Tag: Exercise class 12th Maths NCERT Solutions

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-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

Case	Required Interest	5 % Bond (x)	7 % Bond (y)
(a)	₹ 1 800	₹ 15 000	₹ 15 000
(b)	₹ 2 000	₹ 5 000	₹ 25 000

Tag: Exercise class 12th Maths NCERT Solutions

Miscellaneous Exercises on Chapter 4, Class 12th, Maths

Solution:

Step 1: Expansion along second row

Step 2: Evaluate the 2×2 determinants

Step 3: Substitute back

Question 3.If

Solution:

Step 1: Find B−1

Step 2: Compute (AB)−1=B−1A−1

Step 3: Matrix multiplication

Step 4: Write the result

Answer & proof

(ii):

Step 1. Find A−1

Step 2. Find (A−1)−1

Step 3. Verify numerically

Solution:

Step 1: Simplify columns

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

Step 3: Compute each 2×2 determinant

Step 4: Substitute back

Solution:

Step 1: Simplify rows

Step 2: Expand along the first column

Solution (matrix method)

Check

Solution

Exercise-3.2, Class 12th, Maths, Chapter 3 – Matrices, NCERT

Question 1

Solution

Final Answers:

Question 2

Solution

Question 3

Solution

Question 4

Question 5

Question 6

Solution

Question 7

Solution

Question 8

Solution

Question 9

Step 1: Expand the scalar multiplication

Step 2: Add the two matrices on the LHS

Step 3: Compare corresponding elements

Question 10

Solution

Question 11

Solution

Question 12

Given equation:

Step 1: Expand the LHS

3[xyzw]=[3x3y3z3w]

Step 2: Add the two matrices on RHS

[x6−12w]+[4x+yz+w3]=[x+46+x+y−1+z+w2w+3]

Step 3: Equate the LHS and RHS

[3x3y3z3w]=[x+46+x+y−1+z+w2w+3]

Step 4: Compare corresponding elements

Solution

Question 14

Given matrices

Step 1: Compute ABAB

Step 2: Compute BABA

Step 3: Compare

15. Find A2−5A+6I, where

16. If A=(102021203)A=​102​020​213​​, prove A3−6A2+7A+2I=0A3−6A2+7A+2I=0.

17.

Given:

A=(3−24−2),I=(1001).

A2=kA−2I.

Step 1: Compute A2A2

A2=(3−24−2)(3−24−2)=(3⋅3+(−2)⋅43⋅(−2)+(−2)⋅(−2)4⋅3+(−2)⋅44⋅(−2)+(−2)⋅(−2)).

A2=(9−8−6+412−8−8+4)=(1−24−4)

Step 2: Write the given relation

A2=kA−2I.

(1−24−4)=k(3−24−2)−2(1001).

(3k−2−2k4k−2k−2).

Question 3.
If

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

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

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

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

Step 3: Compute each $2 \times 2$ determinant

Question 9

Step 1: Expand the scalar multiplication

Step 2: Add the two matrices on the LHS

Step 3: Compare corresponding elements

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

Step 1: Compute $A B$

Step 2: Compute $B A$

15. Find $A^{2} - 5 A + 6 I$ , where

16. If $A = (\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array})$ , prove $A^{3} - 6 A^{2} + 7 A + 2 I = 0$ .

$A = (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}), I = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) .$

$A^{2} = k A - 2 I .$

Step 1: Compute $A^{2}$

$A^{2} = (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}) (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}) = (\begin{array}{cc} 3 \cdot 3 + (- 2) \cdot 4 & 3 \cdot (- 2) + (- 2) \cdot (- 2) \\ 4 \cdot 3 + (- 2) \cdot 4 & 4 \cdot (- 2) + (- 2) \cdot (- 2) \end{array}) .$

$A^{2} = (\begin{array}{cc} 9 - 8 & - 6 + 4 \\ 12 - 8 & - 8 + 4 \end{array}) = (\begin{array}{cc} 1 & - 2 \\ 4 & - 4 \end{array})$

$A^{2} = k A - 2 I .$

$(\begin{array}{cc} 1 & - 2 \\ 4 & - 4 \end{array}) = k (\begin{array}{cc} 3 & - 2 \\ 4 & - 2 \end{array}) - 2 (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) .$

$(\begin{array}{cc} 3 k - 2 & - 2 k \\ 4 k & - 2 k - 2 \end{array}) .$

${\begin{cases} 3 k - 2 = 1, \\ - 2 k = - 2, \\ 4 k = 4, \\ - 2 k - 2 = - 4. \end{cases}$ Now solve any one (they should all give the same $k$ ):

$k = 1.$

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

Compute $A^{- 1}$

Step 3: Multiply to find $x$ and $y$

(a) For $I = 1800$

(b) For $I = 2000$

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 $p \times k$ :

22. If $n = p$ , then order of $7 X - 5 Z$ where $X$ is $2 \times n$ and $Z$ is $2 \times p$ ?