Tag: Class 12th Chapter 3 Matrices Solutions NCERT

Miscellaneous Exercise on Chapter 3, Class 12th, Maths, NCERT
Question 1:

$If A and B are symmetric matrices, prove that A B - B A is a skew-symmetric matrix.$

Answer (proof):

Recall a matrix $M$ is skew-symmetric if $M^{T} = - M$

Given $A$ and $B$ are symmetric, so

$A^{T} = A, B^{T} = B .$

Consider the transpose of $A B - B A$ :

$(A B - B A)^{T} = (A B)^{T} - (B A)^{T} = B^{T} A^{T} - A^{T} B^{T} .$

Using symmetry of $A$ and $B$ we get

$(A B - B A)^{T} = B A - A B = - (A B - B A) .$

Thus $(A B - B A)^{T} = - (A B - B A)$ , so $A B - B A$ is skew-symmetric.

(As a remark, every skew-symmetric matrix has zeros on its diagonal, so the diagonal entries of $A B - B A$ are all zero.)

Question 2:

$Show that the matrix B^{'} A B is symmetric according as A is symmetric or skew-symmetric.$

Answer (proof):

Let $A$ and $B$ be square matrices of the same order, and let $B^{'}$ denote the transpose of $B$ .
We need to show that:
- If $A$ is symmetric, then $B^{'} A B$ is symmetric.
- If $A$ is skew-symmetric, then $B^{'} A B$ is skew-symmetric.
Case 1: $A$ is symmetric

If $A$ is symmetric, then $A^{'} = A$ .

Consider $(B^{'} A B)^{'}$ :

$(B^{'} A B)^{'} = B^{'} A^{'} B .$

But since $A^{'} = A$ ,

$(B^{'} A B)^{'} = B^{'} A B .$

Hence, $B^{'} A B$ is symmetric.

Case 2: $A$ is skew-symmetric

If $A$ is skew-symmetric, then $A^{'} = - A$ .

Now, take the transpose of $B^{'} A B$ :

$(B^{'} A B)^{'} = B^{'} A^{'} B = B^{'} (- A) B = - B^{'} A B .$

Thus, $(B^{'} A B)^{'} = - B^{'} A B$ ,
which means $B^{'} A B$ is skew-symmetric.

Hence proved:
The matrix $B^{'} A B$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.

Question 3:

$Find the values of x, y, z if A = (\begin{array}{ccc} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{array}) satisfies A^{T} A = I .$

Answer (solution):

Compute $A^{T} A$ . A straightforward multiplication gives

$A^{T} A = (\begin{array}{ccc} 2 x^{2} & 0 & 0 \\ 0 & 6 y^{2} & 0 \\ 0 & 0 & 3 z^{2} \end{array}) .$

For $A^{T} A = I$ we must have the diagonal entries equal to 1, hence

$2 x^{2} = 1, 6 y^{2} = 1, 3 z^{2} = 1$

Therefore

$x^{2} = \frac{1}{2} ⟹ x = \pm \frac{1}{\sqrt{2}}, y^{2} = \frac{1}{6} ⟹ y = \pm \frac{1}{\sqrt{6}}, z^{2} = \frac{1}{3} ⟹ z = \pm \frac{1}{\sqrt{3}} .$

All choices of independent signs are allowed, so the solutions are

$(x, y, z) = (\pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{3}}),$

(8 sign-combinations in total).

Question 4:

$For what values of x does [1 2 1] (\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}) (\begin{array}{c} 0 \\ 2 \\ x \end{array}) = 0 ?$

Answer (solution):

First compute the product of the matrix with the column vector:

$(\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}) (\begin{array}{c} 0 \\ 2 \\ x \end{array}) = (\begin{array}{c} 1 \cdot 0 + 2 \cdot 2 + 0 \cdot x \\ 2 \cdot 0 + 0 \cdot 2 + 1 \cdot x \\ 1 \cdot 0 + 0 \cdot 2 + 2 \cdot x \end{array}) = (\begin{array}{c} 4 \\ x \\ 2 x \end{array})$

Now left-multiply by $[1 2 1]$ :

$[1 2 1] (\begin{array}{c} 4 \\ x \\ 2 x \end{array}) = 1 \cdot 4 + 2 \cdot x + 1 \cdot 2 x = 4 + 4 x$

Set equal to zero:

$4 + 4 x = 0 \Rightarrow x = - 1$

Question 5:

$If A = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}), show that A^{2} - 5 A + 7 I = 0$

Answer (solution):

Given

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

Step 1: Compute $A^{2}$

$A^{2} = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) = (\begin{array}{cc} 3 (3) + 1 (- 1) & 3 (1) + 1 (2) \\ (- 1) (3) + 2 (- 1) & (- 1) (1) + 2 (2) \end{array}) = (\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}) .$

Step 2: Compute $5 A$

$5 A = 5 (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) = (\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array})$

Step 3: Compute $7 I$

$7 I = 7 (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) = (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array})$

Step 4: Substitute into $A^{2} - 5 A + 7 I$

$A^{2} - 5 A + 7 I = (\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}) - (\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array}) + (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array})$

Compute step-by-step:

$A^{2} - 5 A = (\begin{array}{cc} 8 - 15 & 5 - 5 \\ - 5 - (- 5) & 3 - 10 \end{array}) = (\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array})$

Now add $7 I$ :

$A^{2} - 5 A + 7 I = (\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array}) + (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}) = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Hence proved:

$A^{2} - 5 A + 7 I = 0$

Question 6:

$Find x if [x - 5 - 1] (\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}) (\begin{array}{c} x \\ 4 \\ 1 \end{array}) = 0.$

Solution:

We will simplify step-by-step.

Step 1: Multiply the matrix with the column vector

$(\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}) (\begin{array}{c} x \\ 4 \\ 1 \end{array}) = (\begin{array}{c} 1 \cdot x + 0 \cdot 4 + 2 \cdot 1 \\ 0 \cdot x + 2 \cdot 4 + 1 \cdot 1 \\ 2 \cdot x + 0 \cdot 4 + 3 \cdot 1 \end{array}) = (\begin{array}{c} x + 2 \\ 9 \\ 2 x + 3 \end{array})$

Step 2: Multiply the row vector $[x - 5 - 1]$ with the result

$[x - 5 - 1] (\begin{array}{c} x + 2 \\ 9 \\ 2 x + 3 \end{array}) = x (x + 2) + (- 5) (9) + (- 1) (2 x + 3)$

Simplify:

$= x^{2} + 2 x - 45 - 2 x - 3 = x^{2} - 48.$

Step 3: Set equal to zero

$x^{2} - 48 = 0 \Rightarrow x^{2} = 48$

$x= \pm 4 \sqrt{3}$

Question 7

$A manufacturer produces three products x, y, z sold in two markets with annual sales$ $S = (\begin{array}{ccc} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{array}) (rows: Market I, Market II; columns: x, y, z).$ $(a) Unit sale prices p = (\begin{array}{c} 2.50 \\ 1.50 \\ 1.00 \end{array}) . (b) Unit costs c = (\begin{array}{c} 2.00 \\ 1.00 \\ 0.50 \end{array}) .$ $Find (a) total revenue in each market, (b) gross profit in each market.$

Solution

Use matrix multiplication. Revenue vector $R$ (marketwise) is

$R = S p .$

Compute component-wise:

Market I revenue

$10000 (2.5) + 2000 (1.5) + 18000 (1) = 25000 + 3000 + 18000 = 46000.$

Market II revenue

$6000 (2.5) + 20000 (1.5) + 8000 (1) = 15000 + 30000 + 8000 = 53000.$

So

$R = (\begin{array}{c} 46000 \\ 53000 \end{array}) rupees.$

(b) Total cost vector $C = S c$ :

Market I cost

$10000 (2) + 2000 (1) + 18000 (0.5) = 20000 + 2000 + 9000 = 31000.$

Market II cost

$6000 (2) + 20000 (1) + 8000 (0.5) = 12000 + 20000 + 4000 = 36000.$

So

$C = (\begin{array}{c} 31000 \\ 36000 \end{array}) rupees.$

Gross profit for each market $G = R - C$ :

$G = (\begin{array}{c} 46000 - 31000 \\ 53000 - 36000 \end{array}) = (\begin{array}{c} 15000 \\ 17000 \end{array}) rupees$ Final answers

(a) Total revenue — Market I: Rs. 46,000; Market II: Rs. 53,000.
(b) Gross profit — Market I: Rs. 15,000; Market II: Rs. 17,000.

(Also note matrix forms: $R = S p, C = S c, G = S (p - c)$

Question 8:

$Find the matrix X such that X (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) .$

Solution:

We are given:

$X (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array})$

Let:

$A = (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}), B = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) .$

We need $X$ such that $X A = B$ .

To isolate $X$ , multiply both sides on the right by $A^{T} (A A^{T})^{- 1}$ :
(since $A$ is $2 \times 3$ , not square, we use this formula)

$X = B A^{T} (A A^{T})^{- 1} .$

Step 1: Compute $A A^{T}$

$A A^{T} = (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) (\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}) = (\begin{array}{cc} 1^{2} + 2^{2} + 3^{2} & 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 \\ 4 \cdot 1 + 5 \cdot 2 + 6 \cdot 3 & 4^{2} + 5^{2} + 6^{2} \end{array}) = (\begin{array}{cc} 14 & 32 \\ 32 & 77 \end{array})$

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

First find determinant:

$∣ A A^{T} ∣ = (14) (77) - (32) (32) = 1078 - 1024 = 54$

So,

$(A A^{T})^{- 1} = \frac{1}{54} (\begin{array}{cc} 77 & - 32 \\ - 32 & 14 \end{array}) .$

Step 3: Compute $B A^{T}$

$B A^{T} = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) (\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array})$

$= (\begin{array}{cc} (- 7) (1) + (- 8) (2) + (- 9) (3) & (- 7) (4) + (- 8) (5) + (- 9) (6) \\ (2) (1) + (4) (2) + (6) (3) & (2) (4) + (4) (5) + (6) (6) \end{array})$

Compute each entry:

$B A^{T} = (\begin{array}{cc} - 7 - 16 - 27 & - 28 - 40 - 54 \\ 2 + 8 + 18 & 8 + 20 + 36 \end{array}) = (\begin{array}{cc} - 50 & - 122 \\ 28 & 64 \end{array})$

Step 4: Compute $X = B A^{T} (A A^{T})^{- 1}$

$X = \frac{1}{54} (\begin{array}{cc} - 50 & - 122 \\ 28 & 64 \end{array}) (\begin{array}{cc} 77 & - 32 \\ - 32 & 14 \end{array}) .$

Compute the product:

First row:

$(- 50) (77) + (- 122) (- 32) = - 3850 + 3904 = 54,$ $(- 50) (- 32) + (- 122) (14) = 1600 - 1708 = - 108.$

Second row:

$(28) (77) + (64) (- 32) = 2156 - 2048 = 108,$ $(28) (- 32) + (64) (14) = - 896 + 896 = 0$

So:

$X = \frac{1}{54} (\begin{array}{cc} 54 & - 108 \\ 108 & 0 \end{array}) = (\begin{array}{cc} 1 & - 2 \\ 2 & 0 \end{array}) .$

Choose the correct answer in the following questions:

Question 9:

$If A = (\begin{array}{cc} α & β \\ γ & - α \end{array}) is such that A^{2} = I, then find the correct relation among α, β, γ .$

Options:
(A) $1 + α^{2} + β γ = 0$
(B) $1 - α^{2} + β γ = 0$
(C) $1 - α^{2} - β γ = 0$
(D) $1 + α^{2} - β γ = 0$

Solution:

Given

$A = (\begin{array}{cc} α & β \\ γ & - α \end{array})$

Compute $A^{2}$ :

$A^{2} = (\begin{array}{cc} α & β \\ γ & - α \end{array}) (\begin{array}{cc} α & β \\ γ & - α \end{array}) = (\begin{array}{cc} α^{2} + β γ & α β - α β \\ α γ - α γ & γ β + α^{2} \end{array}) = (\begin{array}{cc} α^{2} + β γ & 0 \\ 0 & α^{2} + β γ \end{array})$

Thus,

$A^{2} = (α^{2} + β γ) I .$

We are told $A^{2} = I$ , so:

$(α^{2} + β γ) I = I$

That means:

$α^{2} + β γ = 1$

Rearranging:

$1 - α^{2} - β γ = 0$

Correct Option:

$(C) 1 - α^{2} - β γ = 0.$

Question 10:

$If the matrix A is both symmetric and skew-symmetric, then:$

Options:
(A) $A$ is a diagonal matrix
(B) $A$ is a zero matrix
(C) $A$ is a square matrix
(D) None of these

Solution:

By definition:
- $A$ is symmetric if $A^{T} = A$
- $A$ is skew-symmetric if $A^{T} = - A$
If $A$ is both, then:

$A = A^{T} = - A$

This implies:

$A = - A \Rightarrow 2 A = 0 \Rightarrow A = 0$

That means all elements of $A$ are zero.

Correct Option:

$(B) A is a zero matrix.$

Question 11:

$If A is a square matrix such that A^{2} = A, then find (I + A)^{3} - 7 A .$

Options:
(A) $A$
(B) $I - A$
(C) $I$
(D) $3 A$

Solution:

We are given that:

$A^{2} = A .$

Let’s expand $(I + A)^{3}$ :

$(I + A)^{3} = I^{3} + 3 I^{2} A + 3 I A^{2} + A^{3}$

Simplify each term using $I^{2} = I$ and $A^{2} = A$ :

$(I + A)^{3} = I + 3 A + 3 A + A^{3}$

Now, compute $A^{3}$ :

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

Substitute this back:

$(I + A)^{3} = I + 3 A + 3 A + A = I + 7 A$

Now compute:

$(I + A)^{3} - 7 A = (I + 7 A) - 7 A = I$

Final Answer:

$(C) I$
October 28, 2025

Exercise-3.3, Class 12th, Maths, Chapter 3, NCERT

Q1. (i)

Matrix $A = [\begin{array}{c} 5 \\ \frac{1}{2} \\ - 1 \end{array}]$

This is a column matrix (3×1).

Transpose:

To find $A^{'}$ (or $A^{T}$ ), we interchange rows and columns.

So,

$A^{'} = [5 \frac{1}{2} - 1]$

Answer:

$A^{'} = [5 \frac{1}{2} - 1]$

(This is now a row matrix (1×3).)

Q1- (ii).

Matrix $B = [\begin{array}{cc} 1 & - 1 \\ 2 & 3 \end{array}]$

This is a 2×2 matrix.

Transpose:

We swap rows with columns:

$B^{'} = [\begin{array}{cc} 1 & 2 \\ - 1 & 3 \end{array}]$

Answer:

$B^{'} = [\begin{array}{cc} 1 & 2 \\ - 1 & 3 \end{array}]$

Q1 (iii)

Matrix

$A = [\begin{array}{ccc} - 1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & - 1 \end{array}]$

This is a 3×3 matrix (3 rows, 3 columns).

To find: $A^{'}$ (the transpose)

To find the transpose, we interchange rows and columns — that is, the first row becomes the first column, the second row becomes the second column, and so on.

Step-by-step:

Row	Becomes Column
Row 1 = $(- 1, 5, 6)$	Column 1 = $[\begin{array}{c} - 1 \\ 5 \\ 6 \end{array}]$
Row 2 = $(\sqrt{3}, 5, 6)$	Column 2 = $[\begin{array}{c} \sqrt{3} \\ 5 \\ 6 \end{array}]$
Row 3 = $(2, 3, - 1)$	Column 3 = $[\begin{array}{c} 2 \\ 3 \\ - 1 \end{array}]$

Transpose:

$A^{'} = [\begin{array}{ccc} - 1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & - 1 \end{array}]$

$A^{'} = [5 \frac{1}{2} - 1]$

(This is now a row matrix (1×3).)

Question 2.

Given:

$A = [\begin{array}{ccc} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{array}], B = [\begin{array}{ccc} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array}]$

We have to verify:

1️⃣ $(A + B)^{'} = A^{'} + B^{'}$

2️⃣ $(A - B)^{'} = A^{'} - B^{'}$

Step 1: Find A + B

Add corresponding elements:

$A + B = [\begin{array}{ccc} (- 1) + (- 4) & 2 + 1 & 3 + (- 5) \\ 5 + 1 & 7 + 2 & 9 + 0 \\ (- 2) + 1 & 1 + 3 & 1 + 1 \end{array}] = [\begin{array}{ccc} - 5 & 3 & - 2 \\ 6 & 9 & 9 \\ - 1 & 4 & 2 \end{array}]$

Now find (A + B)′

Take the transpose — interchange rows and columns:

$(A + B)^{'} = [\begin{array}{ccc} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{array}]$

Step 2: Find A′ and B′ separately

Transpose of A:

$A^{'} = [\begin{array}{ccc} - 1 & 5 & - 2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array}]$

Transpose of B:

$B^{'} = [\begin{array}{ccc} - 4 & 1 & 1 \\ 1 & 2 & 3 \\ - 5 & 0 & 1 \end{array}]$

Step 3: Find A′ + B′

$A^{'} + B^{'} = [\begin{array}{ccc} (- 1) + (- 4) & 5 + 1 & (- 2) + 1 \\ 2 + 1 & 7 + 2 & 1 + 3 \\ 3 + (- 5) & 9 + 0 & 1 + 1 \end{array}] = [\begin{array}{ccc} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{array}]$

This is exactly equal to $(A + B)^{'}$ .

Hence,

$(A + B)^{'} = A^{'} + B^{'}$

Step 4: Now find A – B

Subtract corresponding elements:

$A - B = [\begin{array}{ccc} (- 1) - (- 4) & 2 - 1 & 3 - (- 5) \\ 5 - 1 & 7 - 2 & 9 - 0 \\ (- 2) - 1 & 1 - 3 & 1 - 1 \end{array}] = [\begin{array}{ccc} 3 & 1 & 8 \\ 4 & 5 & 9 \\ - 3 & - 2 & 0 \end{array}]$

Now find (A – B)′

$(A - B)^{'} = [\begin{array}{ccc} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{array}]$

Step 5: Find A′ – B′

$A^{'} - B^{'} = [\begin{array}{ccc} (- 1) - (- 4) & 5 - 1 & (- 2) - 1 \\ 2 - 1 & 7 - 2 & 1 - 3 \\ 3 - (- 5) & 9 - 0 & 1 - 1 \end{array}] = [\begin{array}{ccc} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{array}]$

This is exactly equal to $(A - B)^{'}$ .

Hence,

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

Question 3.

Given

$A^{'} = [\begin{array}{cc} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{array}]$

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

, then verify that

(i) (A + B)′ = A′ + B′

(ii) (A – B)′ = A′ – B′

Determine $A$ from $A^{'}$

Remember: $A^{'}$ is the transpose of $A$ .
So to get $A$ , we transpose $A^{'}$ again.

$A = (A^{'})^{'} = [\begin{array}{ccc} 3 & - 1 & 0 \\ 4 & 2 & 1 \end{array}]$

Hence

$A = [\begin{array}{ccc} 3 & - 1 & 0 \\ 4 & 2 & 1 \end{array}], B = [\begin{array}{ccc} - 1 & 2 & 1 \\ 1 & 2 & 3 \end{array}]$

Compute $A + B$

Add corresponding elements:

$A + B = [\begin{array}{ccc} 3 + (- 1) & - 1 + 2 & 0 + 1 \\ 4 + 1 & 2 + 2 & 1 + 3 \end{array}] = [\begin{array}{ccc} 2 & 1 & 1 \\ 5 & 4 & 4 \end{array}]$

Compute $(A + B)^{'}$

Transpose the above matrix:

$(A + B)^{'} = [\begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array}]$

Compute $A^{'} + B^{'}$

We already know:

$A^{'} = [\begin{array}{cc} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{array}], B^{'} = [\begin{array}{cc} - 1 & 1 \\ 2 & 2 \\ 1 & 3 \end{array}]$

Now add elementwise:

$A^{'} + B^{'} = [\begin{array}{cc} 3 + (- 1) & 4 + 1 \\ - 1 + 2 & 2 + 2 \\ 0 + 1 & 1 + 3 \end{array}] = [\begin{array}{cc} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{array}]$

Therefore

$(A + B)^{'} = A^{'} + B^{'}$

Compute $A - B$

$A - B = [\begin{array}{ccc} 3 - (- 1) & - 1 - 2 & 0 - 1 \\ 4 - 1 & 2 - 2 & 1 - 3 \end{array}] = [\begin{array}{ccc} 4 & - 3 & - 1 \\ 3 & 0 & - 2 \end{array}]$ Compute $(A - B)^{'}$

$(A - B)^{'} = [\begin{array}{cc} 4 & 3 \\ - 3 & 0 \\ - 1 & - 2 \end{array}]$

Compute $A^{'} - B^{'}$

$A^{'} - B^{'} = [\begin{array}{cc} 3 - (- 1) & 4 - 1 \\ - 1 - 2 & 2 - 2 \\ 0 - 1 & 1 - 3 \end{array}] = [\begin{array}{cc} 4 & 3 \\ - 3 & 0 \\ - 1 & - 2 \end{array}]$

Therefore

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

Question 4.

Given:

$A^{'} = [\begin{array}{cc} - 2 & 3 \\ 1 & 2 \end{array}], B = [\begin{array}{cc} - 1 & 0 \\ 1 & 2 \end{array}]$

We need to find

$(A + 2 B)^{'}$

Find $A$ from $A^{'}$

Since $A^{'} = [\begin{array}{cc} - 2 & 3 \\ 1 & 2 \end{array}]$ ,

Transpose it to get $A$ :

$A = (A^{'})^{'} = [\begin{array}{cc} - 2 & 1 \\ 3 & 2 \end{array}]$

Compute $2 B$

Multiply each element of $B$ by 2:

$2 B = [\begin{array}{cc} 2 \times (- 1) & 2 \times 0 \\ 2 \times 1 & 2 \times 2 \end{array}] = [\begin{array}{cc} - 2 & 0 \\ 2 & 4 \end{array}]$

Compute $A + 2 B$

$A + 2 B = [\begin{array}{cc} - 2 & 1 \\ 3 & 2 \end{array}] + [\begin{array}{cc} - 2 & 0 \\ 2 & 4 \end{array}] = [\begin{array}{cc} - 4 & 1 \\ 5 & 6 \end{array}]$

Find $(A + 2 B)^{'}$

Take the transpose:

$(A + 2 B)^{'} = [\begin{array}{cc} - 4 & 5 \\ 1 & 6 \end{array}]$

Question 5.

For the matrices A and B, verify that (AB)′ = B′A′, where

Concept Recap

For any conformable matrices $A$ and $B$ :

$(A B)^{'} = B^{'} A^{'}$

That is, the transpose of a product = product of transposes in reverse order.

Now, let’s check numerically.

(i)

$A = [\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}], B = [\begin{array}{ccc} - 1 & 2 & 1 \end{array}]$

Step 1️⃣: Compute $A B$

Since $A$ is 3×1 and $B$ is 1×3,
$A B$ will be a 3×3 matrix.

$A B = [\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}] [\begin{array}{ccc} - 1 & 2 & 1 \end{array}] = [\begin{array}{ccc} 1 (- 1) & 1 (2) & 1 (1) \\ - 4 (- 1) & - 4 (2) & - 4 (1) \\ 3 (- 1) & 3 (2) & 3 (1) \end{array}] = [\begin{array}{ccc} - 1 & 2 & 1 \\ 4 & - 8 & - 4 \\ - 3 & 6 & 3 \end{array}]$

Find $(A B)^{'}$

Transpose means interchange rows and columns:

$(A B)^{'} = [\begin{array}{ccc} - 1 & 4 & - 3 \\ 2 & - 8 & 6 \\ 1 & - 4 & 3 \end{array}]$

Compute $B^{'} A^{'}$

Now find $B^{'}$ and $A^{'}$ .

$B^{'} = [\begin{array}{c} - 1 \\ 2 \\ 1 \end{array}], A^{'} = [\begin{array}{ccc} 1 & - 4 & 3 \end{array}]$

Multiply $B^{'}$ (3×1) with $A^{'}$ (1×3):

$B^{'} A^{'} = [\begin{array}{c} - 1 \\ 2 \\ 1 \end{array}] [\begin{array}{ccc} 1 & - 4 & 3 \end{array}] = [\begin{array}{ccc} (- 1) (1) & (- 1) (- 4) & (- 1) (3) \\ (2) (1) & (2) (- 4) & (2) (3) \\ (1) (1) & (1) (- 4) & (1) (3) \end{array}]$

$= [\begin{array}{ccc} - 1 & 4 & - 3 \\ 2 & - 8 & 6 \\ 1 & - 4 & 3 \end{array}]$

Hence,

$(A B)^{'} = B^{'} A^{'}$

(ii)

$A = [\begin{array}{c} 0 \\ 1 \\ 2 \end{array}], B = [\begin{array}{ccc} 1 & 5 & 7 \end{array}]$

Compute $A B$

$A B = [\begin{array}{c} 0 \\ 1 \\ 2 \end{array}] [\begin{array}{ccc} 1 & 5 & 7 \end{array}] = [\begin{array}{ccc} 0 (1) & 0 (5) & 0 (7) \\ 1 (1) & 1 (5) & 1 (7) \\ 2 (1) & 2 (5) & 2 (7) \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{array}]$

Find $(A B)^{'}$

$(A B)^{'} = [\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array}]$

Compute $B^{'} A^{'}$

$B^{'} = [\begin{array}{c} 1 \\ 5 \\ 7 \end{array}], A^{'} = [\begin{array}{ccc} 0 & 1 & 2 \end{array}]$

Multiply:

$B^{'} A^{'} = [\begin{array}{c} 1 \\ 5 \\ 7 \end{array}] [\begin{array}{ccc} 0 & 1 & 2 \end{array}] = [\begin{array}{ccc} 1 (0) & 1 (1) & 1 (2) \\ 5 (0) & 5 (1) & 5 (2) \\ 7 (0) & 7 (1) & 7 (2) \end{array}] = [\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{array}]$

Hence,

$(A B)^{'} = B^{'} A^{'}$

Question 6.

verify that A′ A = I

(i)

$A = [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}]$

Find $A^{'}$

$A^{'} = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$

Compute $A^{'} A$

$A^{'} A = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}]$ $A^{'} A = [\begin{array}{cc} (\cos^{2} α + \sin^{2} α) & (\cos α \sin α - \sin α \cos α) \\ (\sin α \cos α - \cos α \sin α) & (\sin^{2} α + \cos^{2} α) \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = I$

Hence, $A^{'} A = I$

(ii)

$A = [\begin{array}{cc} \sin α & \cos α \\ - \cos α & \sin α \end{array}]$

Find $A^{'}$

$A^{'} = [\begin{array}{cc} \sin α & - \cos α \\ \cos α & \sin α \end{array}]$

Compute $A^{'} A$

$A^{'} A = [\begin{array}{cc} \sin α & - \cos α \\ \cos α & \sin α \end{array}] [\begin{array}{cc} \sin α & \cos α \\ - \cos α & \sin α \end{array}]$ $A^{'} A = [\begin{array}{cc} (\sin^{2} α + \cos^{2} α) & (\sin α \cos α - \cos α \sin α) \\ (\cos α \sin α - \sin α \cos α) & (\cos^{2} α + \sin^{2} α) \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = I$

Hence, $A^{'} A = I$

Q7.

Recall:

A matrix $A$ is symmetric if $A^{'} = A$ .
A matrix $A$ is skew-symmetric if $A^{'} = - A$ . (Note diagonal entries of a skew-symmetric matrix must be $0$ .)

(i) Show $A$ is symmetric

$A = [\begin{array}{ccc} 1 & - 1 & 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{array}]$

Find the transpose $A^{'}$ (swap rows and columns):

$A^{'} = [\begin{array}{ccc} 1 & - 1 & 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{array}]$

We see $A^{'} = A$ .
Therefore $A$ is symmetric.

(ii) Show $A$ is skew-symmetric

$A = [\begin{array}{ccc} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ 1 & - 1 & 0 \end{array}]$

Compute the transpose:

$A^{'} = [\begin{array}{ccc} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ - 1 & 1 & 0 \end{array}]$

Compute $- A$ :

$- A = [\begin{array}{ccc} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ - 1 & 1 & 0 \end{array}]$

We have $A^{'} = - A$ . Also all diagonal entries are $0$ , as required.

Therefore $A$ is skew-symmetric.

Question 8.

(i) (A + A′) is a symmetric matrix

(ii) (A – A′) is a skew symmetric matrix

Given

$A = [\begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array}] .$

First find the transpose:

$A^{'} = [\begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array}] .$

(i) $A + A^{'}$

Compute:

$A + A^{'} = [\begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array}] + [\begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array}] = [\begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array}]$

Take its transpose:

$(A + A^{'})^{'} = [\begin{array}{cc} 2 & 11 \\ 11 & 14 \end{array}]$ Since $(A + A^{'})^{'} = A + A^{'}$ , the matrix $A + A^{'}$ is symmetric.

(ii) $A - A^{'}$

Compute:

$A - A^{'} = [\begin{array}{cc} 1 & 5 \\ 6 & 7 \end{array}] - [\begin{array}{cc} 1 & 6 \\ 5 & 7 \end{array}] = [\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$

Take its transpose:

$(A - A^{'})^{'} = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}] = - (A - A^{'})$

Since $(A - A^{'})^{'} = - (A - A^{'})$ , the matrix $A - A^{'}$ is skew-symmetric.

Question 9. Find

$\frac{1}{2} (A + A^{'}) and \frac{1}{2} (A - A^{'})$

We are given:

$A = [\begin{array}{ccc} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{array}]$

Find $A^{'}$

Transpose means interchange rows and columns:

$A^{'} = [\begin{array}{ccc} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{array}]$ Compute $A + A^{'}$

Add corresponding entries of $A$ and $A^{'}$ :

$A + A^{'} = [\begin{array}{ccc} 0 + 0 & a + (- a) & b + (- b) \\ (- a) + a & 0 + 0 & c + (- c) \\ (- b) + b & (- c) + c & 0 + 0 \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Compute $A - A^{'}$

Subtract corresponding entries:

$A - A^{'} = [\begin{array}{ccc} 0 - 0 & a - (- a) & b - (- b) \\ (- a) - a & 0 - 0 & c - (- c) \\ (- b) - b & (- c) - c & 0 - 0 \end{array}] = [\begin{array}{ccc} 0 & 2 a & 2 b \\ - 2 a & 0 & 2 c \\ - 2 b & - 2 c & 0 \end{array}]$

Multiply by ½

$\frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$ $\frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{ccc} 0 & 2 a & 2 b \\ - 2 a & 0 & 2 c \\ - 2 b & - 2 c & 0 \end{array}] = [\begin{array}{ccc} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{array}]$

Final Answers

$\frac{1}{2} (A + A^{'}) = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}], \frac{1}{2} (A - A^{'}) = [\begin{array}{ccc} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{array}]$

Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

we’ll write each matrix $A$ as

$A = \frac{1}{2} (A + A^{'}) + \frac{1}{2} (A - A^{'}),$

where $S = \frac{1}{2} (A + A^{'})$ is symmetric and $K = \frac{1}{2} (A - A^{'})$ is skew-symmetric. I’ll give $S$ and $K$ for each part.

(i) $A = [\begin{array}{cc} 3 & 5 \\ 1 & - 1 \end{array}]$

$A^{'} = [\begin{array}{cc} 3 & 1 \\ 5 & - 1 \end{array}]$ $S = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{cc} 6 & 6 \\ 6 & - 2 \end{array}] = [\begin{array}{cc} 3 & 3 \\ 3 & - 1 \end{array}]$

$K = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{cc} 0 & 4 \\ - 4 & 0 \end{array}] = [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$

So $A = S + K$ with $S$ symmetric and $K$ skew.

(ii) $A = [\begin{array}{ccc} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{array}]$

This matrix is already symmetric (check $A^{'} = A$ ), so

$S = \frac{1}{2} (A + A^{'}) = A = [\begin{array}{ccc} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{array}], K = \frac{1}{2} (A - A^{'}) = 0 = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] .$

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

First compute $A^{'} = [\begin{array}{ccc} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{array}]$

Then

$S = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{ccc} 6 & 1 & - 5 \\ 1 & - 4 & - 4 \\ - 5 & - 4 & 4 \end{array}] = [\begin{array}{ccc} 3 & \frac{1}{2} & - \frac{5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ - \frac{5}{2} & - 2 & 2 \end{array}],$

which is symmetric, and

$K = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{ccc} 0 & 5 & 3 \\ - 5 & 0 & 6 \\ - 3 & - 6 & 0 \end{array}] = [\begin{array}{ccc} 0 & \frac{5}{2} & \frac{3}{2} \\ - \frac{5}{2} & 0 & 3 \\ - \frac{3}{2} & - 3 & 0 \end{array}], $

which is skew-symmetric. (You can check $S^{'} = S$ and $K^{'} = - K$ )

(iv) $A = [\begin{array}{cc} 1 & 5 \\ - 1 & 2 \end{array}]$

$A^{'} = [\begin{array}{cc} 1 & - 1 \\ 5 & 2 \end{array}]$ $S = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{array}{cc} 2 & 4 \\ 4 & 4 \end{array}] = [\begin{array}{cc} 1 & 2 \\ 2 & 2 \end{array}],$

$K = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{array}{cc} 0 & 6 \\ - 6 & 0 \end{array}] = [\begin{array}{cc} 0 & 3 \\ - 3 & 0 \end{array}] .$

Question 11

Given:
$A$ and $B$ are symmetric matrices of the same order.
That means:

$A^{'} = A and B^{'} = B$

We need to find the nature of the matrix $A B - B A$

Take transpose of $A B - B A$

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

Since $A^{'} = A$ and $B^{'} = B$ ,

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

Interpretation

A matrix $M$ is skew-symmetric if $M^{'} = - M$ .

Here we found $(A B - B A)^{'} = - (A B - B A)$ Therefore, $A B - B A$ is a skew-symmetric matrix.

Correct Option: (A) Skew symmetric matrix

Question 12

Given:

$A = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}], and A + A^{'} = I$

We must find the value of $α$

Compute $A^{'}$

$A^{'} = [\begin{array}{cc} \cos α & \sin α \\ - \sin α & \cos α \end{array}]$

Compute $A + A^{'}$

$A + A^{'} = [\begin{array}{cc} \cos α + \cos α & - \sin α + \sin α \\ \sin α - \sin α & \cos α + \cos α \end{array}] = [\begin{array}{cc} 2 \cos α & 0 \\ 0 & 2 \cos α \end{array}]$

Given $A + A^{'} = I$

$[\begin{array}{cc} 2 \cos α & 0 \\ 0 & 2 \cos α \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$

This gives:

$2 \cos α = 1 ⟹ \cos α = \frac{1}{2}$

Solve for $α$

$\cos α = \frac{1}{2} \Rightarrow α = \frac{π}{3}$

Final Answers
1️⃣ $A B - B A$ → Skew-symmetric matrix
2️⃣ $A + A^{'} = I \Rightarrow α = \frac{π}{3}$

October 26, 2025

Exercise-3.2, Class 12th, Maths, Chapter 3 – Matrices, NCERT
Question 1

Let

$A = [\begin{array}{cc} 2 & 4 \\ 3 & 2 \end{array}], B = [\begin{array}{cc} 1 & 3 \\ - 2 & 5 \end{array}], C = [\begin{array}{cc} - 2 & 5 \\ 3 & 4 \end{array}]$

Find:
(i) A + B (ii) A − B (iii) 3A − C (iv) AB (v) BA

Solution

(i) A + B =
Add corresponding elements:

$A + B = [\begin{array}{cc} 2 + 1 & 4 + 3 \\ 3 + (- 2) & 2 + 5 \end{array}] = [\begin{array}{cc} 3 & 7 \\ 1 & 7 \end{array}]$

(ii) A − B =

$A - B = [\begin{array}{cc} 2 - 1 & 4 - 3 \\ 3 - (- 2) & 2 - 5 \end{array}] = [\begin{array}{cc} 1 & 1 \\ 5 & - 3 \end{array}]$

(iii) 3A − C =
First, find 3A:

$3 A = [\begin{array}{cc} 6 & 12 \\ 9 & 6 \end{array}]$

Subtract C:

$3 A - C = [\begin{array}{cc} 6 - (- 2) & 12 - 5 \\ 9 - 3 & 6 - 4 \end{array}] = [\begin{array}{cc} 8 & 7 \\ 6 & 2 \end{array}]$

(iv) AB =

$A B = [\begin{array}{cc} 2 (1) + 4 (- 2) & 2 (3) + 4 (5) \\ 3 (1) + 2 (- 2) & 3 (3) + 2 (5) \end{array}] = [\begin{array}{cc} - 6 & 26 \\ - 1 & 19 \end{array}]$

(v) BA =

$B A = [\begin{array}{cc} 1 (2) + 3 (3) & 1 (4) + 3 (2) \\ (- 2) (2) + 5 (3) & (- 2) (4) + 5 (2) \end{array}] = [\begin{array}{cc} 11 & 10 \\ 11 & 2 \end{array}]$

Final Answers:

(i)

$A + B = [\begin{array}{cc} 3 & 7 \\ 1 & 7 \end{array}]$

(ii)

$A - B = [\begin{array}{cc} 1 & 1 \\ 5 & - 3 \end{array}]$

(iii)

$3 A - C = [\begin{array}{cc} 8 & 7 \\ 6 & 2 \end{array}]$

(iv)

$A B = [\begin{array}{cc} - 6 & 26 \\ - 1 & 19 \end{array}]$

(v)

$B A = [\begin{array}{cc} 11 & 10 \\ 11 & 2 \end{array}]$

Question 2

Compute the following:

(i)

$[\begin{array}{cc} a & b \\ - b & a \end{array}] + [\begin{array}{cc} a & b \\ b & a \end{array}]$

(ii)

$[\begin{array}{cc} a^{2} + b^{2} & b^{2} + c^{2} \\ c^{2} + a^{2} & a^{2} + b^{2} \end{array}] + [\begin{array}{cc} c^{2} + 2 a b & 2 b c \\ - 2 a c & - 2 a b \end{array}]$

(iii)

$[\begin{array}{cc} 1 & 4 \\ 8 & 5 \\ 2 & 8 \end{array}] + [\begin{array}{cc} 6 & 12 \\ 5 & 16 \\ 5 & 3 \end{array}] + [\begin{array}{cc} 7 & 6 \\ 8 & 0 \\ 2 & 4 \end{array}]$

(iv)

$[\begin{array}{cc} \cos^{2} x & \sin^{2} x \\ \sin^{2} x & \cos^{2} x \end{array}] + [\begin{array}{cc} \sin^{2} x & \cos^{2} x \\ \cos^{2} x & \sin^{2} x \end{array}]$

Solution

(i) Add element-wise:

$= [\begin{array}{cc} 2 a & 2 b \\ 0 & 2 a \end{array}]$

(ii)

$= [\begin{array}{cc} a^{2} + b^{2} + c^{2} + 2 a b & b^{2} + c^{2} + 2 b c \\ c^{2} + a^{2} - 2 a c & a^{2} + b^{2} - 2 a b \end{array}]$

(iii)
Add element-wise:

$= [\begin{array}{cc} 14 & 22 \\ 21 & 21 \\ 9 & 15 \end{array}]$

(iv)

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

Question 3

Compute the indicated products:

(i)

$[\begin{array}{cc} a & b \\ - b & a \end{array}] [\begin{array}{cc} a & - b \\ b & a \end{array}]$

(ii)

$[\begin{array}{ccc} 2 & 3 & 4 \end{array}] [\begin{array}{c} 1 \\ 2 \\ 3 \end{array}]$

(iii)

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

Solution

(i)

$= [\begin{array}{cc} a^{2} + b^{2} & 0 \\ 0 & a^{2} + b^{2} \end{array}]$

(ii)

$2 \times 1 + 3 \times 2 + 4 \times 3 = 20$

(iii)

$= [\begin{array}{ccc} - 3 & - 4 & 1 \\ 8 & 13 & 9 \end{array}]$

Question 4

If

$A = [\begin{array}{cc} 2 & 1 \\ 5 & 0 \\ - 1 & - 1 \end{array}], B = [\begin{array}{cc} 3 & - 1 \\ 4 & 2 \\ 2 & 0 \end{array}], C = [\begin{array}{cc} 4 & 1 \\ 0 & 3 \\ 1 & - 2 \end{array}]$

Compute (A + B), (B − C), and verify A + (B − C) = (A + B) − C.

A + B =

$[\begin{array}{cc} 5 & 0 \\ 9 & 2 \\ 1 & - 1 \end{array}]$

B − C =

$[\begin{array}{cc} - 1 & - 2 \\ 4 & - 1 \\ 1 & 2 \end{array}]$

A + (B − C) =

$[\begin{array}{cc} 1 & - 1 \\ 9 & - 1 \\ 0 & 1 \end{array}]$

(A + B) − C =

$[\begin{array}{cc} 1 & - 1 \\ 9 & - 1 \\ 0 & 1 \end{array}]$

Hence, verified.

Question 5

$A = [\begin{array}{ccc} 2 & 1 & 1 \\ 3 & 3 & 5 \\ 6 & 2 & 5 \end{array}], B = [\begin{array}{ccc} 2 & 3 & 5 \\ 1 & 1 & 4 \\ 7 & 5 & 6 \end{array}]$

Compute 3A − 5B

Step 1:

$3 A = [\begin{array}{ccc} 6 & 3 & 3 \\ 9 & 9 & 15 \\ 18 & 6 & 15 \end{array}], 5 B = [\begin{array}{ccc} 10 & 15 & 25 \\ 5 & 5 & 20 \\ 35 & 25 & 30 \end{array}]$

Step 2: Subtract:

$3 A - 5 B = [\begin{array}{ccc} - 4 & - 12 & - 22 \\ 4 & 4 & - 5 \\ - 17 & - 19 & - 15 \end{array}]$

Question 6

Simplify:

$[\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}] [\begin{array}{cc} \sin x & - \cos x \\ \cos x & \sin x \end{array}]$

Solution

Multiply:
(1,1):

$\cos x \sin x + \sin x \cos x = \sin (2 x)$

Result:

$[\begin{array}{cc} \sin (2 x) & 0 \\ 0 & \sin (2 x) \end{array}]$

Question 7

Find X and Y if

(i)

$X + Y = [\begin{array}{cc} 7 & 0 \\ 2 & 5 \end{array}], X - Y = [\begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array}]$

(ii)

$2 X + 3 Y = [\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}], 3 X + 2 Y = [\begin{array}{cc} 2 & - 2 \\ - 1 & 5 \end{array}]$

Solution

Case (i):
Add and subtract:

$2 X = (X + Y) + (X - Y) = [\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}] \Rightarrow X = [\begin{array}{cc} 5 & 0 \\ 1 & 4 \end{array}]$

$2 Y = (X + Y) - (X - Y) = [\begin{array}{cc} 4 & 0 \\ 2 & 2 \end{array}] \Rightarrow Y = [\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}]$

Case (ii):
Multiply first by 3, second by 2 and subtract:

$(6 X + 9 Y) - (6 X + 4 Y) = 5 Y = [\begin{array}{cc} 2 & 13 \\ 14 & - 10 \end{array}]$

$Y = [\begin{array}{cc} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & - 2 \end{array}]$

Substitute in

$2 X + 3 Y = [\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}]$

$X = [\begin{array}{cc} \frac{4}{5} & - \frac{3}{5} \\ - \frac{2}{5} & 2 \end{array}]$

Question 8

Find X, if

$Y = [\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}]$

and

$2 X + Y = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}]$

Solution

Given:

$2 X + Y = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}]$

Substitute Y:

$2 X + [\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}] = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}]$

Now, move Y to RHS:

$2 X = [\begin{array}{cc} 1 & 0 \\ - 3 & 2 \end{array}] - [\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}] = [\begin{array}{cc} - 2 & - 2 \\ - 4 & - 2 \end{array}]$

Divide both sides by 2:

$X = [\begin{array}{cc} - 1 & - 1 \\ - 2 & - 1 \end{array}]$

Question 9

Find the values of x and y, if

$2 [\begin{array}{cc} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$

Step 1: Expand the scalar multiplication

$2 [\begin{array}{cc} 1 & 3 \\ 0 & x \end{array}] = [\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}]$

So the equation becomes:

$[\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$

Step 2: Add the two matrices on the LHS

$[\begin{array}{cc} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$

Step 3: Compare corresponding elements
1. From (1,1): $2 + y = 5 \Rightarrow y = 3$
2. From (2,2): $2 x + 2 = 8 \Rightarrow 2 x = 6 \Rightarrow x = 3$
Final Answers:

$x = 3, y = 3$

Question 10

Solve for x, y, z, t if

$2 [\begin{array}{cc} x & z \\ y & t \end{array}] + 3 [\begin{array}{cc} 1 & - 1 \\ 0 & 2 \end{array}] = [\begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array}]$

Solution

Step 1: Multiply 2 and 3 through matrices:

$[\begin{array}{cc} 2 x & 2 z \\ 2 y & 2 t \end{array}] + [\begin{array}{cc} 3 & - 3 \\ 0 & 6 \end{array}] = [\begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array}]$

Step 2: Add LHS:

$[\begin{array}{cc} 2 x + 3 & 2 z - 3 \\ 2 y & 2 t + 6 \end{array}] = [\begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array}]$

Step 3: Equate corresponding elements:
1. $2 x + 3 = 3 \Rightarrow x = 0$
2. $2 z - 3 = 5 \Rightarrow z = 4$
3. $2 y = 4 \Rightarrow y = 2$
4. $2 t + 6 = 6 \Rightarrow t = 0$
Hence

$x = 0, y = 2, z = 4, t = 0$

Question 11

If

$[\begin{array}{c} x \\ 2 \end{array}] + [\begin{array}{c} - 1 \\ y \end{array}] = [\begin{array}{c} 10 \\ 5 \end{array}]$

Find x and y.

Solution

Add LHS element-wise:

$[\begin{array}{c} x - 1 \\ 2 + y \end{array}] = [\begin{array}{c} 10 \\ 5 \end{array}]$

Comparing elements:

$x - 1 = 10 \Rightarrow x = 11$
$2 + y = 5 \Rightarrow y = 3$

Answer:
$x = 11, y = 3$

Question 12

Given equation:

$3 [\begin{array}{cc} x & y \\ z & w \end{array}] = [\begin{array}{cc} x & 6 \\ - 1 & 2 w \end{array}] + [\begin{array}{cc} 4 & x + y \\ z + w & 3 \end{array}]$

We need to find x  and  y.

Step 1: Expand the LHS

$3 [\begin{array}{cc} x & y \\ z & w \end{array}] = [\begin{array}{cc} 3 x & 3 y \\ 3 z & 3 w \end{array}]$

Step 2: Add the two matrices on RHS

$[\begin{array}{cc} x & 6 \\ - 1 & 2 w \end{array}] + [\begin{array}{cc} 4 & x + y \\ z + w & 3 \end{array}] = [\begin{array}{cc} x + 4 & 6 + x + y \\ - 1 + z + w & 2 w + 3 \end{array}]$

Step 3: Equate the LHS and RHS

$[\begin{array}{cc} 3 x & 3 y \\ 3 z & 3 w \end{array}] = [\begin{array}{cc} x + 4 & 6 + x + y \\ - 1 + z + w & 2 w + 3 \end{array}]$

Step 4: Compare corresponding elements
1. From (1, 1): $3 x = x + 4 \Rightarrow 2 x = 4 \Rightarrow x = 2.$
2. From (1, 2): $3 y = 6 + x + y \Rightarrow 2 y = 6 + x \Rightarrow 2 y = 6 + 2 \Rightarrow 2 y = 8 \Rightarrow y = 4.$
3. Other entries (involving z, w) are not required here.
Final Answer:

$x = 2, y = 4$

Question 13

If

$F (x) = [\begin{array}{ccc} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}],$

prove that $F (x) F (y) = F (x + y)$

Solution

Compute $F (x) F (y)$ :

$F (x) F (y) = [\begin{array}{ccc} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} \cos y & - \sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}]$

Multiply the first two rows:

Top-left 2×2 block:

$[\begin{array}{cc} \cos x \cos y - \sin x \sin y & - (\sin x \cos y + \cos x \sin y) \\ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y \end{array}]$

But using angle addition identities:

$\cos (x + y) = \cos x \cos y - \sin x \sin y$ $\sin (x + y) = \sin x \cos y + \cos x \sin y$

So,

$F (x) F (y) = [\begin{array}{ccc} \cos (x + y) & - \sin (x + y) & 0 \\ \sin (x + y) & \cos (x + y) & 0 \\ 0 & 0 & 1 \end{array}] = F (x + y)$

Hence proved.

Question 14

to show that matrix multiplication is not commutative, i.e.

$A B \neq B A .$

Let’s solve step-by-step carefully.

Given matrices

$A = (\begin{array}{cc} 5 & - 1 \\ 6 & 7 \end{array}), B = (\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}) .$

Step 1: Compute $A B$

$A B = (\begin{array}{cc} 5 & - 1 \\ 6 & 7 \end{array}) (\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}) .$

Multiply row by column:

$A B = (\begin{array}{cc} (5) (2) + (- 1) (3) & (5) (1) + (- 1) (4) \\ (6) (2) + (7) (3) & (6) (1) + (7) (4) \end{array}) = (\begin{array}{cc} 10 - 3 & 5 - 4 \\ 12 + 21 & 6 + 28 \end{array}) = (\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array})$

Step 2: Compute $B A$

$B A = (\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}) (\begin{array}{cc} 5 & - 1 \\ 6 & 7 \end{array})$

Multiply:

$B A = (\begin{array}{cc} (2) (5) + (1) (6) & (2) (- 1) + (1) (7) \\ (3) (5) + (4) (6) & (3) (- 1) + (4) (7) \end{array}) = (\begin{array}{cc} 10 + 6 & - 2 + 7 \\ 15 + 24 & - 3 + 28 \end{array}) = (\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array})$

Step 3: Compare

$A B = (\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}), B A = (\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}) .$

Clearly $A B \neq B$

✅ Hence proved:
Matrix multiplication is not commutative in general:

$A B \neq B A .$

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

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

Compute $A^{2}$ :

$A^{2} = (\begin{array}{ccc} 5 & 1 & 2 \\ 9 & 4 & 5 \\ 4 & 1 & 4 \end{array})$

Now $- 5 A = (\begin{array}{ccc} - 10 & 0 & - 5 \\ - 10 & - 5 & - 15 \\ - 5 & - 5 & 0 \end{array})$
and $6 I = (\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array})$

Add them:

$A^{2} - 5 A + 6 I = (\begin{array}{ccc} 5 & 1 & 2 \\ 9 & 4 & 5 \\ 4 & 1 & 4 \end{array}) + (\begin{array}{ccc} - 10 & 0 & - 5 \\ - 10 & - 5 & - 15 \\ - 5 & - 5 & 0 \end{array}) + (\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}) = (\begin{array}{ccc} 1 & 1 & - 3 \\ - 1 & 5 & - 10 \\ - 1 & - 4 & 10 \end{array}) .$

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

Compute (steps shown succinctly):

$A^{2} = (\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}), A^{3} = (\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}) .$

Now form $A^{3} - 6 A^{2} + 7 A + 2 I$ . Term-by-term:
- $- 6 A^{2} = (\begin{array}{ccc} - 30 & 0 & - 48 \\ - 12 & - 24 & - 30 \\ - 48 & 0 & - 78 \end{array})$
- $7 A = (\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array})$
- $2 I = (\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array})$
Adding $A^{3} + (- 6 A^{2}) + 7 A + 2 I$ gives the zero matrix. So the identity holds.

17.

Given:

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

We have to find the scalar $k$ such that

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

Step 1: Compute $A^{2}$

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

Simplify each element:

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

Step 2: Write the given relation

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

Substitute $A^{2}$ , $A$ , and $I$ :

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

Simplify the right-hand side:

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

Step 3: Equate corresponding entries

${\begin{cases} 3 k - 2 = 1, \\ - 2 k = - 2, \\ 4 k = 4, \\ - 2 k - 2 = - 4. \end{cases}$ Now solve any one (they should all give the same $k$ ):
- From the second: $- 2 k = - 2 \Rightarrow k = 1$
- From the first: $3 k - 2 = 1 \Rightarrow k = 1$
- Others also give $k = 1.$
All consistent.

Final Answer:

$k = 1.$

18. If $A = (\begin{array}{cc} 0 & - \tan (\frac{α}{2}) \\ \tan (\frac{α}{2}) & 0 \end{array})$ and $I$ is $2 \times 2$ identity, show

$I + A = (I - A) (\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array})$

(That is the standard Cayley-type relation when $A$ uses $\tan (α / 2)$

Proof sketch (algebraic verification): Put $t = \tan (\frac{α}{2}).$

Then

$I + A = (\begin{array}{cc} 1 & - t \\ t & 1 \end{array}), I - A = (\begin{array}{cc} 1 & t \\ - t & 1 \end{array})$

Use the trig identities

$\cos α = \frac{1 - t^{2}}{1 + t^{2}}, \sin α = \frac{2 t}{1 + t^{2}}$

Multiply $(I - A)$ by the rotation matrix $R (α) = (\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array})$

$(I - A) R (α) = (\begin{array}{cc} c + t s & - s + t c \\ - t c + s & t s + c \end{array}),$

substitute $c = \cos α, s = \sin α$ and the formulas above; each entry simplifies to match $I + A = (\begin{array}{cc} 1 & - t \\ t & 1 \end{array})$ . Thus the identity holds.

(If your printed sign convention for $A$ is different — e.g. the off-diagonal signs reversed — paste the exact matrix and I’ll adapt the algebra.)

19.

A trust fund has ` 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ` 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1800 (b) Rs 2000
Answer –

A trust fund has ₹30,000 to invest in two types of bonds:

Bond 1 → 5 % interest per year

Bond 2 → 7 % interest per year

We must divide ₹30,000 between them so that the annual total interest is:
(a) ₹ 1 800 (b) ₹ 2 000

Step 1: Define variables

Let
Then: ${\begin{cases} x + y = 30000 \\ 0.05 x + 0.07 y = I \end{cases}$
where $I$ = required interest.

Step 2: Matrix form
$\underset{A}{\underset{⏟}{[\begin{array}{cc} 1 & 1 \\ 0.05 & 0.07 \end{array}]}} [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 30000 \\ I \end{array}]$ That is, $A x = b .$
We can find $x = A^{- 1} b$

Compute $A^{- 1}$

For a $2 \times 2$ matrix $A = [\begin{array}{cc} a & b \\ c & d \end{array}]$
$A^{- 1} = \frac{1}{a d - b c} [\begin{array}{cc} d & - b \\ - c & a \end{array}]$
Here
$a = 1, b = 1, c = 0.05, d = 0.07$
$\det (A) = (1) (0.07) - (1) (0.05) = 0.02$
So:
$A^{- 1} = \frac{1}{0.02} [\begin{array}{cc} 0.07 & - 1 \\ - 0.05 & 1 \end{array}] = [\begin{array}{cc} 3.5 & - 50 \\ - 2.5 & 50 \end{array}] .$

Step 3: Multiply to find $x$ and $y$
$[\begin{array}{c} x \\ y \end{array}] = A^{- 1} [\begin{array}{c} 30000 \\ I \end{array}] = [\begin{array}{cc} 3.5 & - 50 \\ - 2.5 & 50 \end{array}] [\begin{array}{c} 30000 \\ I \end{array}]$
Compute components:
$x = 3.5 (30000) - 50 I, y = - 2.5 (30000) + 50 I .$
Simplify:
$x = 105000 - 50 I, y = - 75000 + 50 I .$

(a) For $I = 1800$
$x = 105000 - 50 (1800) = 105000 - 90000 = 15000,$ $y = - 75000 + 50 (1800) = - 75000 + 90000 = 15000.$

(b) For $I = 2000$
$x = 105000 - 50 (2000) = 105000 - 100000 = 5000,$ $y = - 75000 + 50 (2000) = - 75000 + 100000 = 25000.$

✅ Final Answers Summary

Case Required Interest 5 % Bond (x) 7 % Bond (y)

(a) ₹ 1 800 ₹ 15 000 ₹ 15 000

(b) ₹ 2 000 ₹ 5 000 ₹ 25 000

So, by matrix multiplication,
$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 105000 - 50 I \\ - 75000 + 50 I \end{array}],$
20. Bookshop: 10 dozen chemistry, 8 dozen physics, 10 dozen economics books; prices ₹80, ₹60, ₹40. Find total receipt.

First convert dozens to counts: 10 dozen = 120, 8 dozen = 96, 10 dozen = 120. Multiply quantities by unit prices:

$120 \times 80 + 96 \times 60 + 120 \times 40 = 9600 + 5760 + 4800 = ₹ 20,160 .$

(Matrix form: $[120 96 120] (\begin{array}{c} 80 \\ 60 \\ 40 \end{array}) = 20160$

21. (Multiple choice) With $X$ of order $2 \times n$ , $Y$ of order $3 \times k$ , $Z$ of order $2 \times p$ , $W$ of order $n \times 3$ , $P$ of order $p \times k$ :

What restriction on $n, k, p$ so that $P Y + W Y$ is defined?
- For $P Y$ to be defined: $P$ (size $p \times k$ ) times $Y$ (size $3 \times k$ ) requires number of columns of $P$ = number of rows of $Y$ , so $k = 3$ .
- For $W Y$ to be defined: $W$ (size $n \times 3$ ) times $Y$ (size $3 \times k$ ) is defined for any $n$ once $k$ is known.
- For $P Y$ and $W Y$ to be addable, their resulting orders must match: $P Y$ is $p \times k$ and $W Y$ is $n \times k$ . So $p = n$ .
Thus the restriction is $k = 3$ and $p = n$ .

Answer: (A) $k = 3, p = n$ .

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

If $n = p$ then both $X$ and $Z$ are $2 \times n$ . So $7 X - 5 Z$ is of order $2 \times n$ . Answer: (B) $2 \times n$ .
October 26, 2025
Exercise-3.1, Class 12th, Maths, Chapter 3- Matrices, NCERT

Exercise 3.1

Question 1:

In the matrix

$A = [\begin{array}{cccc} 2 & 5 & 19 & - 7 \\ 1 & - 5 & 17 & 3 \\ 1 & - 2 & 12 & 5 \end{array}]$

write:
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

Answer:
(i) Order = 3 × 4
(ii) Number of elements = 3 × 4 = 12
(iii) a₁₃ = 19, a₂₁ = 1, a₃₃ = 12, a₂₄ = 3, a₂₃ = 17.

Question 2:

If a matrix has 24 elements, what are the possible orders it can have? What if it has 13 elements?

Answer:
We know that if a matrix has m × n elements, then total elements = m × n.

(i) For 24 elements:
Possible orders = (1,24), (2,12), (3,8), (4,6), (6,4), (8,3), (12,2), (24,1)

(ii) For 13 elements (prime number):
Possible orders = (1,13), (13,1)

Question 3:

If a matrix has 18 elements, what are its possible orders? What if it has 5 elements?

Answer:
(i) For 18 elements: $m \times n = 18$
Possible orders = (1,18), (2,9), (3,6), (6,3), (9,2), (18,1)

(ii) For 5 elements (prime): (1,5), (5,1)

Question 4:

Construct a 2 × 2 matrix A = [aᵢⱼ] where:

(i) aᵢⱼ = 2i − j
(ii) aᵢⱼ = i² − 3j
(iii) aᵢⱼ = i² / 2j

Answer:
For i, j = 1, 2

(i)

$A = [\begin{array}{cc} 2 (1) - 1 & 2 (1) - 2 \\ 2 (2) - 1 & 2 (2) - 2 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 3 & 2 \end{array}]$

(ii)

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

(iii)

$A = [\begin{array}{cc} \frac{1^{2}}{2 \times 1} & \frac{1^{2}}{2 \times 2} \\ \frac{2^{2}}{2 \times 1} & \frac{2^{2}}{2 \times 2} \end{array}] = [\begin{array}{cc} \frac{1}{2} & \frac{1}{4} \\ 2 & 1 \end{array}]$

Question 5:

Construct a 3 × 4 matrix where:
(i) aᵢⱼ = ½ (i − 3j)
(ii) aᵢⱼ = 2i − j

Answer:

(i)

$A = [\begin{array}{cccc} - 1 & - 2.5 & - 4 & - 5.5 \\ - 0.5 & - 2 & - 3.5 & - 5 \\ 0 & - 1.5 & - 3 & - 4.5 \end{array}]$

(ii)

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

Question 6:

Find the values of x, y, and z from the following equations:

(i)

$[\begin{array}{ccc} x & 5 & z \\ y & 1 & 5 \end{array}] = [\begin{array}{ccc} 2 & 6 & 2 \\ 1 & 5 & 8 \end{array}]$

(ii)

$[\begin{array}{ccc} 5 & z & x + y \\ 5 & 8 & x + y \end{array}] = [\begin{array}{ccc} x + z & y & z \\ x z & y & z \end{array}]$

Answer:

(i) Comparing elements:
x = 2, y = 1, z = 2.

(ii) On comparing:
x = 3, y = 1, z = 2.

Question 7:

Find a, b, c, d from the equation:

$[\begin{array}{cc} a - b & 2 a - b \\ 2 a + c & 3 c - d \end{array}] = [\begin{array}{cc} 0 & - 15 \\ 0 & 13 \end{array}]$

Answer:
Equating elements:
a − b = 0
2a − b = −15
2a + c = 0
3c − d = 13

Solving, we get
a = −15, b = −15, c = 30, d = 77.

Question 8:

A = [aᵢⱼ] is a square matrix if:
(A) m < n (B) m > n (C) m = n (D) None of these

Answer:
(C) m = n

Question 9:

Which values of x and y make the following matrices equal?

$[\begin{array}{cc} 3 x + 7 & 5 \\ y + 1 & 2 \end{array}] = [\begin{array}{cc} - 3 x & 0 \\ - 2 & 8 \end{array}]$

Answer:
3x + 7 = −3x ⇒ 6x = −7 ⇒ x = −7/6
y + 1 = −2 ⇒ y = −3

x = −7/6, y = −3

Question 10:

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512

Answer:
Each entry can be 0 or 1, i.e., 2 possibilities per element.
Total = $2^{3 \times 3} = 2^{9} = 512$

Answer: (D) 512

October 26, 2025

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

Tag: Class 12th Chapter 3 Matrices Solutions NCERT

Miscellaneous Exercise on Chapter 3, Class 12th, Maths, NCERT

Case 1: A is symmetric

Case 2: A is skew-symmetric

Step 1: Compute A2

Step 2: Compute 5A

Step 3: Compute 7I

Step 4: Substitute into A2−5A+7I

Solution:

Step 1: Multiply the matrix with the column vector

Step 2: Multiply the row vector [x −5 −1] with the result

Step 3: Set equal to zero

Solution:

Step 1: Compute AAT

Step 2: Compute (AAT)−1

Step 3: Compute BATBAT

Step 4: Compute X=BAT(AAT)−1

Solution:

Solution:

Solution:

Exercise-3.3, Class 12th, Maths, Chapter 3, NCERT

Q1. (i)

Transpose:

Q1- (ii).

Transpose:

Q1 (iii)

To find: A′A (the transpose)

Step-by-step:

Transpose:

Question 2.

Given:

A=[−123579−211],B=[−41−5120131]

Step 1: Find A + B

A+B=[(−1)+(−4)2+13+(−5)5+17+29+0(−2)+11+31+1]=[−53−2699−142]

Now find (A + B)′

(A+B)′=[−56−1394−292]

Step 2: Find A′ and B′ separately

Transpose of A:

A′=[−15−2271391]

Transpose of B:

B′=[−411123−501]

Step 3: Find A′ + B′

A′+B′=[(−1)+(−4)5+1(−2)+12+17+21+33+(−5)9+01+1]=[−56−1394−292]

(A+B)′=A′+B′

Step 4: Now find A – B

A−B=[(−1)−(−4)2−13−(−5)5−17−29−0(−2)−11−31−1]=[318459−3−20]

Now find (A – B)′

(A−B)′=[34−315−2890]

Step 5: Find A′ – B′

A′−B′=[(−1)−(−4)5−1(−2)−12−17−21−33−(−5)9−01−1]=[34−315−2890]

(A−B)′=A′−B′

Question 3.

Given

Determine A from A′

Compute A+B

Compute (A+B)′

Compute A′+B′

Compute A−B

Compute A′−B′

Question 4.

Given:

Find A from A′

Compute 2B

Compute A+2B

Find (A+2B)′

Question 5.

Concept Recap

(i)

Step 1️⃣: Compute AB

Find (AB)′

Compute B′A′

(ii)

Compute AB

Find (AB)′

Compute B′A′

Question 6.

verify that A′ A = I

(i)

Find A′

Compute A′A

Case 1: $A$ is symmetric

Case 2: $A$ is skew-symmetric

Step 1: Compute $A^{2}$

Step 2: Compute $5 A$

Step 3: Compute $7 I$

Step 4: Substitute into $A^{2} - 5 A + 7 I$

Step 2: Multiply the row vector $[x - 5 - 1]$ with the result

Step 1: Compute $A A^{T}$

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

Step 3: Compute $B A^{T}$

Step 4: Compute $X = B A^{T} (A A^{T})^{- 1}$

To find: $A^{'}$ (the transpose)

$A = [\begin{array}{ccc} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{array}], B = [\begin{array}{ccc} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{array}]$

$A + B = [\begin{array}{ccc} (- 1) + (- 4) & 2 + 1 & 3 + (- 5) \\ 5 + 1 & 7 + 2 & 9 + 0 \\ (- 2) + 1 & 1 + 3 & 1 + 1 \end{array}] = [\begin{array}{ccc} - 5 & 3 & - 2 \\ 6 & 9 & 9 \\ - 1 & 4 & 2 \end{array}]$

$(A + B)^{'} = [\begin{array}{ccc} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{array}]$

$A^{'} = [\begin{array}{ccc} - 1 & 5 & - 2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{array}]$

$B^{'} = [\begin{array}{ccc} - 4 & 1 & 1 \\ 1 & 2 & 3 \\ - 5 & 0 & 1 \end{array}]$

$A^{'} + B^{'} = [\begin{array}{ccc} (- 1) + (- 4) & 5 + 1 & (- 2) + 1 \\ 2 + 1 & 7 + 2 & 1 + 3 \\ 3 + (- 5) & 9 + 0 & 1 + 1 \end{array}] = [\begin{array}{ccc} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{array}]$

$(A + B)^{'} = A^{'} + B^{'}$

$A - B = [\begin{array}{ccc} (- 1) - (- 4) & 2 - 1 & 3 - (- 5) \\ 5 - 1 & 7 - 2 & 9 - 0 \\ (- 2) - 1 & 1 - 3 & 1 - 1 \end{array}] = [\begin{array}{ccc} 3 & 1 & 8 \\ 4 & 5 & 9 \\ - 3 & - 2 & 0 \end{array}]$

$(A - B)^{'} = [\begin{array}{ccc} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{array}]$

$A^{'} - B^{'} = [\begin{array}{ccc} (- 1) - (- 4) & 5 - 1 & (- 2) - 1 \\ 2 - 1 & 7 - 2 & 1 - 3 \\ 3 - (- 5) & 9 - 0 & 1 - 1 \end{array}] = [\begin{array}{ccc} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{array}]$

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

Determine $A$ from $A^{'}$

Compute $A + B$

Compute $(A + B)^{'}$

Compute $A^{'} + B^{'}$

Compute $A - B$

Compute $A^{'} - B^{'}$

Find $A$ from $A^{'}$

Compute $2 B$

Compute $A + 2 B$

Find $(A + 2 B)^{'}$

Step 1️⃣: Compute $A B$

Find $(A B)^{'}$

Compute $B^{'} A^{'}$

Compute $A B$

Find $(A B)^{'}$

Compute $B^{'} A^{'}$

Find $A^{'}$

Compute $A^{'} A$

Find $A^{'}$

Compute $A^{'} A$

(i) Show $A$ is symmetric

(ii) Show $A$ is skew-symmetric

(i) $A + A^{'}$

(ii) $A - A^{'}$

$\frac{1}{2} (A + A^{'}) and \frac{1}{2} (A - A^{'})$

Find $A^{'}$

Compute $A - A^{'}$

Take transpose of $A B - B A$

Compute $A^{'}$

Compute $A + A^{'}$

Given $A + A^{'} = I$

Solve for $α$

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 $B A$