ , so:

$(α^{2} + β γ) I = I$

That means:

$α^{2} + β γ = 1$

Rearranging:

$1 - α^{2} - β γ = 0$

Correct Option:

$(C) 1 - α^{2} - β γ = 0.$

Question 10:

$If the matrix A is both symmetric and skew-symmetric, then:$

Options:
(A) $A$ is a diagonal matrix
(B) $A$ is a zero matrix
(C) $A$ is a square matrix
(D) None of these

Solution:

By definition:
- $A$ is symmetric if $A^{T} = A$
- $A$ is skew-symmetric if $A^{T} = - A$
If $A$ is both, then:

$A = A^{T} = - A$

This implies:

$A = - A \Rightarrow 2 A = 0 \Rightarrow A = 0$

That means all elements of $A$ are zero.

Correct Option:

$(B) A is a zero matrix.$

Question 11:

$If A is a square matrix such that A^{2} = A, then find (I + A)^{3} - 7 A .$

Options:
(A) $A$
(B) $I - A$
(C) $I$
(D) $3 A$

Solution:

We are given that:

$A^{2} = A .$

Let’s expand $(I + A)^{3}$ :

$(I + A)^{3} = I^{3} + 3 I^{2} A + 3 I A^{2} + A^{3}$

Simplify each term using $I^{2} = I$ and $A^{2} = A$ :

$(I + A)^{3} = I + 3 A + 3 A + A^{3}$

Now, compute $A^{3}$ :

$A^{3} = A^{2} A = A \cdot A = A^{2} = A$

Substitute this back:

$(I + A)^{3} = I + 3 A + 3 A + A = I + 7 A$

Now compute:

$(I + A)^{3} - 7 A = (I + 7 A) - 7 A = I$

Final Answer:

$(C) I$
October 28, 2025

Expression	Value
$\lim_{x \to 0} f (x)$	$- 1$
$f (0)$	$- 1$

Exercise-3.3, Class 12th, Maths, Chapter 3, NCERT

Q1. (i)

Matrix $A = [\begin{array}{c} 5 \\ \frac{1}{2} \\ - 1 \end{array}]$

This is a column matrix (3×1).

Transpose:

To find $A^{'}$ (or $A^{T}$ ), we interchange rows and columns.

So,

$A^{'} = [5 \frac{1}{2} - 1]$

Answer:

$A^{'} = [5 \frac{1}{2} - 1]$

(This is now a row matrix (1×3).)

Q1- (ii).

Matrix $B = [\begin{array}{cc} 1 & - 1 \\ 2 & 3 \end{array}]$

This is a 2×2 matrix.

Transpose:

We swap rows with columns:

$B^{'} = [\begin{array}{cc} 1 & 2 \\ - 1 & 3 \end{array}]$

Answer:

$B^{'} = [\begin{array}{cc} 1 & 2 \\ - 1 & 3 \end{array}]$

Q1 (iii)

Matrix

$A = [\begin{array}{ccc} - 1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & - 1 \end{array}]$

This is a 3×3 matrix (3 rows, 3 columns).

To find: $A^{'}$ (the transpose)

To find the transpose, we interchange rows and columns — that is, the first row becomes the first column, the second row becomes the second column, and so on.

Step-by-step:

Row	Becomes Column
Row 1 = $(- 1, 5, 6)$	Column 1 = $[\begin{array}{c} - 1 \\ 5 \\ 6 \end{array}]$
Row 2 = $(\sqrt{3}, 5, 6)$	Column 2 = $[\begin{array}{c} \sqrt{3} \\ 5 \\ 6 \end{array}]$
Row 3 = $(2, 3, - 1)$	Column 3 = $[\begin{array}{c} 2 \\ 3 \\ - 1 \end{array}]$

Transpose:

$A^{'} = [\begin{array}{ccc} - 1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & - 1 \end{array}]$

$A^{'} = [5 \frac{1}{2} - 1]$

(This is now a row matrix (1×3).)

Question 2.

Given:

$A = [\begin{array}{ccc} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{array}], B = [\begin{array}{ccc} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array}]$

We have to verify:

1️⃣ $(A + B)^{'} = A^{'} + B^{'}$

2️⃣ $(A - B)^{'} = A^{'} - B^{'}$

Step 1: Find A + B

Add corresponding elements:

$A + B = [\begin{array}{ccc} (- 1) + (- 4) & 2 + 1 & 3 + (- 5) \\ 5 + 1 & 7 + 2 & 9 + 0 \\ (- 2) + 1 & 1 + 3 & 1 + 1 \end{array}] = [\begin{array}{ccc} - 5 & 3 & - 2 \\ 6 & 9 & 9 \\ - 1 & 4 & 2 \end{array}]$

Now find (A + B)′

Take the transpose — interchange rows and columns:

$(A + B)^{'} = [\begin{array}{ccc} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{array}]$

Step 2: Find A′ and B′ separately

Transpose of A:

$A^{'} = [\begin{array}{ccc} - 1 & 5 & - 2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array}]$

Transpose of B:

$B^{'} = [\begin{array}{ccc} - 4 & 1 & 1 \\ 1 & 2 & 3 \\ - 5 & 0 & 1 \end{array}]$

Step 3: Find A′ + B′

$A^{'} + B^{'} = [\begin{array}{ccc} (- 1) + (- 4) & 5 + 1 & (- 2) + 1 \\ 2 + 1 & 7 + 2 & 1 + 3 \\ 3 + (- 5) & 9 + 0 & 1 + 1 \end{array}] = [\begin{array}{ccc} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{array}]$

This is exactly equal to $(A + B)^{'}$ .

Hence,

$(A + B)^{'} = A^{'} + B^{'}$

Step 4: Now find A – B

Subtract corresponding elements:

$A - B = [\begin{array}{ccc} (- 1) - (- 4) & 2 - 1 & 3 - (- 5) \\ 5 - 1 & 7 - 2 & 9 - 0 \\ (- 2) - 1 & 1 - 3 & 1 - 1 \end{array}] = [\begin{array}{ccc} 3 & 1 & 8 \\ 4 & 5 & 9 \\ - 3 & - 2 & 0 \end{array}]$

Now find (A – B)′

$(A - B)^{'} = [\begin{array}{ccc} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{array}]$

Step 5: Find A′ – B′

$A^{'} - B^{'} = [\begin{array}{ccc} (- 1) - (- 4) & 5 - 1 & (- 2) - 1 \\ 2 - 1 & 7 - 2 & 1 - 3 \\ 3 - (- 5) & 9 - 0 & 1 - 1 \end{array}] = [\begin{array}{ccc} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{array}]$

This is exactly equal to $(A - B)^{'}$ .

Hence,

$(A - B)^{'} = A^{'} - B^{'}$

Question 3.

Given

$A^{'} = [\begin{array}{cc} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{array}]$

$B = [\begin{array}{ccc} - 1 & 2 & 1 \\ 1 & 2 & 3 \end{array}]$

, then verify that

(i) (A + B)′ = A′ + B′

(ii) (A – B)′ = A′ – B′

Determine $A$ from $A^{'}$

Remember: $A^{'}$ is the transpose of $A$ .
So to get $A$ , we transpose $A^{'}$ again.

$A = (A^{'})^{'} = [\begin{array}{ccc} 3 & - 1 & 0 \\ 4 & 2 & 1 \end{array}]$

Hence

$A = [\begin{array}{ccc} 3 & - 1 & 0 \\ 4 & 2 & 1 \end{array}], B = [\begin{array}{ccc} - 1 & 2 & 1 \\ 1 & 2 & 3 \end{array}]$

Compute $A + B$

Add corresponding elements:

$A + B = [\begin{array}{ccc} 3 + (- 1) & - 1 + 2 & 0 + 1 \\ 4 + 1 & 2 + 2 & 1 + 3 \end{array}] = [\begin{array}{ccc} 2 & 1 & 1 \\ 5 & 4 & 4 \end{array}]$

Compute $(A + B)^{'}$

Transpose the above matrix:

$(A + B)^{'} = [\begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array}]$

Compute $A^{'} + B^{'}$

We already know:

$A^{'} = [\begin{array}{cc} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{array}], B^{'} = [\begin{array}{cc} - 1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array}]$

Now add elementwise:

$A^{'} + B^{'} = [\begin{array}{cc} 3 + (- 1) & 4 + 1 \\ - 1 + 2 & 2 + 2 \\ 0 + 1 & 1 + 3 \end{array}] = [\begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array}]$

Therefore

$(A + B)^{'} = A^{'} + B^{'}$

Compute $A - B$

$A - B = [\begin{array}{ccc} 3 - (- 1) & - 1 - 2 & 0 - 1 \\ 4 - 1 & 2 - 2 & 1 - 3 \end{array}] = [\begin{array}{ccc} 4 & - 3 & - 1 \\ 3 & 0 & - 2 \end{array}]$ Compute $(A - B)^{'}$

$(A - B)^{'} = [\begin{array}{cc} 4 & 3 \\ - 3 & 0 \\ - 1 & - 2 \end{array}]$

Compute $A^{'} - B^{'}$

$A^{'} - B^{'} = [\begin{array}{cc} 3 - (- 1) & 4 - 1 \\ - 1 - 2 & 2 - 2 \\ 0 - 1 & 1 - 3 \end{array}] = [\begin{array}{cc} 4 & 3 \\ - 3 & 0 \\ - 1 & - 2 \end{array}]$

Therefore

$(A - B)^{'} = A^{'} - B^{'}$

Question 4.

Given:

$A^{'} = [\begin{array}{cc} - 2 & 3 \\ 1 & 2 \end{array}], B = [\begin{array}{cc} - 1 & 0 \\ 1 & 2 \end{array}]$

We need to find

$(A + 2 B)^{'}$

Find $A$ from $A^{'}$

Since $A^{'} = [\begin{array}{cc} - 2 & 3 \\ 1 & 2 \end{array}]$ ,

Transpose it to get $A$ :

$A = (A^{'})^{'} = [\begin{array}{cc} - 2 & 1 \\ 3 & 2 \end{array}]$

Compute $2 B$

Multiply each element of $B$ by 2:

$2 B = [\begin{array}{cc} 2 \times (- 1) & 2 \times 0 \\ 2 \times 1 & 2 \times 2 \end{array}] = [\begin{array}{cc} - 2 & 0 \\ 2 & 4 \end{array}]$

Compute $A + 2 B$

$A + 2 B = [\begin{array}{cc} - 2 & 1 \\ 3 & 2 \end{array}] + [\begin{array}{cc} - 2 & 0 \\ 2 & 4 \end{array}] = [\begin{array}{cc} - 4 & 1 \\ 5 & 6 \end{array}]$

Find $(A + 2 B)^{'}$

Take the transpose:

$(A + 2 B)^{'} = [\begin{array}{cc} - 4 & 5 \\ 1 & 6 \end{array}]$

Question 5.

For the matrices A and B, verify that (AB)′ = B′A′, where

Concept Recap

For any conformable matrices $A$ and $B$ :

$(A B)^{'} = B^{'} A^{'}$

That is, the transpose of a product = product of transposes in reverse order.

Now, let’s check numerically.

(i)

$A = [\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}], B = [\begin{array}{ccc} - 1 & 2 & 1 \end{array}]$

Step 1️⃣: Compute $A B$

Since $A$ is 3×1 and $B$ is 1×3,
$A B$ will be a 3×3 matrix.

$A B = [\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}] [\begin{array}{ccc} - 1 & 2 & 1 \end{array}] = [\begin{array}{ccc} 1 (- 1) & 1 (2) & 1 (1) \\ - 4 (- 1) & - 4 (2) & - 4 (1) \\ 3 (- 1) & 3 (2) & 3 (1) \end{array}] = [\begin{array}{ccc} - 1 & 2 & 1 \\ 4 & - 8 & - 4 \\ - 3 & 6 & 3 \end{array}]$

Find $(A B)^{'}$

Transpose means interchange rows and columns:

$(A B)^{'} = [\begin{array}{ccc} - 1 & 4 & - 3 \\ 2 & - 8 & 6 \\ 1 & - 4 & 3 \end{array}]$

Compute $B^{'} A^{'}$

Now find $B^{'}$ and $A^{'}$ .

$B^{'} = [\begin{array}{c} - 1 \\ 2 \\ 1 \end{array}], A^{'} = [\begin{array}{ccc} 1 & - 4 & 3 \end{array}]$

Multiply $B^{'}$ (3×1) with $A^{'}$ (1×3):

$B^{'} A^{'} = [\begin{array}{c} - 1 \\ 2 \\ 1 \end{array}] [\begin{array}{ccc} 1 & - 4 & 3 \end{array}] = [\begin{array}{ccc} (- 1) (1) & (- 1) (- 4) & (- 1) (3) \\ (2) (1) & (2) (- 4) & (2) (3) \\ (1) (1) & (1) (- 4) & (1) (3) \end{array}]$

$= [\begin{array}{ccc} - 1 & 4 & - 3 \\ 2 & - 8 & 6 \\ 1 & - 4 & 3 \end{array}]$

Hence,

$(A B)^{'} = B^{'} A^{'}$

(ii)

$A = [\begin{array}{c} 0 \\ 1 \\ 2 \end{array}], B = [\begin{array}{ccc} 1 & 5 & 7 \end{array}]$

Compute $A B$

$A B = [\begin{array}{c} 0 \\ 1 \\ 2 \end{array}] [\begin{array}{ccc} 1 & 5 & 7 \end{array}] = [\begin{array}{ccc} 0 (1) & 0 (5) & 0 (7) \\ 1 (1) & 1 (5) & 1 (7) \\ 2 (1) & 2 (5) & 2 (7) \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{array}]$

Find $(A B)^{'}$

$(A B)^{'} = [\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array}]$

Compute $B^{'} A^{'}$

$B^{'} = [\begin{array}{c} 1 \\ 5 \\ 7 \end{array}], A^{'} = [\begin{array}{ccc} 0 & 1 & 2 \end{array}]$

Multiply:

$B^{'} A^{'} = [\begin{array}{c} 1 \\ 5 \\ 7 \end{array}] [\begin{array}{ccc} 0 & 1 & 2 \end{array}] = [\begin{array}{ccc} 1 (0) & 1 (1) & 1 (2) \\ 5 (0) & 5 (1) & 5 (2) \\ 7 (0) & 7 (1) & 7 (2) \end{array}] = [\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array}]$

Hence,

$(A B)^{'} = B^{'} A^{'}$

Question 6.

verify that A′ A = I

(i)

$A = [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}]$

Find $A^{'}$

$A^{'} = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$

Compute $A^{'} A$

$A^{'} A = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}]$ $A^{'} A = [\begin{array}{cc} (\cos^{2} α + \sin^{2} α) & (\cos α \sin α - \sin α \cos α) \\ (\sin α \cos α - \cos α \sin α) & (\sin^{2} α + \cos^{2} α) \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = I$

Hence, $A^{'} A = I$

(ii)

$A = [\begin{array}{cc} \sin α & \cos α \\ - \cos α & \sin α \end{array}]$

Find $A^{'}$

$A^{'} = [\begin{array}{cc} \sin α & - \cos α \\ \cos α & \sin α \end{array}]$

Compute $A^{'} A$

$A^{'} A = [\begin{array}{cc} \sin α & - \cos α \\ \cos α & \sin α \end{array}] [\begin{array}{cc} \sin α & \cos α \\ - \cos α & \sin α \end{array}]$ $A^{'} A = [\begin{array}{cc} (\sin^{2} α + \cos^{2} α) & (\sin α \cos α - \cos α \sin α) \\ (\cos α \sin α - \sin α \cos α) & (\cos^{2} α + \sin^{2} α) \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = I$

Hence, $A^{'} A = I$

Q7.

Recall:

A matrix $A$ is symmetric if $A^{'} = A$ .
A matrix $A$ is skew-symmetric if $A^{'} = - A$ . (Note diagonal entries of a skew-symmetric matrix must be $0$ .)

(i) Show $A$ is symmetric

$A = [\begin{array}{ccc} 1 & - 1 & 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{array}]$

Find the transpose $A^{'}$ (swap rows and columns):

$A^{'} = [\begin{array}{ccc} 1 & - 1 & 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{array}]$

We see $A^{'} = A$ .
Therefore $A$ is symmetric.

(ii) Show $A$ is skew-symmetric

$A = [\begin{array}{ccc} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ 1 & - 1 & 0 \end{array}]$

Compute the transpose:

$A^{'} = [\begin{array}{ccc} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ - 1 & 1 & 0 \end{array}]$

Compute $- A$ :

$- A = [\begin{array}{ccc} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ - 1 & 1 & 0 \end{array}]$

We have $A^{'} = - A$ . Also all diagonal entries are $0$ , as required.

Therefore $A$ is skew-symmetric.

Question 8.

(i) (A + A′) is a symmetric matrix

(ii) (A – A′) is a skew symmetric matrix

Given

$A = [\begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array}] .$

First find the transpose:

$A^{'} = [\begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array}] .$

(i) $A + A^{'}$

Compute:

$A + A^{'} = [\begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array}] + [\begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array}] = [\begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array}]$

Take its transpose:

$(A + A^{'})^{'} = [\begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array}]$ Since $(A + A^{'})^{'} = A + A^{'}$ , the matrix $A + A^{'}$ is symmetric.

(ii) $A - A^{'}$

Compute:

$A - A^{'} = [\begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array}] - [\begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array}] = [\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$

Take its transpose:

$(A - A^{'})^{'} = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}] = - (A - A^{'})$

Since $(A - A^{'})^{'} = - (A - A^{'})$ , the matrix $A - A^{'}$ is skew-symmetric.

Question 9. Find

$\frac{1}{2} (A + A^{'}) and \frac{1}{2} (A - A^{'})$

We are given:

$A = [\begin{array}{ccc} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{array}]$

Find $A^{'}$

Transpose means interchange rows and columns:

$A^{'} = [\begin{array}{ccc} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{array}]$ Compute $A + A^{'}$

Add corresponding entries of $A$ and $A^{'}$ :

$A + A^{'} = [\begin{array}{ccc} 0 + 0 & a + (- a) & b + (- b) \\ (- a) + a & 0 + 0 & c + (- c) \\ (- b) + b & (- c) + c & 0 + 0 \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Compute $A - A^{'}$

Subtract corresponding entries:

$A - A^{'} = [\begin{array}{ccc} 0 - 0 & a - (- a) & b - (- b) \\ (- a) - a & 0 - 0 & c - (- c) \\ (- b) - b & (- c) - c & 0 - 0 \end{array}] = [\begin{array}{ccc} 0 & 2 a & 2 b \\ - 2 a & 0 & 2 c \\ - 2 b & - 2 c & 0 \end{array}]$

Multiply by ½

$\frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$ $\frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{ccc} 0 & 2 a & 2 b \\ - 2 a & 0 & 2 c \\ - 2 b & - 2 c & 0 \end{array}] = [\begin{array}{ccc} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{array}]$

Final Answers

$\frac{1}{2} (A + A^{'}) = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}], \frac{1}{2} (A - A^{'}) = [\begin{array}{ccc} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{array}]$

Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

we’ll write each matrix $A$ as

$A = \frac{1}{2} (A + A^{'}) + \frac{1}{2} (A - A^{'}),$

where $S = \frac{1}{2} (A + A^{'})$ is symmetric and $K = \frac{1}{2} (A - A^{'})$ is skew-symmetric. I’ll give $S$ and $K$ for each part.

(i) $A = [\begin{array}{cc} 3 & 5 \\ 1 & - 1 \end{array}]$

$A^{'} = [\begin{array}{cc} 3 & 1 \\ 5 & - 1 \end{array}]$ $S = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{cc} 6 & 6 \\ 6 & - 2 \end{array}] = [\begin{array}{cc} 3 & 3 \\ 3 & - 1 \end{array}]$

$K = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{cc} 0 & 4 \\ - 4 & 0 \end{array}] = [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$

So $A = S + K$ with $S$ symmetric and $K$ skew.

(ii) $A = [\begin{array}{ccc} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{array}]$

This matrix is already symmetric (check $A^{'} = A$ ), so

$S = \frac{1}{2} (A + A^{'}) = A = [\begin{array}{ccc} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{array}], K = \frac{1}{2} (A - A^{'}) = 0 = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] .$

(iii) $A = [\begin{array}{ccc} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{array}]$

First compute $A^{'} = [\begin{array}{ccc} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{array}]$

Then

$S = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{ccc} 6 & 1 & - 5 \\ 1 & - 4 & - 4 \\ - 5 & - 4 & 4 \end{array}] = [\begin{array}{ccc} 3 & \frac{1}{2} & - \frac{5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ - \frac{5}{2} & - 2 & 2 \end{array}],$

which is symmetric, and

$K = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{ccc} 0 & 5 & 3 \\ - 5 & 0 & 6 \\ - 3 & - 6 & 0 \end{array}] = [\begin{array}{ccc} 0 & \frac{5}{2} & \frac{3}{2} \\ - \frac{5}{2} & 0 & 3 \\ - \frac{3}{2} & - 3 & 0 \end{array}], $

which is skew-symmetric. (You can check $S^{'} = S$ and $K^{'} = - K$ )

(iv) $A = [\begin{array}{cc} 1 & 5 \\ - 1 & 2 \end{array}]$

$A^{'} = [\begin{array}{cc} 1 & - 1 \\ 5 & 2 \end{array}]$ $S = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{cc} 2 & 4 \\ 4 & 4 \end{array}] = [\begin{array}{cc} 1 & 2 \\ 2 & 2 \end{array}],$

$K = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{cc} 0 & 6 \\ - 6 & 0 \end{array}] = [\begin{array}{cc} 0 & 3 \\ - 3 & 0 \end{array}] .$

Question 11

Given:
$A$ and $B$ are symmetric matrices of the same order.
That means:

$A^{'} = A and B^{'} = B$

We need to find the nature of the matrix $A B - B A$

Take transpose of $A B - B A$

$(A B - B A)^{'} = (A B)^{'} - (B A)^{'} = B^{'} A^{'} - A^{'} B^{'}$

Since $A^{'} = A$ and $B^{'} = B$ ,

$(A B - B A)^{'} = B A - A B = - (A B - B A)$

Interpretation

A matrix $M$ is skew-symmetric if $M^{'} = - M$ .

Here we found $(A B - B A)^{'} = - (A B - B A)$ Therefore, $A B - B A$ is a skew-symmetric matrix.

Correct Option: (A) Skew symmetric matrix

Question 12

Given:

$A = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}], and A + A^{'} = I$

We must find the value of $α$

Compute $A^{'}$

$A^{'} = [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}]$

Compute $A + A^{'}$

$A + A^{'} = [\begin{array}{cc} \cos α + \cos α & - \sin α + \sin α \\ \sin α - \sin α & \cos α + \cos α \end{array}] = [\begin{array}{cc} 2 \cos α & 0 \\ 0 & 2 \cos α \end{array}]$

Given $A + A^{'} = I$

$[\begin{array}{cc} 2 \cos α & 0 \\ 0 & 2 \cos α \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$

This gives:

$2 \cos α = 1 ⟹ \cos α = \frac{1}{2}$

Solve for $α$

$\cos α = \frac{1}{2} \Rightarrow α = \frac{π}{3}$

Final Answers
1️⃣ $A B - B A$ → Skew-symmetric matrix
2️⃣ $A + A^{'} = I \Rightarrow α = \frac{π}{3}$

October 26, 2025

Exercise-3.2, Class 12th, Maths, Chapter 3 – Matrices, NCERT
Question 1

Let

$A = [\begin{array}{cc} 2 & 4 \\ 3 & 2 \end{array}], B = [\begin{array}{cc} 1 & 3 \\ - 2 & 5 \end{array}], C = [\begin{array}{cc} - 2 & 5 \\ 3 & 4 \end{array}]$

Find:
(i) A + B (ii) A − B (iii) 3A − C (iv) AB (v) BA

Solution

(i) A + B =
Add corresponding elements:

$A + B = [\begin{array}{cc} 2 + 1 & 4 + 3 \\ 3 + (- 2) & 2 + 5 \end{array}] = [\begin{array}{cc} 3 & 7 \\ 1 & 7 \end{array}]$

(ii) A − B =

$A - B = [\begin{array}{cc} 2 - 1 & 4 - 3 \\ 3 - (- 2) & 2 - 5 \end{array}] = [\begin{array}{cc} 1 & 1 \\ 5 & - 3 \end{array}]$

(iii) 3A − C =
First, find 3A:

$3 A = [\begin{array}{cc} 6 & 12 \\ 9 & 6 \end{array}]$

Subtract C:

$3 A - C = [\begin{array}{cc} 6 - (- 2) & 12 - 5 \\ 9 - 3 & 6 - 4 \end{array}] = [\begin{array}{cc} 8 & 7 \\ 6 & 2 \end{array}]$

(iv) AB =

$A B = [\begin{array}{cc} 2 (1) + 4 (- 2) & 2 (3) + 4 (5) \\ 3 (1) + 2 (- 2) & 3 (3) + 2 (5) \end{array}] = [\begin{array}{cc} - 6 & 26 \\ - 1 & 19 \end{array}]$

(v) BA =

$B A = [\begin{array}{cc} 1 (2) + 3 (3) & 1 (4) + 3 (2) \\ (- 2) (2) + 5 (3) & (- 2) (4) + 5 (2) \end{array}] = [\begin{array}{cc} 11 & 10 \\ 11 & 2 \end{array}]$

Final Answers:

(i)

$A + B = [\begin{array}{cc} 3 & 7 \\ 1 & 7 \end{array}]$

(ii)

$A - B = [\begin{array}{cc} 1 & 1 \\ 5 & - 3 \end{array}]$

(iii)

$3 A - C = [\begin{array}{cc} 8 & 7 \\ 6 & 2 \end{array}]$

(iv)

$A B = [\begin{array}{cc} - 6 & 26 \\ - 1 & 19 \end{array}]$

(v)

$B A = [\begin{array}{cc} 11 & 10 \\ 11 & 2 \end{array}]$

Question 2

Compute the following:

(i)

$[\begin{array}{cc} a & b \\ - b & a \end{array}] + [\begin{array}{cc} a & b \\ b & a \end{array}]$

(ii)

$[\begin{array}{cc} a^{2} + b^{2} & b^{2} + c^{2} \\ c^{2} + a^{2} & a^{2} + b^{2} \end{array}] + [\begin{array}{cc} c^{2} + 2 a b & 2 b c \\ - 2 a c & - 2 a b \end{array}]$

(iii)

$[\begin{array}{cc} 1 & 4 \\ 8 & 5 \\ 2 & 8 \end{array}] + [\begin{array}{cc} 6 & 12 \\ 5 & 16 \\ 5 & 3 \end{array}] + [\begin{array}{cc} 7 & 6 \\ 8 & 0 \\ 2 & 4 \end{array}]$

(iv)

$[\begin{array}{cc} \cos^{2} x & \sin^{2} x \\ \sin^{2} x & \cos^{2} x \end{array}] + [\begin{array}{cc} \sin^{2} x & \cos^{2} x \\ \cos^{2} x & \sin^{2} x \end{array}]$

Solution

(i) Add element-wise:

$= [\begin{array}{cc} 2 a & 2 b \\ 0 & 2 a \end{array}]$

(ii)

$= [\begin{array}{cc} a^{2} + b^{2} + c^{2} + 2 a b & b^{2} + c^{2} + 2 b c \\ c^{2} + a^{2} - 2 a c & a^{2} + b^{2} - 2 a b \end{array}]$

(iii)
Add element-wise:

$= [\begin{array}{cc} 14 & 22 \\ 21 & 21 \\ 9 & 15 \end{array}]$

(iv)

$= [\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}]$

Question 3

Compute the indicated products:

(i)

$[\begin{array}{cc} a & b \\ - b & a \end{array}] [\begin{array}{cc} a & - b \\ b & a \end{array}]$

(ii)

$[\begin{array}{ccc} 2 & 3 & 4 \end{array}] [\begin{array}{c} 1 \\ 2 \\ 3 \end{array}]$

(iii)

$[\begin{array}{cc} 1 & - 2 \\ 2 & 3 \end{array}] [\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}]$

Solution

(i)

$= [\begin{array}{cc} a^{2} + b^{2} & 0 \\ 0 & a^{2} + b^{2} \end{array}]$

(ii)

$2 \times 1 + 3 \times 2 + 4 \times 3 = 20$

(iii)

$= [\begin{array}{ccc} - 3 & - 4 & 1 \\ 8 & 13 & 9 \end{array}]$

Question 4

If

$A = [\begin{array}{cc} 2 & 1 \\ 5 & 0 \\ - 1 & - 1 \end{array}], B = [\begin{array}{cc} 3 & - 1 \\ 4 & 2 \\ 2 & 0 \end{array}], C = [\begin{array}{cc} 4 & 1 \\ 0 & 3 \\ 1 & - 2 \end{array}]$

Compute (A + B), (B − C), and verify A + (B − C) = (A + B) − C.

A + B =

$[\begin{array}{cc} 5 & 0 \\ 9 & 2 \\ 1 & - 1 \end{array}]$

B − C =

$[\begin{array}{cc} - 1 & - 2 \\ 4 & - 1 \\ 1 & 2 \end{array}]$

A + (B − C) =

$[\begin{array}{cc} 1 & - 1 \\ 9 & - 1 \\ 0 & 1 \end{array}]$

(A + B) − C =

$[\begin{array}{cc} 1 & - 1 \\ 9 & - 1 \\ 0 & 1 \end{array}]$

Hence, verified.

Question 5

$A = [\begin{array}{ccc} 2 & 1 & 1 \\ 3 & 3 & 5 \\ 6 & 2 & 5 \end{array}], B = [\begin{array}{ccc} 2 & 3 & 5 \\ 1 & 1 & 4 \\ 7 & 5 & 6 \end{array}]$

Compute 3A − 5B

Step 1:

$3 A = [\begin{array}{ccc} 6 & 3 & 3 \\ 9 & 9 & 15 \\ 18 & 6 & 15 \end{array}], 5 B = [\begin{array}{ccc} 10 & 15 & 25 \\ 5 & 5 & 20 \\ 35 & 25 & 30 \end{array}]$

Step 2: Subtract:

$3 A - 5 B = [\begin{array}{ccc} - 4 & - 12 & - 22 \\ 4 & 4 & - 5 \\ - 17 & - 19 & - 15 \end{array}]$

Question 6

Simplify:

$[\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}] [\begin{array}{cc} \sin x & - \cos x \\ \cos x & \sin x \end{array}]$

Solution

Multiply:
(1,1):

$\cos x \sin x + \sin x \cos x = \sin (2 x)$

Result:

$[\begin{array}{cc} \sin (2 x) & 0 \\ 0 & \sin (2 x) \end{array}]$

Question 7

Find X and Y if

(i)

$X + Y = [\begin{array}{cc} 7 & 0 \\ 2 & 5 \end{array}], X - Y = [\begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array}]$

(ii)

$2 X + 3 Y = [\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}], 3 X + 2 Y = [\begin{array}{cc} 2 & - 2 \\ - 1 & 5 \end{array}]$

Solution

Case (i):
Add and subtract:

$2 X = (X + Y) + (X - Y) = [\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}] \Rightarrow X = [\begin{array}{cc} 5 & 0 \\ 1 & 4 \end{array}]$

$2 Y = (X + Y) - (X - Y) = [\begin{array}{cc} 4 & 0 \\ 2 & 2 \end{array}] \Rightarrow Y = [\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}]$

Case (ii):
Multiply first by 3, second by 2 and subtract:

$(6 X + 9 Y) - (6 X + 4 Y) = 5 Y = [\begin{array}{cc} 2 & 13 \\ 14 & - 10 \end{array}]$

$Y = [\begin{array}{cc} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & - 2 \end{array}]$

Substitute in

$2 X + 3 Y = [\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}]$

$X = [\begin{array}{cc} \frac{4}{5} & - \frac{3}{5} \\ - \frac{2}{5} & 2 \end{array}]$

Question 8

Find X, if

$Y = [\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}]$

and

$2 X + Y = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}]$

Solution

Given:

$2 X + Y = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}]$

Substitute Y:

$2 X + [\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}] = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}]$

Now, move Y to RHS:

$2 X = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}] - [\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}] = [\begin{array}{cc} - 2 & - 2 \\ - 4 & - 2 \end{array}]$

Divide both sides by 2:

$X = [\begin{array}{cc} - 1 & - 1 \\ - 2 & - 1 \end{array}]$

Question 9

Find the values of x and y, if

$2 [\begin{array}{cc} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$

Step 1: Expand the scalar multiplication

$2 [\begin{array}{cc} 1 & 3 \\ 0 & x \end{array}] = [\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}]$

So the equation becomes:

$[\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$

Step 2: Add the two matrices on the LHS

$[\begin{array}{cc} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$

Step 3: Compare corresponding elements
1. From (1,1): $2 + y = 5 \Rightarrow y = 3$
2. From (2,2): $2 x + 2 = 8 \Rightarrow 2 x = 6 \Rightarrow x = 3$
Final Answers:

$x = 3, y = 3$

Question 10

Solve for x, y, z, t if

$2 [\begin{array}{cc} x & z \\ y & t \end{array}] + 3 [\begin{array}{cc} 1 & - 1 \\ 0 & 2 \end{array}] = [\begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array}]$

Solution

Step 1: Multiply 2 and 3 through matrices:

$[\begin{array}{cc} 2 x & 2 z \\ 2 y & 2 t \end{array}] + [\begin{array}{cc} 3 & - 3 \\ 0 & 6 \end{array}] = [\begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array}]$

Step 2: Add LHS:

$[\begin{array}{cc} 2 x + 3 & 2 z - 3 \\ 2 y & 2 t + 6 \end{array}] = [\begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array}]$

Step 3: Equate corresponding elements:
1. $2 x + 3 = 3 \Rightarrow x = 0$
2. $2 z - 3 = 5 \Rightarrow z = 4$
3. $2 y = 4 \Rightarrow y = 2$
4. $2 t + 6 = 6 \Rightarrow t = 0$
Hence

$x = 0, y = 2, z = 4, t = 0$

Question 11

If

$[\begin{array}{c} x \\ 2 \end{array}] + [\begin{array}{c} - 1 \\ y \end{array}] = [\begin{array}{c} 10 \\ 5 \end{array}]$

Find x and y.

Solution

Add LHS element-wise:

$[\begin{array}{c} x - 1 \\ 2 + y \end{array}] = [\begin{array}{c} 10 \\ 5 \end{array}]$

Comparing elements:

$x - 1 = 10 \Rightarrow x = 11$
$2 + y = 5 \Rightarrow y = 3$

Answer:
$x = 11, y = 3$

Question 12

Given equation:

$3 [\begin{array}{cc} x & y \\ z & w \end{array}] = [\begin{array}{cc} x & 6 \\ - 1 & 2 w \end{array}] + [\begin{array}{cc} 4 & x + y \\ z + w & 3 \end{array}]$

We need to find x  and  y.

Step 1: Expand the LHS

$3 [\begin{array}{cc} x & y \\ z & w \end{array}] = [\begin{array}{cc} 3 x & 3 y \\ 3 z & 3 w \end{array}]$

Step 2: Add the two matrices on RHS

$[\begin{array}{cc} x & 6 \\ - 1 & 2 w \end{array}] + [\begin{array}{cc} 4 & x + y \\ z + w & 3 \end{array}] = [\begin{array}{cc} x + 4 & 6 + x + y \\ - 1 + z + w & 2 w + 3 \end{array}]$

Step 3: Equate the LHS and RHS

$[\begin{array}{cc} 3 x & 3 y \\ 3 z & 3 w \end{array}] = [\begin{array}{cc} x + 4 & 6 + x + y \\ - 1 + z + w & 2 w + 3 \end{array}]$

Step 4: Compare corresponding elements
1. From (1, 1): $3 x = x + 4 \Rightarrow 2 x = 4 \Rightarrow x = 2.$
2. From (1, 2): $3 y = 6 + x + y \Rightarrow 2 y = 6 + x \Rightarrow 2 y = 6 + 2 \Rightarrow 2 y = 8 \Rightarrow y = 4.$
3. Other entries (involving z, w) are not required here.
Final Answer:

$x = 2, y = 4$

Question 13

If

$F (x) = [\begin{array}{ccc} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}],$

prove that $F (x) F (y) = F (x + y)$

Solution

Compute $F (x) F (y)$ :

$F (x) F (y) = [\begin{array}{ccc} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} \cos y & - \sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}]$

Multiply the first two rows:

Top-left 2×2 block:

$[\begin{array}{cc} \cos x \cos y - \sin x \sin y & - (\sin x \cos y + \cos x \sin y) \\ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y \end{array}]$

But using angle addition identities:

$\cos (x + y) = \cos x \cos y - \sin x \sin y$ $\sin (x + y) = \sin x \cos y + \cos x \sin y$

So,

$F (x) F (y) = [\begin{array}{ccc} \cos (x + y) & - \sin (x + y) & 0 \\ \sin (x + y) & \cos (x + y) & 0 \\ 0 & 0 & 1 \end{array}] = F (x + y)$

Hence proved.

Question 14

to show that matrix multiplication is not commutative, i.e.

$A B \neq B A .$

Let’s solve step-by-step carefully.

Given matrices

$A = (\begin{array}{cc} 5 & - 1 \\ 6 & 7 \end{array}), B = (\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}) .$

Step 1: Compute $A B$

$A B = (\begin{array}{cc} 5 & - 1 \\ 6 & 7 \end{array}) (\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}) .$

Multiply row by column:

$A B = (\begin{array}{cc} (5) (2) + (- 1) (3) & (5) (1) + (- 1) (4) \\ (6) (2) + (7) (3) & (6) (1) + (7) (4) \end{array}) = (\begin{array}{cc} 10 - 3 & 5 - 4 \\ 12 + 21 & 6 + 28 \end{array}) = (\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array})$

Step 2: Compute $B A$

$B A = (\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}) (\begin{array}{cc} 5 & - 1 \\ 6 & 7 \end{array})$

Multiply:

$B A = (\begin{array}{cc} (2) (5) + (1) (6) & (2) (- 1) + (1) (7) \\ (3) (5) + (4) (6) & (3) (- 1) + (4) (7) \end{array}) = (\begin{array}{cc} 10 + 6 & - 2 + 7 \\ 15 + 24 & - 3 + 28 \end{array}) = (\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array})$

Step 3: Compare

$A B = (\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}), B A = (\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}) .$

Clearly $A B \neq B$

✅ Hence proved:
Matrix multiplication is not commutative in general:

$A B \neq B A .$

15. Find $A^{2} - 5 A + 6 I$ , where

$A = (\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{array})$

Compute $A^{2}$ :

$A^{2} = (\begin{array}{ccc} 5 & 1 & 2 \\ 9 & 4 & 5 \\ 4 & 1 & 4 \end{array})$

Now $- 5 A = (\begin{array}{ccc} - 10 & 0 & - 5 \\ - 10 & - 5 & - 15 \\ - 5 & - 5 & 0 \end{array})$
and $6 I = (\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array})$

Add them:

$A^{2} - 5 A + 6 I = (\begin{array}{ccc} 5 & 1 & 2 \\ 9 & 4 & 5 \\ 4 & 1 & 4 \end{array}) + (\begin{array}{ccc} - 10 & 0 & - 5 \\ - 10 & - 5 & - 15 \\ - 5 & - 5 & 0 \end{array}) + (\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}) = (\begin{array}{ccc} 1 & 1 & - 3 \\ - 1 & 5 & - 10 \\ - 1 & - 4 & 10 \end{array}) .$

16. If $A = (\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array})$ , prove $A^{3} - 6 A^{2} + 7 A + 2 I = 0$ .

Compute (steps shown succinctly):

$A^{2} = (\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}), A^{3} = (\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}) .$

Now form $A^{3} - 6 A^{2} + 7 A + 2 I$ . Term-by-term:
- $- 6 A^{2} = (\begin{array}{ccc} - 30 & 0 & - 48 \\ - 12 & - 24 & - 30 \\ - 48 & 0 & - 78 \end{array})$
- $7 A = (\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array})$
- $2 I = (\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array})$
Adding $A^{3} + (- 6 A^{2}) + 7 A + 2 I$ gives the zero matrix. So the identity holds.

17.

Given:

$A = (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}), I = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) .$

We have to find the scalar $k$ such that

$A^{2} = k A - 2 I .$

Step 1: Compute $A^{2}$

$A^{2} = (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}) (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}) = (\begin{array}{cc} 3 \cdot 3 + (- 2) \cdot 4 & 3 \cdot (- 2) + (- 2) \cdot (- 2) \\ 4 \cdot 3 + (- 2) \cdot 4 & 4 \cdot (- 2) + (- 2) \cdot (- 2) \end{array}) .$

Simplify each element:

$A^{2} = (\begin{array}{cc} 9 - 8 & - 6 + 4 \\ 12 - 8 & - 8 + 4 \end{array}) = (\begin{array}{cc} 1 & - 2 \\ 4 & - 4 \end{array})$

Step 2: Write the given relation

$A^{2} = k A - 2 I .$

Substitute $A^{2}$ , $A$ , and $I$ :

$(\begin{array}{cc} 1 & - 2 \\ 4 & - 4 \end{array}) = k (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}) - 2 (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) .$

Simplify the right-hand side:

$(\begin{array}{cc} 3 k - 2 & - 2 k \\ 4 k & - 2 k - 2 \end{array}) .$

Step 3: Equate corresponding entries

${\begin{cases} 3 k - 2 = 1, \\ - 2 k = - 2, \\ 4 k = 4, \\ - 2 k - 2 = - 4. \end{cases}$ Now solve any one (they should all give the same $k$ ):
- From the second: $- 2 k = - 2 \Rightarrow k = 1$
- From the first: $3 k - 2 = 1 \Rightarrow k = 1$
- Others also give $k = 1.$
All consistent.

Final Answer:

$k = 1.$

18. If $A = (\begin{array}{cc} 0 & - \tan (\frac{α}{2}) \\ \tan (\frac{α}{2}) & 0 \end{array})$ and $I$ is $2 \times 2$ identity, show

$I + A = (I - A) (\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array})$

(That is the standard Cayley-type relation when $A$ uses $\tan (α / 2)$

Proof sketch (algebraic verification): Put $t = \tan (\frac{α}{2}).$

Then

$I + A = (\begin{array}{cc} 1 & - t \\ t & 1 \end{array}), I - A = (\begin{array}{cc} 1 & t \\ - t & 1 \end{array})$

Use the trig identities

$\cos α = \frac{1 - t^{2}}{1 + t^{2}}, \sin α = \frac{2 t}{1 + t^{2}}$

Multiply $(I - A)$ by the rotation matrix $R (α) = (\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array})$

$(I - A) R (α) = (\begin{array}{cc} c + t s & - s + t c \\ - t c + s & t s + c \end{array}),$

substitute $c = \cos α, s = \sin α$ and the formulas above; each entry simplifies to match $I + A = (\begin{array}{cc} 1 & - t \\ t & 1 \end{array})$ . Thus the identity holds.

(If your printed sign convention for $A$ is different — e.g. the off-diagonal signs reversed — paste the exact matrix and I’ll adapt the algebra.)

19.

A trust fund has ` 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ` 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1800 (b) Rs 2000
Answer –

A trust fund has ₹30,000 to invest in two types of bonds:

Bond 1 → 5 % interest per year

Bond 2 → 7 % interest per year

We must divide ₹30,000 between them so that the annual total interest is:
(a) ₹ 1 800 (b) ₹ 2 000

Step 1: Define variables

Let
Then: ${\begin{cases} x + y = 30000 \\ 0.05 x + 0.07 y = I \end{cases}$
where $I$ = required interest.

Step 2: Matrix form
$\underset{A}{\underset{⏟}{[\begin{array}{cc} 1 & 1 \\ 0.05 & 0.07 \end{array}]}} [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 30000 \\ I \end{array}]$ That is, $A x = b .$
We can find $x = A^{- 1} b$

Compute $A^{- 1}$

For a $2 \times 2$ matrix $A = [\begin{array}{cc} a & b \\ c & d \end{array}]$
$A^{- 1} = \frac{1}{a d - b c} [\begin{array}{cc} d & - b \\ - c & a \end{array}]$
Here
$a = 1, b = 1, c = 0.05, d = 0.07$
$\det (A) = (1) (0.07) - (1) (0.05) = 0.02$
So:
$A^{- 1} = \frac{1}{0.02} [\begin{array}{cc} 0.07 & - 1 \\ - 0.05 & 1 \end{array}] = [\begin{array}{cc} 3.5 & - 50 \\ - 2.5 & 50 \end{array}] .$

Step 3: Multiply to find $x$ and $y$
$[\begin{array}{c} x \\ y \end{array}] = A^{- 1} [\begin{array}{c} 30000 \\ I \end{array}] = [\begin{array}{cc} 3.5 & - 50 \\ - 2.5 & 50 \end{array}] [\begin{array}{c} 30000 \\ I \end{array}]$
Compute components:
$x = 3.5 (30000) - 50 I, y = - 2.5 (30000) + 50 I .$
Simplify:
$x = 105000 - 50 I, y = - 75000 + 50 I .$

(a) For $I = 1800$
$x = 105000 - 50 (1800) = 105000 - 90000 = 15000,$ $y = - 75000 + 50 (1800) = - 75000 + 90000 = 15000.$

(b) For $I = 2000$
$x = 105000 - 50 (2000) = 105000 - 100000 = 5000,$ $y = - 75000 + 50 (2000) = - 75000 + 100000 = 25000.$

✅ Final Answers Summary

Case Required Interest 5 % Bond (x) 7 % Bond (y)

(a) ₹ 1 800 ₹ 15 000 ₹ 15 000

(b) ₹ 2 000 ₹ 5 000 ₹ 25 000

So, by matrix multiplication,
$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 105000 - 50 I \\ - 75000 + 50 I \end{array}],$
20. Bookshop: 10 dozen chemistry, 8 dozen physics, 10 dozen economics books; prices ₹80, ₹60, ₹40. Find total receipt.

First convert dozens to counts: 10 dozen = 120, 8 dozen = 96, 10 dozen = 120. Multiply quantities by unit prices:

$120 \times 80 + 96 \times 60 + 120 \times 40 = 9600 + 5760 + 4800 = ₹ 20,160 .$

(Matrix form: $[120 96 120] (\begin{array}{c} 80 \\ 60 \\ 40 \end{array}) = 20160$

21. (Multiple choice) With $X$ of order $2 \times n$ , $Y$ of order $3 \times k$ , $Z$ of order $2 \times p$ , $W$ of order $n \times 3$ , $P$ of order $p \times k$ :

What restriction on $n, k, p$ so that $P Y + W Y$ is defined?
- For $P Y$ to be defined: $P$ (size $p \times k$ ) times $Y$ (size $3 \times k$ ) requires number of columns of $P$ = number of rows of $Y$ , so $k = 3$ .
- For $W Y$ to be defined: $W$ (size $n \times 3$ ) times $Y$ (size $3 \times k$ ) is defined for any $n$ once $k$ is known.
- For $P Y$ and $W Y$ to be addable, their resulting orders must match: $P Y$ is $p \times k$ and $W Y$ is $n \times k$ . So $p = n$ .
Thus the restriction is $k = 3$ and $p = n$ .

Answer: (A) $k = 3, p = n$ .

22. If $n = p$ , then order of $7 X - 5 Z$ where $X$ is $2 \times n$ and $Z$ is $2 \times p$ ?

If $n = p$ then both $X$ and $Z$ are $2 \times n$ . So $7 X - 5 Z$ is of order $2 \times n$ . Answer: (B) $2 \times n$ .
October 26, 2025
Exercise-3.1, Class 12th, Maths, Chapter 3- Matrices, NCERT

Exercise 3.1

Question 1:

In the matrix

$A = [\begin{array}{cccc} 2 & 5 & 19 & - 7 \\ 1 & - 5 & 17 & 3 \\ 1 & - 2 & 12 & 5 \end{array}]$

write:
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

Answer:
(i) Order = 3 × 4
(ii) Number of elements = 3 × 4 = 12
(iii) a₁₃ = 19, a₂₁ = 1, a₃₃ = 12, a₂₄ = 3, a₂₃ = 17.

Question 2:

If a matrix has 24 elements, what are the possible orders it can have? What if it has 13 elements?

Answer:
We know that if a matrix has m × n elements, then total elements = m × n.

(i) For 24 elements:
Possible orders = (1,24), (2,12), (3,8), (4,6), (6,4), (8,3), (12,2), (24,1)

(ii) For 13 elements (prime number):
Possible orders = (1,13), (13,1)

Question 3:

If a matrix has 18 elements, what are its possible orders? What if it has 5 elements?

Answer:
(i) For 18 elements: $m \times n = 18$
Possible orders = (1,18), (2,9), (3,6), (6,3), (9,2), (18,1)

(ii) For 5 elements (prime): (1,5), (5,1)

Question 4:

Construct a 2 × 2 matrix A = [aᵢⱼ] where:

(i) aᵢⱼ = 2i − j
(ii) aᵢⱼ = i² − 3j
(iii) aᵢⱼ = i² / 2j

Answer:
For i, j = 1, 2

(i)

$A = [\begin{array}{cc} 2 (1) - 1 & 2 (1) - 2 \\ 2 (2) - 1 & 2 (2) - 2 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 3 & 2 \end{array}]$

(ii)

$A = [\begin{array}{cc} 1^{2} - 3 (1) & 1^{2} - 3 (2) \\ 2^{2} - 3 (1) & 2^{2} - 3 (2) \end{array}] = [\begin{array}{cc} - 2 & - 5 \\ 1 & - 2 \end{array}]$

(iii)

$A = [\begin{array}{cc} \frac{1^{2}}{2 \times 1} & \frac{1^{2}}{2 \times 2} \\ \frac{2^{2}}{2 \times 1} & \frac{2^{2}}{2 \times 2} \end{array}] = [\begin{array}{cc} \frac{1}{2} & \frac{1}{4} \\ 2 & 1 \end{array}]$

Question 5:

Construct a 3 × 4 matrix where:
(i) aᵢⱼ = ½ (i − 3j)
(ii) aᵢⱼ = 2i − j

Answer:

(i)

$A = [\begin{array}{cccc} - 1 & - 2.5 & - 4 & - 5.5 \\ - 0.5 & - 2 & - 3.5 & - 5 \\ 0 & - 1.5 & - 3 & - 4.5 \end{array}]$

(ii)

$A = [\begin{array}{cccc} 1 & 0 & - 1 & - 2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{array}]$

Question 6:

Find the values of x, y, and z from the following equations:

(i)

$[\begin{array}{ccc} x & 5 & z \\ y & 1 & 5 \end{array}] = [\begin{array}{ccc} 2 & 6 & 2 \\ 1 & 5 & 8 \end{array}]$

(ii)

$[\begin{array}{ccc} 5 & z & x + y \\ 5 & 8 & x + y \end{array}] = [\begin{array}{ccc} x + z & y & z \\ x z & y & z \end{array}]$

Answer:

(i) Comparing elements:
x = 2, y = 1, z = 2.

(ii) On comparing:
x = 3, y = 1, z = 2.

Question 7:

Find a, b, c, d from the equation:

$[\begin{array}{cc} a - b & 2 a - b \\ 2 a + c & 3 c - d \end{array}] = [\begin{array}{cc} 0 & - 15 \\ 0 & 13 \end{array}]$

Answer:
Equating elements:
a − b = 0
2a − b = −15
2a + c = 0
3c − d = 13

Solving, we get
a = −15, b = −15, c = 30, d = 77.

Question 8:

A = [aᵢⱼ] is a square matrix if:
(A) m < n (B) m > n (C) m = n (D) None of these

Answer:
(C) m = n

Question 9:

Which values of x and y make the following matrices equal?

$[\begin{array}{cc} 3 x + 7 & 5 \\ y + 1 & 2 \end{array}] = [\begin{array}{cc} - 3 x & 0 \\ - 2 & 8 \end{array}]$

Answer:
3x + 7 = −3x ⇒ 6x = −7 ⇒ x = −7/6
y + 1 = −2 ⇒ y = −3

x = −7/6, y = −3

Question 10:

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512

Answer:
Each entry can be 0 or 1, i.e., 2 possibilities per element.
Total = $2^{3 \times 3} = 2^{9} = 512$

Answer: (D) 512

October 26, 2025
Miscellaneous Exercise on Chapter – 2, Class 12th, Maths, NCERT

1. $\cos^{- 1} (\cos (13 π / 6))$

We know that $\cos^{- 1} (\cos θ) = θ$ if $θ \in [0, π]$ .

$13 π / 6 = 2 π + π / 6 ⟹ \cos (13 π / 6) = \cos (π / 6)$

Thus,

$\cos^{- 1} (\cos (13 π / 6)) = π / 6$

2. $\tan^{- 1} (\tan (7 π / 6))$

$\tan^{- 1} (\tan θ) = θ$ if $θ \in (- π / 2, π / 2)$ .

$7 π / 6 = π + π / 6 ⟹ \tan (7 π / 6) = \tan (π / 6)$

So,

$\tan^{- 1} (\tan (7 π / 6)) = π / 6 - π = - 5 π / 6$

Since range is $(- π / 2, π / 2)$ , principal value is $π / 6 - π = - π / 6$

3. Prove $2 \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{24}{7})$

Let $\sin^{- 1} \frac{3}{5} = θ$ , so $\sin θ = \frac{3}{5}$ .
Then, $\cos θ = \frac{4}{5}$

$\tan (2 θ) = \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{2 (\frac{3}{4})}{1 - (\frac{3}{4})^{2}} = \frac{24}{7}$

Hence,

$2 \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{24}{7})$

4. Prove $\sin^{- 1} \frac{8}{17} + \sin^{- 1} \frac{3}{5} = \tan^{- 1} \frac{77}{36}$

Let $\sin^{- 1} \frac{8}{17} = α$ and $\sin^{- 1} \frac{3}{5} = β$

$\tan α = \frac{8}{15}, \tan β = \frac{3}{4}$

Thus,

$\tan (α + β) = \frac{8 / 15 + 3 / 4}{1 - (8 / 15) (3 / 4)} = \frac{77}{36}$

Hence proved.

5. Prove $\cos^{- 1} \frac{4}{5} + \cos^{- 1} \frac{12}{13} = \cos^{- 1} \frac{33}{65}$

Let $A = \cos^{- 1} \frac{4}{5}, B = \cos^{- 1} \frac{12}{13}$ .

$\cos (A + B) = \cos A \cos B - \sin A \sin B$ $= \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{33}{65}$

Hence proved.

6. $\cos^{- 1} \frac{12}{13} + \sin^{- 1} \frac{3}{5} = \sin^{- 1} \frac{56}{65}$

Let $A = \cos^{- 1} \frac{12}{13}, B = \sin^{- 1} \frac{3}{5}$
Then $\sin A = \frac{5}{13}, \cos B = \frac{4}{5}$

$\sin (A + B) = \sin A \cos B + \cos A \sin B = \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5} = \frac{56}{65}$

Hence proved.

7. $\tan^{- 1} \frac{63}{16} = \sin^{- 1} \frac{5}{13} + \cos^{- 1} \frac{3}{5}$

Let $\sin^{- 1} \frac{5}{13} = A$ , $\cos^{- 1} \frac{3}{5} = B$ $\tan (A + B) = \frac{5 / 12 + 4 / 3}{1 - \frac{5}{12} \cdot \frac{4}{3}} = \frac{63}{16}$

Hence proved.

8. Prove $\tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} = \frac{1}{2} \cos^{- 1} x, x \in (0, 1)$

Put $x = \cos 2 θ$ , then

$\sqrt{\frac{1 - x}{1 + x}} = \sqrt{\frac{1 - \cos 2 θ}{1 + \cos 2 θ}} = \tan θ$

Thus,

$\tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} = θ = \frac{1}{2} \cos^{- 1} x$

9. Prove $\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}$ Let $t = \tan \frac{x}{2}$ , then:

$\sin x = \frac{2 t}{1 + t^{2}}$

After rationalizing and simplifying, the expression equals $\cot^{- 1} (\frac{1}{\tan (x / 2)}) = \frac{x}{2}$

10. Prove $\tan^{- 1} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x$

Let $x = \cos 2 θ$ , then

$\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} = \frac{\cos θ - \sin θ}{\cos θ + \sin θ} = \tan (\frac{π}{4} - θ)$

Hence,

$\tan^{- 1} (LHS) = \frac{π}{4} - θ = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x$

11. Solve $2 \tan^{- 1} (\cos x) = \tan^{- 1} (2 \csc x)$

Let $\tan^{- 1} (\cos x) = θ$

Then $\tan (2 θ) = \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{2 \cos x}{1 - \cos^{2} x} = 2 \csc x$

Hence proven.

12. Prove $\tan^{- 1} \frac{1 - x}{1 + x} = \frac{π}{4} - \tan^{- 1} x$

Use the tangent subtraction identity:

$\tan (\frac{π}{4} - \tan^{- 1} x) = \frac{1 - x}{1 + x}$

Taking $\tan^{- 1}$ both sides gives the result.

13. $\sin (\tan^{- 1} x) = \frac{x}{\sqrt{1 + x^{2}}}$

Let $\tan^{- 1} x = θ \Rightarrow \tan θ = x$

So a right-angle triangle gives $\sin θ = \frac{x}{\sqrt{1 + x^{2}}}$

Hence,

$\sin (\tan^{- 1} x) = \frac{x}{\sqrt{1 + x^{2}}}$

14. If $\sin^{- 1} (1 - 2 x^{2}) = 2 \sin^{- 1} x$

Use double angle formula:

$\sin (2 \sin^{- 1} x) = 2 x \sqrt{1 - x^{2}}$

Equating $1 - 2 x^{2} = 2 x \sqrt{1 - x^{2}}$ and solving gives $x = 0$ or $x = 1$ .

October 25, 2025
Exercise-2.1, Class 12th, Maths, Chapter – 2, NCERT
EXERCISE 2.1 — Solutions

1. Find the principal value of $\sin^{- 1} ⁣ (\frac{1}{2})$ .

Solution. Principal branch of $\sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ .
$\sin \frac{π}{6} = \frac{1}{2}$ . $\frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ .

$\sin^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{6}$

2. Find the principal value of $\cos^{- 1} ⁣ (\frac{\sqrt{3}}{2})$ .

Solution. Principal branch of $\cos^{- 1}$ is $[0, π]$ .
$\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\frac{π}{6} \in [0, π]$ .

$\cos^{- 1} ⁣ (\frac{\sqrt{3}}{2}) = \frac{π}{6}$

3. Find the principal value of ${cosec}^{- 1} (2)$ .

Solution. If $y = {cosec}^{- 1} 2$ then $cosec y = 2 \Rightarrow \sin y = \frac{1}{2}$ . Principal branch for ${cosec}^{- 1}$ (as used in the book) corresponds to $y \in [- \frac{π}{2}, \frac{π}{2}]$ excluding 0 (or equivalent principal branch placing the value at $π / 6$ or $π - π / 6$ ; standard choice giving principal value in $[- \frac{π}{2}, \frac{π}{2}] ∖ {0}$ yields $y = \frac{π}{6}$ . So

${cosec}^{- 1} (2) = \frac{π}{6}$

(Interpretation note: many texts give ${cosec}^{- 1} 2 = π - \frac{π}{6} = \frac{5 π}{6}$ if using a different branch — but the book’s principal branch choice yields $π / 6$ .)

4. Find the principal value of $\tan^{- 1} (\sqrt{3})$ .

Solution. $\tan \frac{π}{3} = \sqrt{3}$ . Principal branch of $\tan^{- 1}$ is $(- \frac{π}{2}, \frac{π}{2})$ , and $\frac{π}{3} \in (- \frac{π}{2}, \frac{π}{2})$ .

$\tan^{- 1} (\sqrt{3}) = \frac{π}{3}$

5. Find the principal value of $\cos^{- 1} ⁣ (- \frac{1}{2})$ .

Solution. $\cos \frac{2 π}{3} = - \frac{1}{2}$ , and $\frac{2 π}{3} \in [0, π]$ . So

$\cos^{- 1} ⁣ (- \frac{1}{2}) = \frac{2 π}{3}$

6. Find the principal value of $\tan^{- 1} (- 1)$ .

Solution. $\tan (- \frac{π}{4}) = - 1$ . Principal branch is $(- \frac{π}{2}, \frac{π}{2})$ .

$\tan^{- 1} (- 1) = - \frac{π}{4}$

7. Find the principal value of $\sec^{- 1} ⁣ (\frac{2}{\sqrt{3}})$ .

Solution. $\sec θ = \frac{2}{\sqrt{3}}$ ⇒ $\cos θ = \frac{\sqrt{3}}{2}$ ⇒ $θ = \frac{π}{6}$ (principal $\sec^{- 1}$ branch is $[0, π] ∖ {\frac{π}{2}}$ ).

$\sec^{- 1} ⁣ (\frac{2}{\sqrt{3}}) = \frac{π}{6}$

8. Find the principal value of $\cot^{- 1} (\sqrt{3})$ .

Solution. $\cot θ = \sqrt{3} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow θ = \frac{π}{6}$ . Principal branch of $\cot^{- 1}$ in the book is $(0, π)$ , and $\frac{π}{6} \in (0, π)$ .

$\cot^{- 1} (\sqrt{3}) = \frac{π}{6}$

9. Find the principal value of $\cos^{- 1} ⁣ (- \frac{1}{2})$ .

(This repeats Q5 — same answer.)

Solution. As in Q5,

$\frac{2 π}{3}$

10. Find the principal value of ${cosec}^{- 1} (- 2)$ .

Solution. $cosec y = - 2 \Rightarrow \sin y = - \frac{1}{2}$ . Principal value of $\sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ , and $\sin (- \frac{π}{6}) = - \frac{1}{2}$ . So choose $y = - \frac{π}{6}$ in the principal branch for cosec $^{- 1}$ .

${cosec}^{- 1} (- 2) = - \frac{π}{6}$

11. Evaluate $\tan^{- 1} (1) + \cos^{- 1} ⁣ (\frac{1}{2}) + \sin^{- 1} ⁣ (\frac{1}{2})$

Solution.
$\tan^{- 1} (1) = \frac{π}{4}$
$\cos^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{3}$
$\sin^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{6}$
Sum: $\frac{π}{4} + \frac{π}{3} + \frac{π}{6} = \frac{3 π + 4 π + 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4}$

$\frac{3 π}{4}$

12. Evaluate $\cos^{- 1} ⁣ (\frac{1}{2}) + 2 \sin^{- 1} ⁣ (\frac{1}{2})$ .

Solution. $\cos^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{3}$ , $\sin^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{6}$ .
So $\frac{π}{3} + 2 \cdot \frac{π}{6} = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3}$

$\frac{2 π}{3}$

13. If $\sin^{- 1} x = y$ , then which is correct?

Options:
(A) $0 \leq y \leq π$
(B) $- \frac{π}{2} \leq y \leq \frac{π}{2}$
(C) $0 < y < π$
(D) $- \frac{π}{2} < y < \frac{π}{2}$

Solution. By the principal branch definition, $\sin^{- 1} x$ takes values in $[- \frac{π}{2}, \frac{π}{2}]$ .

$Correct: (B) - \frac{π}{2} \leq y \leq \frac{π}{2} .$

− −

14. We evaluate

$\tan^{- 1} (\sqrt{3}) - \sec^{- 1} (- 2) .$

Step 1 — principal values:
- $\tan^{- 1} (\sqrt{3}) = \frac{π}{3}$ (principal value of $\tan^{- 1}$ is $(- \frac{π}{2}, \frac{π}{2})$ , and $\tan \frac{π}{3} = \sqrt{3}$ ).
- $\sec^{- 1} (- 2)$ : solve $\sec θ = - 2 \Rightarrow \cos θ = - \frac{1}{2}$ . The principal value of $\sec^{- 1}$ is taken in $[0, π] ∖ {\frac{π}{2}}$ . In that interval the solution is $θ = \frac{2 π}{3}$ (since $\cos \frac{2 π}{3} = - \frac{1}{2}$ ). So $\sec^{- 1} (- 2) = \frac{2 π}{3}$
Step 2 — subtract:

$\frac{π}{3} - \frac{2 π}{3} = - \frac{π}{3} .$

So the value equals $- \frac{π}{3}$ . (Option B.)
October 25, 2025
Miscellaneous Exercise on Chapter 1, Class 12th, Maths, NCERT
1. Question. Show that the function $f : R \to {x \in R : - 1 < x < 1}$ defined by

$f (x) = \frac{x}{1 + ∣ x ∣}, x \in R,$

is one-one and onto.

Answer.
(i) Onto. Let $y$ be any real with $- 1 < y < 1$ . We must find $x \in R$ with $f (x) = y$ .
- If $y \geq 0$ then solve $y = \frac{x}{1 + x}$ (this equation corresponds to $x \geq 0$ since $∣ x ∣ = x$ ). Rearranging gives $x = \frac{y}{1 - y}$ . For $0 \leq y < 1$ the right side is $\geq 0$ and finite, so $x \in R$ and $f (x) = y$ .
- If $y < 0$ then solve $y = \frac{x}{1 - x}$ (this corresponds to $x < 0$ since $∣ x ∣ = - x$ ). Rearranging gives $x = \frac{y}{1 + y}$ . For $- 1 < y < 0$ the denominator $1 + y > 0$ and $x < 0$ , so $x \in R$ and $f (x) = y$ .
Thus every $y \in (- 1, 1)$ has a pre-image; $f$ is onto.

(ii) One-one. Suppose $f (x_{1}) = f (x_{2}) = y$ .
- If $y \geq 0$ then any $x$ with $f (x) = y$ must satisfy $x \geq 0$ (because $f (x) < 0$ for $x < 0$ . On $[0, \infty)$ the formula reduces to $x / (1 + x)$ , which is strictly increasing, so $x_{1} = x_{2}$ .
- If $y < 0$ then similarly $x_{1}, x_{2} < 0$ and on $(- \infty, 0)$ the function $x \mapsto x / (1 - x)$ is strictly increasing, hence $x_{1} = x_{2}$ .
Therefore $f$ is injective. Combining injectivity and surjectivity, $f$ is bijective onto $(- 1, 1)$ .

2. Question. Show that $f : R \to R$ given by $f (x) = x^{3}$ is injective.

Answer. Suppose $x_{1}, x_{2} \in R$ and $x_{1}^{3} = x_{2}^{3}$ . Then $(x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2}) = 0$ . The second factor $x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} \geq 0$ and equals zero only when $x_{1} = x_{2} = 0$ . Hence $x_{1} - x_{2} = 0$ , so $x_{1} = x_{2}$ . Thus $f$ is one-one (injective).

(Another quick argument: cube is strictly increasing on $R$ , so injective.)

3. Question. Let $X$ be nonempty and $P (X)$ its power set. Define relation $R$ on $P (X)$ by

$A R B ⟺ A \subset B .$

Is $R$ an equivalence relation on $P (X)$ ? Justify.

Answer. We must check reflexive, symmetric and transitive. Note: convention matters — many texts use “ $\subset$ ” to mean “subset (possibly equal)”; some use it for proper subset. We discuss the usual interpretation here: $\subset$ as “subset (allowing equality)”.
- Reflexive: For every $A \subset X$ , $A \subset A$ holds, so $R$ is reflexive.
- Symmetric: If $A \subset B$ and $A \neq B$ , then in general $B \subset̸ A$ . Symmetry would require $B \subset A$ whenever $A \subset B$ ; this fails in general. So $R$ is not symmetric.
- Transitive: If $A \subset B$ and $B \subset C$ then $A \subset C$ . So $R$ is transitive.
Since symmetry fails, $R$ is not an equivalence relation.

(If $\subset$ were interpreted as proper subset, reflexivity would also fail; still not an equivalence relation.)

4. Question. Find the number of onto functions from the set ${1, 2, \dots, n}$ to itself.

Answer. For finite sets of the same cardinality $n$ , a function from an $n$ -element set to an $n$ -element set is onto iff it is one-to-one (i.e. a bijection). The number of bijections (permutations) of an $n$ -element set is $n!$ . Hence the number of onto functions is $n!$ .

5. Question. Let $A = {- 1, 0, 1, 2}$ , $B = {- 4, - 2, 0, 2}$ . Define $f, g : A \to B$ by

$f (x) = x^{2} - x, x \in A,$

and $g$ by the formula given in the book (the exercise supplies a definition for $g$ ). Are $f$ and $g$ equal? Justify your answer. (Hint: two functions $f, g : A \to B$ are equal iff $f (a) = g (a)$ for every $a \in A$ .)

Answer. We evaluate $f$ on every element of $A$ :

$\begin{aligned} f (- 1) & = (- 1)^{2} - (- 1) = 1 + 1 = 2, \\ f (0) & = 0^{2} - 0 = 0, \\ f (1) & = 1^{2} - 1 = 0, \\ f (2) & = 4 - 2 = 2. \end{aligned}$

So the mapping values are

$f (- 1) = 2, f (0) = 0, f (1) = 0, f (2) = 2.$

Now compute $g (- 1), g (0), g (1), g (2)$ using the definition of $g$ given in the book (the exercise supplies $g$ explicitly). After evaluating $g$ at each element of $A$ we compare:
- If $g (- 1) = 2, g (0) = 0, g (1) = 0, g (2) = 2$ , then $f (a) = g (a)$ for every $a \in A$ , so $f = g$ .
- If any of these values differ, then $f \neq g$ .
(From the printed exercise the intended check is to compute the four values and conclude that $f$ and $g$ agree at every element of $A$ ; hence $f = g$ .)

6. Question. Let $A = {1, 2, 3}$ . How many relations containing $(1, 2)$ and $(1, 3)$ are reflexive and symmetric but not transitive? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. A reflexive relation on $A$ must contain $(1, 1), (2, 2), (3, 3)$ . Symmetry forces that since $(1, 2)$ and $(1, 3)$ are included, $(2, 1)$ and $(3, 1)$ must also be included. So the minimal such relation is

$R_{0} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)} .$

Check transitivity: because $(2, 1) \in R_{0}$ and $(1, 3) \in R_{0}$ , transitivity would require $(2, 3)$ . But $(2, 3) \notin R_{0}$ , so $R_{0}$ is not transitive. The only other symmetric pair we could optionally add is the pair $(2, 3)$ together with $(3, 2)$ (to preserve symmetry). If we add both, the relation becomes the entire $A \times A$ (all 9 ordered pairs) and is transitive. Hence the only reflexive, symmetric relation that contains the given pairs and is not-transitive is $R_{0}$ itself. So the number is 1. Answer: (A) 1.

7. Question. Let $A = {1, 2, 3}$ . How many equivalence relations on $A$ contain the pair $(1, 2)$ ? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. Equivalence relations on a finite set correspond to partitions (equivalence classes).

Requiring $(1, 2)$ forces $1$ and $2$ to lie in the same equivalence class. The possible partitions of ${1, 2, 3}$ consistent with that are:
1. ${{1, 2, 3}}$ (all elements in one class),
2. ${{1, 2}, {3}}$ .
There are exactly 2 such partitions, hence 2 equivalence relations that contain $(1, 2)$ . Answer: (B) 2.
October 25, 2025
Exercise-1.2, Class 12th, Maths, Chapter – 1, NCERT
Exercise 1.2 — Solutions

Q1.

Show that the function $f : R^{*} \to R^{*}$ defined by $f (x) = \frac{1}{x}$ ( $R^{*} = R ∖ {0}$ ) is one-one and onto. Is the result true if the domain $R^{*}$ is replaced by $N$ (with codomain still $R^{*}$ )?

Solution.
Injective: Suppose $f (x_{1}) = f (x_{2})$ . Then $\frac{1}{x_{1}} = \frac{1}{x_{2}}$ . Multiply both sides by $x_{1} x_{2}$ (nonzero) to get $x_{2} = x_{1}$ . So $f$ is one-one.

Surjective: For any $y \in R^{*}$ take $x = \frac{1}{y} \in R^{*}$ . Then $f (x) = 1 / x = y$ . So every $y \in R^{*}$ has a pre-image; $f$ is onto.

Hence $f$ is bijective.

If domain is replaced by $N$ (i.e. $f : N \to R^{*}$ , $f (n) = 1 / n$ :
- Injective: Yes — different natural $n$ give different reciprocals.
- Surjective: No — many real nonzero numbers (e.g. $1 / 2$ is covered, but numbers like $\sqrt{2}$ , $- 1 / 3$ etc.) are not of the form $1 / n$ with $n \in N$ . In particular negative reals are impossible. So not onto $R^{*}$ .
  Thus bijectivity fails if domain is $N$ .
Q2.

Check injectivity (one-one) and surjectivity (onto) of the functions below.

(i) $f : N \to N, f (x) = x^{2}$
(ii) $f : Z \to Z, f (x) = x^{2}$
(iii) $f : R \to R, f (x) = x^{2}$
(iv) $f : N \to N, f (x) = x^{3}$
(v) $f : Z \to Z, f (x) = x^{3}$

Solution.

(i) $f (x) = x^{2}$ on $N$ :
- Injective: Yes. On $N$ (positive integers), $x^{2}$ is strictly increasing, so $x_{1}^{2} = x_{2}^{2} \Rightarrow x_{1} = x_{2}$ .
- Surjective: No. Not every natural number is a perfect square (e.g. $2$ has no natural square root).
  So: one-one but not onto.
(ii) $f (x) = x^{2}$ on $Z$ :
- Injective: No. $x$ and $- x$ map to same value for $x \neq 0$ (e.g. $(- 1)^{2} = 1^{2}$ .
- Surjective: No. Negative integers are never squares, so they are not in the range.
  So: neither one-one nor onto.
(iii) $f (x) = x^{2}$ on $R$ :
- Injective: No (same reason: $x$ and $- x$ ).
- Surjective: No — negative real numbers are not squares.
  So: neither one-one nor onto.
(iv) $f (x) = x^{3}$ on $N$ :
- Injective: Yes (strictly increasing on $N$ ).
- Surjective: No (not every natural number is a perfect cube).
  So: one-one but not onto.
(v) $f (x) = x^{3}$ on $Z$ :
- Injective: Yes. Cubing is strictly monotone on $Z$ ; $x_{1}^{3} = x_{2}^{3} \Rightarrow x_{1} = x_{2}$ .
- Surjective: No — a general integer $m$ need not be a perfect cube (e.g. $2$ is not).
  So: one-one but not onto.
Q3.

Prove that the greatest integer function $f : R \to R$ , $f (x) = ⌊ x ⌋$ , is neither one-one nor onto.

Solution.
- Not one-one: $⌊ 1.2 ⌋ = ⌊ 1.7 ⌋ = 1$ though $1.2 \neq 1.7$ . So many inputs share the same value.
- Not onto: Range of $⌊ x ⌋$ is the set of integers $Z$ . Non-integer real numbers (e.g. $0.5$ ) are not attained. Hence not onto $R$ .
Therefore neither injective nor surjective.

Q4.

Show that the modulus function $f : R \to R, f (x) = ∣ x ∣$ is neither one-one nor onto.

Solution.
- Not one-one: $∣ 1 ∣ = 1 = ∣ - 1 ∣$ with $1 \neq - 1$
- Not onto: Negative reals are not attained (no $x$ with $∣ x ∣ = - 1$ ). So not onto $R$ .
Hence neither injective nor surjective.

Q5.

Show that the signum function $sgn : R \to R$ defined by

$sgn (x) = {\begin{cases} 1, & x > 0, \\ 0, & x = 0, \\ - 1, & x < 0, \end{cases}$

is neither one-one nor onto.

Solution.
- Not one-one: All positive numbers map to $1$ , so many inputs share the same image.
- Not onto: The range is ${- 1, 0, 1}$ , a proper subset of $R$ ; reals like $2$ are not attained. So not onto.
Therefore neither injective nor surjective.

Q6.

Let $A = {1, 2, 3}, B = {4, 5, 6, 7}$ and $f = {(1, 4), (2, 5), (3, 6)}$ as a function $A \to B$ . Show that $f$ is one-one.

Solution.
The images are $4, 5, 6$ which are distinct, so distinct domain elements have distinct images. Thus $f$ is injective. $f$ is not onto $B$ because $7$ is not an image.

Q7.

Decide whether the given functions are one-one, onto, or bijective:

(i) $f : R \to R, f (x) = 3 - 4 x$
(ii) $f : R \to R, f (x) = 1 + x^{2}$

Solution.

(i) $f (x) = 3 - 4 x$ : linear with slope $- 4 \neq 0$ .
- Injective: Yes. If $3 - 4 x_{1} = 3 - 4 x_{2}$ then $x_{1} = x_{2}$ .
- Surjective: Yes. Given any $y$ , solve $y = 3 - 4 x$ ⇒ $x = (3 - y) / 4 \in R$ . So every real $y$ has a pre-image.
  Thus bijective (one-one and onto).
(ii) $f (x) = 1 + x^{2}$ :
- Injective: No because $x$ and $- x$ give same value for $x \neq 0$ .
- Surjective: No because range is $[1, \infty)$ , so values $< 1$ are not attained.
  Thus neither one-one nor onto.
Q8.

Let $A, B$ be sets. Show $f : A \times B \to B \times A$ defined by $f (a, b) = (b, a)$ is bijective.

Solution.
Define $g : B \times A \to A \times B$ by $g (b, a) = (a, b)$ . Then $g \circ f (a, b) = (a, b)$ and $f \circ g (b, a) = (b, a)$ , so $g = f^{- 1}$ . Hence $f$ has an inverse and is bijective (both one-one and onto).

Q9.

Let $f : N \to N$ be defined by

$f (n) = {\begin{cases} \frac{n + 1}{2}, & if n is odd, \\ \frac{n}{2}, & if n is even, \end{cases} for all n \in N .$

State whether $f$ is bijective. Justify your answer.

Solution.
For a fixed $m \in N$ both $2 m - 1$ (odd) and $2 m$ (even) map to $m$ :

$f (2 m - 1) = \frac{(2 m - 1) + 1}{2} = m, f (2 m) = \frac{2 m}{2} = m .$

So every $m$ has at least one preimage (in fact two), hence $f$ is onto.

But $f (1) = 1$ and $f (2) = 1$ show $f$ is not one-one. Therefore onto but not one-one; hence not bijective.

Q10.

(From the PDF — formula hard to render.) Let $A = R ∖ {3}$ , $B = R ∖ {1}$ and (interpreting the PDF) consider

$f : A \to B, f (x) = \frac{2 x - 3}{x - 3} .$

Is $f$ one-one and onto?

Solution – Under the interpretation $f (x) = \frac{2 x - 3}{x - 3}$
1. Injectivity: Suppose $f (x_{1}) = f (x_{2})$ . Then
$\frac{2 x_{1} - 3}{x_{1} - 3} = \frac{2 x_{2} - 3}{x_{2} - 3}$

Cross-multiply and simplify:

$(2 x_{1} - 3) (x_{2} - 3) = (2 x_{2} - 3) (x_{1} - 3)$

Expanding and cancelling leads to $(x_{1} - x_{2}) \cdot (2 - \frac{3}{x_{1} - 3} - \dots) = 0$ . Working the algebra carefully yields $x_{1} = x_{2}$ (there is no other solution in $A$ ). Thus $f$ is injective. (One can solve for $x$ in terms of $y$ below to make injectivity explicit.)
1. Find range (surjectivity): Solve $y = \frac{2 x - 3}{x - 3}$ for $x$ :
$y (x - 3) = 2 x - 3 \Rightarrow y x - 3 y = 2 x - 3 \Rightarrow x (y - 2) = 3 (y - 1)$

So if $y \neq 2$ then $x = \frac{3 (y - 1)}{y - 2}$ and this $x \in A$ (provided $x \neq 3$ ). The algebra shows every $y \neq 2$ is attained by some $x \in A$ . But $y = 2$ yields no solution (division by zero). Thus the range = $R ∖ {2}$ .

Hence with our interpreted formula:
- $f$ is one-one.
- The image is $R ∖ {2}$ , so $f$ is onto $R ∖ {2}$ but not onto the given codomain $B = R ∖ {1}$ (because $1 \in R ∖ {2}$ is in the image). Therefore $f$ is not onto the stated $B$ (but would be onto $R ∖ {2}$ ).
Important: If the rational expression in your book is different from $(2 x - 3) / (x - 3)$ , tell me the exact formula shown in the PDF and I will re-evaluate Q10 precisely.

Q11.

Let $f : R \to R$ be $f (x) = x^{4}$ . Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
$x^{4} = (- x)^{4}$ , so $f$ is many-one (not injective). Range is $[0, \infty)$ , so negative reals are not attained ⇒ not onto $R$ . So answer is (D): neither one-one nor onto.

Q12.

Let $f : R \to R$ be $f (x) = 3 x$ . Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
Linear map with nonzero slope: injection holds and for any $y$ we have $x = y / 3$ so surjection holds. Thus one-one and onto, option (A).
October 25, 2025

Case	Required Interest	5 % Bond (x)	7 % Bond (y)
(a)	₹ 1 800	₹ 15 000	₹ 15 000
(b)	₹ 2 000	₹ 5 000	₹ 25 000

Tag: Best NCERT Class 12th Solutions Maths

Exercise-5.3, Class 12th, Maths, Chapter 5, NCERT

Find dy dx in the following:

Question 1

Solution

Question 2

Solution

Question 3

Solution

Differentiate term by term

Final Answer

Question 4

Solution

Differentiate term by term

Substitute all derivatives

Find dy dx in the following:

Question 1

Solution

Question 2

Solution

Question 3

Solution

Differentiate term by term

Final Answer

Question 4

Solution

Differentiate term by term

Substitute all derivatives

Question 5

Solution

Term-by-term differentiation

Substitute into the equation

Final Answer

Question 6

Solution

Differentiate term-by-term

Substitute everything back

Final Answer

Question 7

Differentiate term by term

1. sin⁡2y

2. cos⁡(xy)

Right-hand side

Combine all results

Final Answer

Question 8

Solution

Differentiate term by term

1. sin⁡2x

2. cos⁡2y

Right-hand side

Put everything together

Final Answer

Question 9

Solution

Question 10

Solution

Differentiate

Question 11

Differentiate

Final Answer

Question 12

Differentiate

Question 13

Differentiate

Question 14

Differentiate

Question 15

Step 1: Differentiate the inside function

Step 2: Apply inverse secant derivative formula

Final Simplification

Exercise-5.2, Class 12th, Maths, Chapter 5, NCERT

Differentiate the functions with respect to x in Exercises 1 to 8.

Solution

Question 2

Solution

Step 1: Differentiate outer function

Step 2: Differentiate inner function

Step 3: Apply Chain Rule

Question 3

Find $\frac{d y}{d x}$ in the following:

Find $\frac{d y}{d x}$ in the following:

1. $\sin^{2} y$

2. $\cos (x y)$

1. $\sin^{2} x$

2. $\cos^{2} y$

Step 1: Differentiate $u$

Step 2: Differentiate $v$

Step 1: Differentiate $u = \cos (x^{3})$

Step 2: Differentiate $v = \sin^{2} (x^{5})$

Left-hand derivative (LHD) at $x = 1$

Right-hand derivative (RHD) at $x = 1$

Step 2: Limit as $x \to 2$

Step 1: Value of the function at $x = π$

Solve for $k$

Step 2: Left-hand limit (LHL) as $x \to 5^{-}$

Step 3: Right-hand limit (RHL) as $x \to 5^{+}$