NCERT Exemplar Solutions for Class 12 Maths Chapter 3 Matrices

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

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Access answers to NCERT Exemplar Solutions For Class 12 Maths Chapter 3 Matrices

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.

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 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 a2 × 2 matrix where

(i) aij = (i – 2j)2/ 2

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

Solution:

We have,

A = [aij]2×2

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

So, the terms of the matrix are

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 2

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

So, the terms of the matrix are

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 3

4. Construct a 3 × 2 matrix whose elements are given by aij = ei.xsin jx

Solution:

Let A be a 3 x 2 matrix

Such that, aij = ei.xsin jx; where where 1 ≤ i ≤ 3; 1 ≤ j ≤ 2

So, the terms are given as

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 4

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 5

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 =
NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 6 and B =
NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 7

Solution:

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 8

7. If

(i) X + Y

(ii) 2X – 3Y

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 9

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 10

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 11

Solution:

Given,

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 12

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.

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 13

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 14

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 15

Hence, from (i) and (ii),

(A + B) (A – B) ≠ A2 – B2

10. Find the value of x if

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 16

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 17

[16 + 2x + 10x + 12 + x2 + 4x] = 0

[x2 + 16x + 28] = 0

x2 + 16x + 28 = 0

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

Therefore, x = -2, -14

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 18

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

Solution:

Given,

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 19

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 20

12. Find the matrix A satisfying the matrix equation:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 21

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 22

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 23

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 24

13. Find A, if

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 25

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]

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 26

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 27

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 28

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 29

– Hence Proved

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 30

17. Given Is (AB)’ = B’ A’ ?

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 31

From (i) and (ii), we have

(AB)’ = B’ A’

18. Solve for x and y:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 32

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 33

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 34

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 35

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 36

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 37

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 38

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

matrix, but B ≠ C.

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 39

From (i) and (ii),

Hence, AB = AC but B ≠ C.

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 40

22. If

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 41

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 42

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 43

23.

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 44

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 45

24. If: = A, find A.

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 46

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 47

25. If verify that

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 48

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 49

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 50

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 51

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 52

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 53

27. If then verify that:

(i) (A¢)¢ = A

(ii) (AB)¢ = B¢A¢

(iii) (kA)¢ = (kA¢).

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 54

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 55

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 56

28. If then verify that:

(i) (2A + B)¢ = 2A¢ + B¢

(ii) (A – B)¢ = A¢ – B¢.

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 57

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 58

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 symmetric matrix for any matrix A.

Now, let Q = AA¢

So, Q¢ = (AA¢)¢ = (A) ¢ = AA¢ = Q

Hence, AA¢ is 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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 59

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) (AT)T = A

(f) (bA)T = b AT

(g) (AB)T = BT AT

(h) (A –B)C = AC – BC

(i) (A – B)T = AT – BT

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 60

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 61

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 62

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 63

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 64

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 65

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 66

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 67

35. Verify that A2 = I when A =

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 68

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 = (Ak) ¢, where k N

And,

P(k + 1): (A¢)k+1 = (A¢)k

= (Ak) ¢A¢

= (AAk) ¢

= (Ak+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 inverse, by elementary row operations (if possible), of the following matrices

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 69

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 70

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 7

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 71

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 72

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

Solution:

Let’s consider a matrix C, such that

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 73

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 74

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 75

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 76

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 77

Solution:

Given,

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 78

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 79

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 80

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,

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 81

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 82

43. If A = find A2 + 2A + 7I.

Solution:

Given,

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 83

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 84

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 85

By using equality of matrices, we get

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

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 86

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

Solution:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 87

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 88

46. If P (x) = then show that

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

Solution:

Given,

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 89

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 90

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,

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 91

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.

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 92

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

Solution:

Matrix A is such that A¢ = A-1

AA¢ = I

NCERT Exemplar Solutions Class 12 Mathematics Chapter 3 - 93

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