The NCERT Exemplar textbooks play a vital role in providing complete and advanced knowledge on concepts of various subjects in the NCERT textbooks for various classes. Students can access the NCERT Exemplar Solutions available subject-wise for getting guidance to learn and solve questions from any chapter. These solutions are created to meet the needs of all students at various levels. Further, all solutions are developed by subject experts, following the latest CBSE patterns, which help to score high marks in the examinations.

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 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.

## Download the PDF of NCERT Exemplar Solutions For Class 12 Maths Chapter 3 Matrices

### Access Answers to NCERT Exemplar 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.

**2. In the matrix A = , write :**

**(i) The order of the matrix A **

**(ii) The number of elements **

**(iii) Write elements a _{23}, a_{31}, a_{12}**

**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: a_{23} = x^{2} â€“ y, a_{31} = 0, a_{12} = 1

**3. Construct a _{2 Ã— 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 â‰¤ i â‰¤ 2; 1 â‰¤ j â‰¤ 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 Ã— 2 matrix whose elements are given by a _{ij} = e^{i.x}sin jx**

**Solution: **

Let A be a 3 x 2 matrix

Such that, a_{ij} = e^{i.x}sin jx; where where 1 â‰¤ i â‰¤ 3; 1 â‰¤ j â‰¤ 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

a_{11} = b_{11}; a + 4 = 2a + 2 â‡’ a = 2

a_{12} = b_{12}; 3b = b^{2} + 2 â‡’ b^{2} â€“ 3b + 2 = 0 â‡’ b = 1, 2

a_{22} = b_{22}; -6 = b^{2} â€“ 5b â‡’ b^{2} â€“ 5b + 6 = 0 â‡’ 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 order of both the matrices is same. Thus, the sum of matrices A and B is not possible.

**7. If **

**(i) X + Y **

**(ii) 2X â€“ 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

Itâ€™s also seen that this value od x also satisfies the equation 3x + 8 = 20 and x^{2} + 8x = 12x.

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

**9. If show that **

**(A + B) (A – B) â‰ A ^{2} â€“ B^{2}**

**Solution: **

Hence, from (i) and (ii),

(A + B) (A – B) â‰ A^{2} â€“ B^{2}

**10. Find the value of x if **

**Solution: **

[16 + 2x + 10x + 12 + x

^{2}+ 4x] = 0 [x

^{2}+ 16x + 28] = 0

x^{2} + 16x + 28 = 0

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

Therefore, x = -2, -14

**11. Show that satisfies the equation A ^{2} â€“ 3A â€“ 7I = 0 and hence find A^{-1}.**

**Solution: **

Given,

Now,

A2 â€“ 3A â€“ 7I = 0

Multiplying both sides with A^{-1}, we get

A^{-1} [A^{2} â€“ 3A â€“ 7I] = A^{-1} 0

A^{-1}. A . A â€“ 3A^{-1}. A â€“ 7A^{-1}. I = 0

I . A â€“ 3I â€“ 7A^{-1} = 0 [As A^{-1}. A = I]

A â€“ 3I â€“ 7A^{-1} = 0

7A^{-1} = A â€“ 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 â‡’x = -1

4y = 8 â‡’ y = 2

And, 4z = 4 â‡’ z = 1

Therefore, A = [-1 2 1]

**14. If then verify **

**(BA) ^{2} â‰ B^{2} A^{2}**

**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, A^{2} and B^{2} are not defined as the orders donâ€™t satisfy the multiplication condition.

Hence, (BA)^{2} â‰ B^{2} A^{2}

**15. If possible, find BA and AB where**

**Solution: **

**16. Show by an example that for A â‰ 0, B â‰ 0, AB = 0. **

**Solution: **

– Hence Proved

**17. Given Is (AB)â€™ = Bâ€™ Aâ€™ ?**

**Solution: **

From (i) and (ii), we have

(AB)â€™ = Bâ€™ Aâ€™

**18. Solve for x and y: **

** Solution: **

Now, we have

2x + 3y â€“ 8 = 0 â€¦.. (1) and

x + 5y â€“ 11 = 0 â€¦.. (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 â‰ C.**

**Solution: **

From (i) and (ii),

Hence, AB = AC but B â‰ 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 A ^{2} + A = A (A + I), where I is 3 Ã— 3 unit matrix.**

**Solution: **

**27. If then verify that:**

**(i) (AÂ¢)Â¢ = A**

**(ii) (AB)Â¢ = BÂ¢AÂ¢**

**(iii) ( kA)Â¢ = (kAÂ¢).**

**Solution: **

**28. If then verify that:**

**(i) (2A + B)Â¢ = 2AÂ¢ + BÂ¢**

**(ii) (A â€“ B)Â¢ = AÂ¢ â€“ BÂ¢. **

**Solution: **

**29. Show that AÂ¢A and AAÂ¢ are both symmetric matrices for any matrix A.**

**Solution:**

Let P = AÂ¢A

So, PÂ¢ = (AÂ¢A) Â¢

= AÂ¢(AÂ¢)Â¢ [As (AB) Â¢ = BÂ¢AÂ¢]

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

Now, let Q = AAÂ¢

So, QÂ¢ = (AAÂ¢)Â¢ = (A) Â¢ = AAÂ¢ = Q

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

**30. Let A and B be square matrices of the order 3 Ã— 3. Is (AB) ^{2} = A^{2} B^{2} ? 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

= A^{2}B^{2}

Thus, (AB)^{2} = A^{2}B^{2} is true only if AB = BA.

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

**(A + B) ^{2} = A^{2} + 2AB + B^{2}.**

**Solution: **

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

So, (A + B)^{2} = (A + B) . (A + B)

= A^{2} + AB + BA + B^{2}

= A^{2} + AB + AB + B^{2} [Since, AB = BA]

= A^{2 }+ 2AB + B^{2}

**32. Let and a = 4, b = â€“2. **

**Show that:**

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

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

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

**(d) a (Câ€“A) = aC â€“ aA**

**(e) (A ^{T})^{T} = A**

**(f) ( bA)^{T} = b A^{T}**

**(g) (AB) ^{T} = B^{T} A^{T}**

**(h) (A â€“B)C = AC â€“ BC**

**(i) (A â€“ B) ^{T} = A^{T} â€“ B^{T}**

**Solution: **

**Solution: **

**34. If and x^{2} = â€“1, then show that (A + B)^{2} = A^{2} + B^{2} **

**Solution: **

**35. Verify that A ^{2} = I when A = **

**Solution: **

**36. Prove by Mathematical Induction that (A**Â¢**)^{n} = (A^{n})** Â¢

**, where**

*n âˆˆ*N for any square matrix A.**Solution: **

Let P(n): (AÂ¢)^{n} = (A^{n}) Â¢

So, P(1): (AÂ¢) = (A) Â¢

AÂ¢ = AÂ¢

Hence, P(1) is true.

Now, let P(k) = (AÂ¢)^{k} = (A^{k}) Â¢, where k *âˆˆ *N

And,

P(k + 1): (AÂ¢)^{k+1} = (AÂ¢)^{k}AÂ¢

= (A^{k}) Â¢AÂ¢

= (AA^{k}) Â¢

= (A^{k+1}) Â¢

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

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

**37. Find 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 â€“ 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: **

Letâ€™s consider a matrix C, such that

**Comparing the terms, **

**2a + 48 = 0 â‡’ a = -24**

**20 + 2b = 0 â‡’ b = -10**

**56 + 2c = 0 â‡’ c = -28 **

**And, **

**76 + 2d = 0 â‡’ d = -38**

**Therefore, the matrix C is **

**40. If , then find A ^{2} â€“ 5A â€“ 14I. Hence, obtain A^{3}. **

**Solution: **

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

**Solution: **

Given,

Now,

3a = a + 4 â‡’ a = 2

3b = 6 + a + b

3b â€“ b = 8

â‡’ b = 4

And,

3d = 3 + 2d

â‡’ d = 3

And,

3c = c + d â€“ 1

2c = 3 â€“ 1 = 2

â‡’ 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 â€“ d = -1 â‡’ d = 2a + 1 = 3;

2b â€“ e = -8 â‡’ e = 2(-2) + 8 = 4

2c â€“ f = -10 â‡’ f = 2c + 10 = 0

Thus,

**43. If A = find A ^{2} + 2A + 7I. **

**Solution: **

Given,

**44. If A = and A ^{-1} = Aâ€™, find value of a.**

**Solution: **

By using equality of matrices, we get

cos^{2} Î± + sin^{2 }Î± = 1, which is true for all real values of Î±.

**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 A ^{2} = 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} = I^{3 }+ 3I^{2}A + 3IA^{2} + A^{3}

= I + 3IA + 3A^{2 }+ AA^{2} (As I^{n }= I, n âˆˆ N)

= I + 3A + 3A + AA

= I + 3A + 3A + A^{2} = 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 AÂ¢BA is skew symmetric.**

**Solution: **

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

So, BÂ¢ = -B

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

(AÂ¢BA) Â¢ = AÂ¢BÂ¢ (AÂ¢)Â¢ [Since, (AB) Â¢ = BÂ¢AÂ¢]

= AÂ¢ (-B)A

= -AÂ¢BA

Hence, AÂ¢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} = A^{n} B^{n}.**

**Solution: **

Let P(n) : (AB)^{n} = A^{n}B^{n}

So, P(1) : (AB)^{1} = A^{1}B^{1}

AB = AB

So, P(1) is true.

Let P(n) is true for some k âˆˆ 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 âˆˆ N.

**50. Find x, y, z if A = satisfies AÂ¢ = A ^{-1}. **

**Solution: **

Matrix A is such that AÂ¢ = A^{-1}

AAÂ¢ = I