Tag: Determinants class 12th Maths 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-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