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

Last Updated : 03 Apr, 2024

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

Solution:

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

A’ = A and B’ = B

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

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

= BA – AB

(AB â€“ BA)’ = – (AB – BA)

Hence, AB â€“ BA is a skew symmetric matrix

### Question 2. Show that the matrix Bâ€²AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.

Solution:

Let’s take A as symmetric matrix

A’ = A

Then,

(Bâ€²AB)’ = {B'(AB)}’

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

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

= B’A B

As, here (Bâ€²AB)’ = B’A B. It is a symmetric matrix.

Let’s take A as skew matrix

A’ = -A

Then,

(Bâ€²AB)’ = {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 (Bâ€²AB)’ = -B’A B. It is a skew matrix.

Hence, we can conclude that Bâ€²AB 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 [Tex]A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}  [/Tex] satisfy the equation Aâ€²A = I

Solution:

[Tex]A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix} [/Tex]

[Tex]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} [/Tex]

A’A = [Tex]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} [/Tex]

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

By evaluating the values, we have

2x2 = 1

x = Â± [Tex]\frac{1}{\sqrt{2}}[/Tex]

6y2 = 1

y = Â± [Tex]\frac{1}{\sqrt{6}}[/Tex]

3z2 = 1

z = Â± [Tex]\frac{1}{\sqrt{3}}[/Tex]

### Question 4: For what values of x : [Tex]\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[/Tex]

Solution:

[Tex]\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[/Tex]

[Tex]\begin{bmatrix} 1+4+1 & 2+0+0 &0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0[/Tex]

[Tex]\begin{bmatrix} 6 & 2 &4 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0[/Tex]

[Tex]\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[/Tex]

### Question 5: If [Tex]A =\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}  [/Tex], show that A2 â€“ 5A + 7I = 0.

Solution:

[Tex]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}[/Tex]

[Tex]5A =5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} =\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} [/Tex]

[Tex]7I =7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}[/Tex]

A2 â€“ 5A + 7I = [Tex]\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}[/Tex]

Hence proved!

### Question 6: Find x, if [Tex]\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[/Tex]

Solution:

[Tex]\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}[/Tex]

### (a) If unit sale prices of x, y and z are â‚¹ 2.50, â‚¹ 1.50 and â‚¹ 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:

[Tex]\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}[/Tex]

After multiplication, we get

[Tex]\begin{bmatrix} 25,000 + 3,000 +18,000\\15,000 + 30,000 +8,000 \end{bmatrix}=\begin{bmatrix} 46,000\\53,000 \end{bmatrix}[/Tex]

Hence, the total revenue in Market I and market II are â‚¹ 46,000 and â‚¹ 53,000 respectively.

### (b) If the unit costs of the above three commodities are â‚¹ 2.00, â‚¹ 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:

[Tex]\begin{bmatrix} 10,000 & 2,000 &18,000\\6,000 & 20,000 &8,000 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\0.5 \end{bmatrix}[/Tex]

After multiplication, we get

[Tex]\begin{bmatrix} 20,000 + 2,000 +9,000\\12,000 + 20,000 +4,000 \end{bmatrix}=\begin{bmatrix} 31,000\\36,000 \end{bmatrix}[/Tex]

As, Profit earned = Total revenue – Cost price

Profit earned [Tex]=\begin{bmatrix} 46,000\\53,000 \end{bmatrix}-\begin{bmatrix} 31,000\\36,000 \end{bmatrix}[/Tex]

Profit earned = [Tex]=\begin{bmatrix} 15,000\\17,000 \end{bmatrix}[/Tex]

Hence, profit earned in Market I and market II are â‚¹ 15,000 and â‚¹ 17,000 respectively. Which is equal to â‚¹ 32,000

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

Solution:

[Tex]X\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}[/Tex]

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

Let’s take X as,

[Tex]X= \begin{bmatrix} p & q \\ r & s \end{bmatrix}[/Tex]

Now solving the matrix, we have

[Tex]\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}\\[/Tex]

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

[Tex]X= \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}[/Tex]

### Question 9: If [Tex]A= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}  [/Tex] is such that AÂ² = I, then

(A) 1 + Î±Â² + Î²Î³ = 0

(B) 1 â€“ Î±Â² + Î²Î³ = 0

(C) 1 â€“ Î±Â² â€“ Î²Î³ = 0

(D) 1 + Î±Â² â€“ Î²Î³ = 0

Solution:

[Tex]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} [/Tex]

As, A2 = I

[Tex]\begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \beta\gamma+\alpha^2 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}[/Tex]

Î±Â² + Î²Î³ = 1

1 – Î±Â² – Î²Î³ = 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)Â³ â€“ 7 A is equal to

(A) A

(B) I â€“ A

(C) I

(D) 3A

Solution:

(I + A)Â³ â€“ 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.

## Deleted Questions

### Let [Tex]A =\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}  [/Tex], show that (aI + bA)n = an I + nan â€“ 1 bA, where I is the identity matrix of order 2 and n âˆˆ N.

Solution:

Using mathematical induction,

Step 1: Let’s check for n=1

(aI + bA)n = (aI + bA)1 = (aI + bA)

anI + nan â€“ 1 bA = aI + 1a1 â€“ 1 bA = (aI + bA)

It is true for P(1)

Step 2: Now take n=k

(aI + bA)k = akI + kak â€“ 1 bA …………………(1)

Step 3: Let’s check whether, its true for n = k+1

(aI + bA)k+1 = (aI + bA)k (aI + bA)

= (akI + kak â€“ 1 bA) (aI + bA)

= ak+1IÃ—I + kak bAI + ak bAI + kak-1 b2AA

AA = [Tex]\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} = 0 [/Tex]

= ak+1IÃ—I + kak bAI + ak bAI + 0

= ak+1I + (k+1)ak+1-1 bA

= P(k+1)

Hence, P(n) is true.

### If [Tex]A =\begin{bmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{bmatrix}  [/Tex], prove that [Tex]A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1}\\ 3^{n-1} & 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} ,n\in N[/Tex]

Solution:

Using mathematical induction,

Step 1: Let’s check for n=1

[Tex]A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1}\\ 3^{1-1} & 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1} & 3^{1-1} \end{bmatrix}=\begin{bmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{bmatrix}[/Tex]

It is true for P(1)

Step 2: Now take n=k

[Tex]A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1} \end{bmatrix}[/Tex]

Step 3: Let’s check whether, its true for n = k+1

[Tex]A^{k+1}=A^kA=\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1} \end{bmatrix} \begin{bmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{bmatrix}\\ A^{k+1} = \begin{bmatrix} 3^k & 3^k &3^k\\ 3^k & 3^k &3^k\\ 3^k & 3^k &3^k \end{bmatrix}\\ A^{k+1} = \begin{bmatrix} 3^{k+1-1} & 3^{k+1-1} &3^{k+1-1}\\ 3^{k+1-1} & 3^{k+1-1} &3^{k+1-1}\\ 3^{k+1-1} & 3^{k+1-1} &3^{k+1-1} \end{bmatrix}[/Tex]

= P(k+1)

Hence, P(n) is true.

### If [Tex]A =\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}[/Tex], prove that [Tex]A^n =\begin{bmatrix} 1+2n & -4n\\ n & 1-2n \end{bmatrix}  [/Tex] ,where n is any positive integer.

Solution:

Using mathematical induction,

Step 1: Let’s check for n=1

[Tex]A^1 = A =\begin{bmatrix} 1+2(1) & -4(1)\\ n & 1-2(1) \end{bmatrix}=\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}  [/Tex]

It is true for P(1)

Step 2: Now take n=k

[Tex]A^k =\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}[/Tex]

Step 3: Let’s check whether, its true for n = k+1

[Tex]A^{k+1} = A^kA =\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}\\ = \begin{bmatrix} 3(1+2k)+1(-4k) & -4(1+2k)+(-1)(-4k)\\ 3k+1(1-2k) & (-4)(k)+(-1)(1-2k) \end{bmatrix}\\ = \begin{bmatrix} 3+6k-4k & -4-8k+4k\\ 3k+1-2k & -4k-1+2k \end{bmatrix}\\ = \begin{bmatrix} 3+2k & -4-4k\\ k+1 & -2k-1 \end{bmatrix}\\ = \begin{bmatrix} 1+2(k+1) & -4(k+1)\\ k+1 & 1-2(k+1) \end{bmatrix}\\[/Tex]

= P(k+1)

Hence, P(n) is true.

### If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n âˆˆ N.

Solution:

Using mathematical induction,

Step 1: Let’s check for n=1

ABn = AB1 = AB

BnA = B1A = BA

It is true for P(1)

Step 2: Now take n=k

ABk = BkA

Step 3: Let’s check whether, its true for n = k+1

AB(k+1) = ABkB

= BkAB

= Bk+1 A

= P(k+1)

Hence, P(n) is true.

Now, for (AB)n = AnBn

Using mathematical induction,

Step 1: Let’s check for n=1

(AB)1 = AB

B1A1 = BA

It is true for P(1)

Step 2: Now take n=k

(AB)k = AkBk

Step 3: Let’s check whether, its true for n = k+1

(AB)(k+1) = (AB)k(AB)

= AkBk AB

= Ak+1 Bk+1

= (AB)k+1

= P(k+1)

Hence, P(n) is true.

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