NCERT Solutions Class 12 – Mathematics Part I – Chapter 3 Matrices – Miscellaneous Exercise on Chapter 3

Last Updated : 01 May, 2024

Question 1. If A and B are symmetric matrices, prove that AB – BA is a skew-symmetric matrix.

Solution:

As, it is mentioned that A and B are symmetric matrices,

A’ = A and B’ = B

(AB – BA)’ = (AB)’ – (BA)’ (using, (A-B)’ = A’ – B’)

= B’A’ – A’B’ (using, (AB)’ = B’A’)

= BA – AB

(AB – BA)’ = – (AB – BA)

Hence, AB – BA is a skew symmetric matrix

Question 2. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.

Solution:

Let’s take A as symmetric matrix

A’ = A

Then,

(B′AB)’ = {B'(AB)}’

= (AB)’ (B’)’ (using, (AB)’ = B’A’)

= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)

= B’A B

As, here (B′AB)’ = B’A B. It is a symmetric matrix.

Let’s take A as skew matrix

A’ = -A

Then,

(B′AB)’ = {B'(AB)}’

= (AB)’ (B’)’ (using, (AB)’ = B’A’)

= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)

= B'(-A) B

= – B’A B

As, here (B′AB)’ = -B’A B. It is a skew matrix.

Hence, we can conclude that B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Question 3. Find the values of x, y, z if the matrix $A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}$ satisfy the equation A′A = I

Solution:

$A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}$

$A' =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}^T=\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix}$

A’A = $I =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix}$

$\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix} = \begin{bmatrix} 1 & 0 &0\\ 0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}$

$\begin{bmatrix} 0+x^2+x^2 & 0+xy-xy &0-xz+xz\\ 0+xy-xy & 4y^2+y^2+y^2 &2yz-yz-yz\\ 0-zx+zx & 2yz-yz-yz &z^2+z^2+z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 &0\\ 0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}$

$\begin{bmatrix} 2x^2 & 0 &0\\ 0 & 6y^2 &0\\ 0 & 0 &3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 &0\\ 0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}$

By evaluating the values, we have

2x² = 1

x = ± $\frac{1}{\sqrt{2}}$

6y² = 1

y = ± $\frac{1}{\sqrt{6}}$

3z² = 1

z = ± $\frac{1}{\sqrt{3}}$

Question 4: For what values of x : $\begin{bmatrix} 1 & 2 &1 \end{bmatrix} \begin{bmatrix} 1 & 2 &0\\ 2 & 0 &1\\ 1 & 0 &2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0$

Solution:

$\begin{bmatrix} 1 & 2 &1 \end{bmatrix} \begin{bmatrix} 1 & 2 &0\\ 2 & 0 &1\\ 1 & 0 &2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0$

$\begin{bmatrix} 1+4+1 & 2+0+0 &0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0$

$\begin{bmatrix} 6 & 2 &4 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0$

$\begin{bmatrix} 6(0) + 2(2) +4(x) \end{bmatrix}= 0\\ \begin{bmatrix} 0 + 4 +4x \end{bmatrix}= 0\\ 4(x+1) = 0\\ x+1 = 0\\ x = -1$

Question 5: If $A =\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ , show that A² – 5A + 7I = 0.

Solution:

$A^2 = AA =\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\\ = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix}\\ = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

$5A =5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} =\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$

$7I =7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

A² – 5A + 7I = $\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+ \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\\ =\begin{bmatrix} 8-15+7 & 3-3+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}\\ = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Hence proved!

Question 6: Find x, if $\begin{bmatrix} x & -5 &-1 \end{bmatrix} \begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1\\ 2 & 0 &3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

Solution:

$\begin{bmatrix} x & -5 &-1 \end{bmatrix} \begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1\\ 2 & 0 &3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0\\ \begin{bmatrix} x+0-2 & 0-10+0 &2x-5-3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0\\ \begin{bmatrix} x-2 & -10 &2x-8 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0\\ \begin{bmatrix} x(x-2) + (-10)(4) +1(2x-8) \end{bmatrix}= 0\\ \begin{bmatrix} x^2-2x -40+2x-8 \end{bmatrix}= 0\\ \begin{bmatrix} x^2-48 \end{bmatrix}= 0\\ x^2 = 48\\ x = \pm \sqrt{48}\\ x = \pm 4\sqrt{3}$

Question 7: A manufacturer produces three products x, y, z which he sells in two markets.

Annual sales are indicated below:

Market	Products
I	10,000	2,000	18,000
II	6,000	20,000	8,000

(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Solution:

Total revenue in market I and II can be arranged from given data as follows:

$\begin{bmatrix} 10,000 & 2,000 &18,000\\6,000 & 20,000 &8,000 \end{bmatrix} \begin{bmatrix} 2.5 \\ 1.5 \\1 \end{bmatrix}$

After multiplication, we get

$\begin{bmatrix} 25,000 + 3,000 +18,000\\15,000 + 30,000 +8,000 \end{bmatrix}=\begin{bmatrix} 46,000\\53,000 \end{bmatrix}$

Hence, the total revenue in Market I and market II are ₹ 46,000 and ₹ 53,000 respectively.

(b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.

Solution:

Total cost prices of all the products in market I and market II can be arranged from given data as follows:

$\begin{bmatrix} 10,000 & 2,000 &18,000\\6,000 & 20,000 &8,000 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\0.5 \end{bmatrix}$

After multiplication, we get

$\begin{bmatrix} 20,000 + 2,000 +9,000\\12,000 + 20,000 +4,000 \end{bmatrix}=\begin{bmatrix} 31,000\\36,000 \end{bmatrix}$

As, Profit earned = Total revenue – Cost price

Profit earned $=\begin{bmatrix} 46,000\\53,000 \end{bmatrix}-\begin{bmatrix} 31,000\\36,000 \end{bmatrix}$

Profit earned = $=\begin{bmatrix} 15,000\\17,000 \end{bmatrix}$

Hence, profit earned in Market I and market II are ₹ 15,000 and ₹ 17,000 respectively. Which is equal to ₹ 32,000

Question 8. Find the matrix X so that $X\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}$

Solution:

$X\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}$

Here, the RHS is a 2×3 matrix and LHS is 2×3. So, X will be 2×2 matrix.

Let’s take X as,

$X= \begin{bmatrix} p & q \\ r & s \end{bmatrix}$

Now solving the matrix, we have

$\begin{bmatrix} p & q\\ r & s \end{bmatrix}\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}\\ \begin{bmatrix} p+4q & 2p+5qb &3p+6q\\ r+4s & 2r+5s &3r+6s \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}\\$

Equating each of them, we get

p+4q = -7 ………..(1)

2p+5q = -8 ………….(2)

3p + 6q = -9

r + 4s = 2 …………(3)

2r + 5s = 4 ……………(5)

3r + 6s = 6

Solving (1) and (2), we get

p = 1 and q = -2

Solving (3) and (4), we get

r = 2 and s = 0

Hence, matrix X is

$X= \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$

Choose the correct answer in the following questions:

Question 9: If $A= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that A² = I, then

(A) 1 + α² + βγ = 0

(B) 1 – α² + βγ = 0

(C) 1 – α² – βγ = 0

(D) 1 + α² – βγ = 0

Solution:

$A^2 = AA= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\\ = \begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \beta\gamma+\alpha^2 \end{bmatrix}$

As, A² = I

$\begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \beta\gamma+\alpha^2 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

α² + βγ = 1

1 – α² – βγ = 0

Hence, Option (C) is correct.

Question 10. If the matrix A is both symmetric and skew symmetric, then

(A) A is a diagonal matrix

(B) A is a zero matrix

(C) A is a square matrix

(D) None of these

Solution:

If the matrix A is both symmetric and skew symmetric, then

A = A’

and A = -A

Only zero matrix satisfies both the conditions.

Hence, Option (B) is correct.

Question 11. If A is square matrix such that A² = A, then (I + A)³ – 7 A is equal to

(A) A

(B) I – A

(C) I

(D) 3A

Solution:

(I + A)³ – 7 A = I³ + A³ + 3A^2 + 3AI^2 – 7A

= I3 + A3 + 3A² + 3A – 7A

= I + A³ + 3A² – 4A

As, A² = A

A3 = A²A = AA = A

So, I + A3 + 3A2 – 4A = I + A + 3A – 4A = I

Hence, Option (C) is correct.

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NCERT Solutions Class 12 – Mathematics Part I – Chapter 3 Matrices – Miscellaneous Exercise on Chapter 3

Question 1. If A and B are symmetric matrices, prove that AB – BA is a skew-symmetric matrix.

Question 2. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.

Question 3. Find the values of x, y, z if the matrix satisfy the equation A′A = I

Question 4: For what values of x :

Question 5: If , show that A2 – 5A + 7I = 0.

Question 6: Find x, if

Question 7: A manufacturer produces three products x, y, z which he sells in two markets.

Annual sales are indicated below:

(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.

Question 8. Find the matrix X so that

Choose the correct answer in the following questions:

Question 9: If is such that A² = I, then

Question 10. If the matrix A is both symmetric and skew symmetric, then

Question 11. If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to

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Question 3. Find the values of x, y, z if the matrix $A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}$ satisfy the equation A′A = I

Question 4: For what values of x : $\begin{bmatrix} 1 & 2 &1 \end{bmatrix} \begin{bmatrix} 1 & 2 &0\\ 2 & 0 &1\\ 1 & 0 &2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0$

Question 5: If $A =\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ , show that A² – 5A + 7I = 0.

Question 6: Find x, if $\begin{bmatrix} x & -5 &-1 \end{bmatrix} \begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1\\ 2 & 0 &3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

Question 8. Find the matrix X so that $X\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}$

Question 9: If $A= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that A² = I, then

Question 11. If A is square matrix such that A² = A, then (I + A)³ – 7 A is equal to