NCERT Solutions Class 12 Maths (matrices) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-3 (matrices) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

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

Solution:

Using mathematical induction,

Step 1: Let’s check for n=1

(aI + bA)ⁿ = (aI + bA)¹ = (aI + bA)

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

It is true for P(1)

Step 2: Now take n=k

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

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

(aI + bA)^k+1 = (aI + bA)^k (aI + bA)

= (a^kI + ka^{k – 1} bA) (aI + bA)

= a^k+1I×I + ka^k bAI + a^k bAI + ka^k-1 b²AA

AA = 

= a^k+1I×I + ka^k bAI + a^k bAI + 0

= a^k+1I + (k+1)a^k+1-1 bA

= P(k+1)

Hence, P(n) is true.

Question 2: If , prove that 

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 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 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 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

2x2 = 1

x = ± 

6y2 = 1

y = ± 

3z2 = 1

z = ± 

Question 7: For what values of x: 

Solution:

Question 8: If ,show that A² – 5A + 7I = 0.

Solution:

A2 – 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 ABⁿ = BⁿA. Further, prove that (AB)ⁿ = AⁿBⁿ for all n ∈ 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

AB^k = B^kA

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

AB^(k+1) = AB^kB

= B^kAB

= B^k+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 = A^kB^k

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

(AB)^(k+1) = (AB)^k(AB)

= A^kB^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, A2 = I

α² + 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 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.