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

### Question 1: Let , show that (aI + bA)^{n} = a^{n} I + na^{n – 1} bA, where I is the identity matrix of order 2 and **n ∈ N**.

**Solution:**

Using mathematical induction,

Let’s check for n=1Step 1:(aI + bA)

^{n}= (aI + bA)^{1}= (aI + bA)a

^{n}I + na^{n – 1}bA = aI + 1a^{1 – 1}bA = (aI + bA)It is true for P(1)

Now take n=kStep 2:(aI + bA)

^{k}= a^{k}I + ka^{k – 1}bA …………………(1)

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

^{k+1}= (aI + bA)^{k}(aI + bA)= (a

^{k}I + ka^{k – 1}bA) (aI + bA)= a

^{k+1}I×I + ka^{k}bAI + a^{k}bAI + ka^{k-1}b^{2}AAAA =

= a

^{k+1}I×I + ka^{k}bAI + a^{k}bAI + 0= a

^{k+1}I + (k+1)a^{k+1-1}bA= P(k+1)

Hence, P(n) is true.

### Question 2: If , prove that

**Solution:**

Using mathematical induction,

: Let’s check for n=1Step 1It is true for P(1)

: Now take n=kStep 2

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

Hence, P(n) is true.

### Question 3: If , prove that ,where n is any positive integer.

**Solution:**

Using mathematical induction,

Step 1: Let’s check for n=1

It is true for P(1)

Step 2: Now take n=k

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

= P(k+1)

Hence, P(n) is true.

### Question 4. 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 5. 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 matrixA’ = 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 matrixA’ = -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 6. Find the values of x, y, z if the matrix satisfy the equation A′A = I

**Solution:**

A’A =

By evaluating the values, we have

2x

^{2}= 1x = ±

6y

^{2}= 1y = ±

3z

^{2}= 1z = ±

### Question 7: For what values of x :

**Solution:**

### Question 8: If , show that A^{2} – 5A + 7I = 0.

**Solution:**

A

^{2}– 5A + 7I =Hence proved!

### Question 9: Find x, if

**Solution:**

### Question 10: 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:

After multiplication, we get

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:

After multiplication, we get

As, Profit earned = Total revenue – Cost price

Profit earned

Profit earned =

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

### Question 11. Find the matrix X so that

**Solution:**

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,

Now solving the matrix, we have

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

### Question 12: If A and B are square matrices of the same order such that AB = BA, then prove by induction that AB^{n} = B^{n}A. Further, prove that (AB)^{n} = A^{n}B^{n} for all n ∈ N.

**Solution:**

Using mathematical induction,

Let’s check for n=1Step 1:AB

^{n}= AB^{1}= ABB

^{n}A = B^{1}A = BAIt is true for P(1)

Now take n=kStep 2:AB

^{k}= B^{k}A

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

^{(k+1)}= AB^{k}B= B

^{k}AB= B

^{k+1}A= P(k+1)

Hence, P(n) is true.

Now, for

(AB)^{n}= A^{n}B^{n}Using mathematical induction,

: Let’s check for n=1Step 1(AB)

^{1}= ABB

^{1}A^{1}= BAIt is true for P(1)

: Now take n=kStep 2(AB)

^{k}= A^{k}B^{k}

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

^{(k+1)}= (AB)^{k}(AB)= A

^{k}B^{k}AB= A

^{k+1}B^{k+1}= (AB)

^{k+1}= P(k+1)

Hence, P(n) is true.

### Choose the correct answer in the following questions:

### Question 13: If 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:**

As, A

^{2}= Iα² + βγ = 1

1 – α² – βγ = 0

Hence, Option (C) is correct.

### Question 14. 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 15. If A is square matrix such that A^{2} = 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 = I

^{3}+ A^{3}+ 3A^2 + 3AI^2 – 7A= I3 + A3 + 3A

^{2}+ 3A – 7A= I + A

^{3}+ 3A^{2}– 4AAs, A

^{2}= AA3 = A

^{2}A = AA = ASo, I + A3 + 3A2 – 4A = I + A + 3A – 4A = I

Hence, Option (C) is correct.