NCERT Exemplar Solutions for Class 12 Maths Chapter 3 Matrices

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Chapter 3 of the NCERT Exemplar Solutions for Class 12 Mathematics is Matrices. The topics covered in this chapter are the introduction to the matrix and its order, its types, properties, operations performed on matrices, multiplication of two matrices, symmetric and skew-symmetric matrices, transpose of the matrix, invertible matrix, etc. Students can download the solutions PDF of NCERT Exemplar Solutions for Class 12 Maths Chapter 3 Matrices from the links given below to learn the right methods of solving the exercise problems of this chapter.

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Exercise 3.3 Page No: 52

Short Answer (S.A.)

1. If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

Solution:

For a given matrix of order m x n, it has mn elements, where m and n are natural numbers.

Here we have, m x n = 28

(m, n) = {(1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1)}

So, the possible orders are 1 x 28, 2 x 14, 4 x 7, 7 x 4, 14 x 2, 28 x 1.

Also, if it has 13 elements, then m x n = 13

(m, n) = {(1, 13), (13, 1)}

Thus, the possible orders are 1 x 13, 13 x 1.

2. In the matrix A = , write :

(i) The order of the matrix A

(ii) The number of elements

(iii) Write elements a₂₃, a₃₁, a₁₂

Solution:

For the given matrix,

(i) The order of the matrix A is 3 x 3.

(ii) The number of elements of the matrix = 3 x 3 = 9

(iii) Elements: a₂₃ = x² – y, a₃₁ = 0, a₁₂ = 1

3. Construct a_{2 × 2} matrix where

(i) a_ij = (i – 2j)²/ 2

(ii) a_ij = |-2i + 3j|

Solution:

We have,

A = [a_ij]_2×2

(i) Such that, a_ij = (i – 2j)²/ 2; where 1 ≤ i ≤ 2; 1 ≤ j ≤ 2

So, the terms of the matrix are

(ii) Here, a_ij = |-2i + 3j|

So, the terms of the matrix are

4. Construct a 3 × 2 matrix whose elements are given by a_ij = e^i.xsin jx

Solution:

Let A be a 3 x 2 matrix

Such that, a_ij = e^i.xsin jx; where where 1 ≤ i ≤ 3; 1 ≤ j ≤ 2

So, the terms are given as

5. Find values of a and b if A = B, where

Solution:

Given, matrix A = matrix B

Then their corresponding elements are equal.

So, we have

a₁₁ = b₁₁; a + 4 = 2a + 2 ⇒ a = 2

a₁₂ = b₁₂; 3b = b² + 2 ⇒ b² – 3b + 2 = 0 ⇒ b = 1, 2

a₂₂ = b₂₂; -6 = b² – 5b ⇒ b² – 5b + 6 = 0 ⇒ b = 2, 3

Hence, a = 2 and b = 2 (common value)

6. If possible, find the sum of the matrices A and B, where A =
and B =

Solution:

The given two matrices, A and B, are of different orders. Two matrices can be added only if the order of both matrices is the same. Thus, the sum of matrices A and B is not possible.

7. If

(i) X + Y

(ii) 2X – 3Y

(iii) A matrix Z such that X + Y + Z is a zero matrix.

Solution:

8. Find non-zero values of x satisfying the matrix equation:

Solution:

Given,

On comparing the corresponding elements, we get

2x + 10x = 48

12x = 48

Thus, x = 4

It’s also seen that this value od x also satisfies the equation 3x + 8 = 20 and x² + 8x = 12x.

Therefore, x = 4 (common) is the solution of the given matrix equation.

9. Show that
(A + B) (A – B) ≠ A² – B²

Solution:

Hence, from (i) and (ii),

(A + B) (A – B) ≠ A² – B²

10. Find the value of x if

Solution:

x² + 16x + 28 = 0

(x + 2) (x + 14) = 0

Therefore, x = -2, -14

11. Show that satisfies the equation A² – 3A – 7I = 0 and hence find A^-1.

Solution:

Given,

Now,

A² – 3A – 7I = 0

Multiplying both sides with A^-1, we get

A^-1 [A² – 3A – 7I] = A^-1 0

A^-1. A . A – 3A^-1. A – 7A^-1. I = 0

I . A – 3I – 7A^-1 = 0 [As A^-1. A = I]

A – 3I – 7A^-1 = 0

7A^-1 = A – 3I

12. Find the matrix A satisfying the matrix equation:

Solution:

Now,

P^-1 PAQ = P^-1 I

So, IAQ = P^-1

AQ = P^-1

AQQ^-1 = P^-1 Q^-1

AI = P^-1 Q^-1

A = P^-1 Q^-1

13. Find A, if

Solution:

On comparing elements of both sides, we have

4x = -4 ⇒x = -1

4y = 8 ⇒ y = 2

And, 4z = 4 ⇒ z = 1

Therefore, A = [-1 2 1]

14. If then verify
(BA)² ≠ B² A²

Solution:

The given matrices A has order 3 x 2 and B has order 2 x 3.

So, BA is defined and will have order 3 x 3.

But, A² and B² are not defined as the orders don’t satisfy the multiplication condition.

Hence, (BA)² ≠ B² A²

15. If possible, find BA and AB where

Solution:

16. Show by an example that for A ≠ 0, B ≠ 0, AB = 0.

Solution:

– Hence Proved

17. Given Is (AB)’ = B’ A’ ?

Solution:

From (i) and (ii), we have

(AB)’ = B’ A’

18. Solve for x and y:

Solution:

Now, we have

2x + 3y – 8 = 0 ….. (1) and

x + 5y – 11 = 0 ….. (2)

On solving the equations (1) and (2), we get

x = 1 and y = 2

19. If X and Y are 2 x 2 matrices, then solve the following matrix equations for X and Y

Solution:

20. If A = [3 5], B = [7 3], then find a non-zero matrix C such that AB = AC.

Solution:

Given, A = [3 5]_1×2 and B = [7 3]_1×2

For AC = BC

We have order of C = 2 x n

21. Give an example of matrices A, B and C such that AB = AC, where A is non-zero

matrix, but B ≠ C.

Solution:

From (i) and (ii),

Hence, AB = AC but B ≠ C.

22. If

(i) (AB) C = A (BC) (ii) A (B + C) = AB + AC.

Solution:

23.

Solution:

24. If: = A, find A.

Solution:

25. If verify that

A (B + C) = (AB + AC).

Solution:

26. If then verify that A² + A = A (A + I), where I is 3 × 3 unit matrix.

Solution:

27. If then verify that:

(i) (A¢)¢ = A

(ii) (AB)¢ = B¢A¢

(iii) (kA)¢ = (kA¢).

Solution:

28. If then verify that:

(i) (2A + B)¢ = 2A¢ + B¢

(ii) (A – B)¢ = A¢ – B¢.

Solution:

29. Show that A¢A and AA¢ are both symmetric matrices for any matrix A.

Solution:

Let P = A¢A

So, P¢ = (A¢A) ¢

= A¢(A¢)¢ [As (AB) ¢ = B¢A¢]

Hence, A¢A is the symmetric matrix for any matrix A.

Now, let Q = AA¢

So, Q¢ = (AA¢)¢ = (A) ¢ = AA¢ = Q

Hence, AA¢ is the symmetric matrix for any matrix A.

30. Let A and B be square matrices of the order 3 × 3. Is (AB)² = A² B²? Give reasons.

Solution:

As, A and B be square matrices of order 3 x 3.

We have, (AB)² = AB . AB

= A(BA)B

= A(AB)B [If AB = BA]

= AABB

= A²B²

Thus, (AB)² = A²B² is true only if AB = BA.

31. Show that if A and B are square matrices such that AB = BA, then

(A + B)² = A² + 2AB + B².

Solution:

Given, A and B are square matrices such that AB = BA.

So, (A + B)² = (A + B) . (A + B)

= A² + AB + BA + B²

= A² + AB + AB + B² [Since, AB = BA]

= A² + 2AB + B²

32. Let and a = 4, b = –2.

Show that:

(a) A + (B + C) = (A + B) + C

(b) A (BC) = (AB) C

(c) (a + b)B = aB + bB

(d) a (C–A) = aC – aA

(e) (A^T)^T = A

(f) (bA)^T = b A^T

(g) (AB)^T = B^T A^T

(h) (A –B)C = AC – BC

(i) (A – B)^T = A^T – B^T

Solution:

Solution:

34. If and x² = –1, then show that (A + B)² = A² + B²

Solution:

35. Verify that A² = I when A =

Solution:

36. Prove by Mathematical Induction that (A¢)ⁿ = (Aⁿ) ¢, where n ∈ N for any square matrix A.

Solution:

Let P(n): (A¢)ⁿ = (Aⁿ) ¢

So, P(1): (A¢) = (A) ¢

A¢ = A¢

Hence, P(1) is true.

Now, let P(k) = (A¢)^k = (A^k) ¢, where k ∈ N

And,

P(k + 1): (A¢)^k+1 = (A¢)^kA¢

= (A^k) ¢A¢

= (AA^k) ¢

= (A^k+1) ¢

Hence, P(1) is true and whenever P(k) is true P(k + 1) is true.

Therefore, P(n) is true for all n ∈ N.

37. Find the inverse, by elementary row operations (if possible), of the following matrices

Solution:

As we obtain all the zeroes in a row of the matrix A on the L.H.S., A^-1 does not exist.

38. If then find values of x, y, z and w.

Solution:

In the given matrix equation,

Comparing the corresponding elements, we get

x + y = 6,

xy = 8,

z + 6 = 0 and

w = 4

From the first two equations, we have

(6 – y) . y = 8

y² – 6y + 8 = 0

(y – 2) (y – 4) = 0

y = 2 or y = 4

Hence, x = 4 and x = 2

Also, z + 6 = 0

z = -6 and w = 4

Therefore,

x = 2, y = 4 or x = 4, y = 2, z = -6 and w = 4

39. If find a matrix C such that 3A + 5B + 2C is a null matrix.

Solution:

Let’s consider a matrix C, such that

Comparing the terms,

2a + 48 = 0 ⇒ a = -24

20 + 2b = 0 ⇒ b = -10

56 + 2c = 0 ⇒ c = -28

And,

76 + 2d = 0 ⇒ d = -38

Therefore, the matrix C is

40. If , then find A² – 5A – 14I. Hence, obtain A³.

Solution:

41. Find the value of a, b, c and d, if

Solution:

Given,

Now,

3a = a + 4 ⇒ a = 2

3b = 6 + a + b

3b – b = 8

⇒ b = 4

And,

3d = 3 + 2d

⇒ d = 3

And,

3c = c + d – 1

2c = 3 – 1 = 2

⇒ c = 1

Hence,

a = 2, b = 4, c = 1 and d = 3

42. Find the matrix A such that

Solution:

Let,

Now, by equality of matrices, we get

a = 1, b = -2, c = -5

And,

2a – d = -1 ⇒ d = 2a + 1 = 3;

2b – e = -8 ⇒ e = 2(-2) + 8 = 4

2c – f = -10 ⇒ f = 2c + 10 = 0

Thus,

43. If A = find A² + 2A + 7I.

Solution:

Given,

44. If A = and A^-1 = A’, find the value of a.

Solution:

By using the equality of matrices, we get

cos² α + sin² α = 1, which is true for all real values of α.

45. If the matrix is a skew symmetric matrix, find the values of a, b and c.

Solution:

46. If P (x) = then show that

P (x) . P (y) = P (x + y) = P (y) . P (x).

Solution:

Given,

47. If A is square matrix such that A² = A, show that (I + A)³ = 7A + I.

Solution:

We know that,

A . I = I . A

So, A and I are commutative.

Thus, we can expand (I + A)³ like real numbers expansion.

So, (I + A)³ = I³ + 3I²A + 3IA² + A³

= I + 3IA + 3A² + AA² (As Iⁿ = I, n ∈ N)

= I + 3A + 3A + AA

= I + 3A + 3A + A² = I + 3A + 3A + A = I + 7A

48. If A, B are square matrices of same order and B is a skew-symmetric matrix,

show that A¢BA is skew symmetric.

Solution:

Given, A and B are square matrices such that B is a skew-symmetric matrix

So, B¢ = -B

Now, we have to prove that A¢BA is a skew-symmetric matrix.

(A¢BA) ¢ = A¢B¢ (A¢)¢ [Since, (AB) ¢ = B¢A¢]

= A¢ (-B)A

= -A¢BA

Hence, A¢BA is a skew-symmetric matrix.

Long Answer (L.A)

49. If AB = BA for any two square matrices, prove by mathematical induction that (AB)ⁿ = Aⁿ Bⁿ.

Solution:

Let P(n) : (AB)ⁿ = AⁿBⁿ

So, P(1) : (AB)¹ = A¹B¹

AB = AB

So, P(1) is true.

Let P(n) is true for some k ∈ N

Now,

Hence, P(1) is true and whenever P(k) is true P(k + 1) is true.

Thus, P(n) is true for all n ∈ N.

50. Find x, y, z if A = satisfies A¢ = A^-1.

Solution:

Matrix A is such that A¢ = A^-1

AA¢ = I