# 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 + nan – 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 + nan – 1 bA = aI + 1a1 – 1 bA = (aI + bA)

It is true for P(1)

Step 2: Now take n=k

(aI + bA)k = akI + kak – 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 – 1 bA) (aI + bA)

= ak+1I×I + kak bAI + ak bAI + kak-1 b2AA

AA = = ak+1I×I + kak bAI + ak bAI + 0

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

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.

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.