NCERT Solutions for Class 12 Maths Chapter 3: Matrices (NCERT 2026–27)

These Class 12 Maths Chapter 3 solutions cover Matrices from the NCERT textbook (Part I, Reprint 2026–27). Every question of Exercise 3.1, 3.2, 3.3 and 3.4 and the complete Miscellaneous Exercise is solved step by step, with each matrix shown as a clear grid and each answer cross-checked against the textbook answer key. Use these solutions to master order of a matrix, addition, scalar multiplication, matrix multiplication, transpose, and symmetric & skew-symmetric matrices.

Class: 12 Subject: Mathematics Book: Mathematics (Part I) Chapter: 3 – Matrices Exercises: 3.1, 3.2, 3.3, 3.4, Miscellaneous Session: 2026–27
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Chapter 3 Overview

Chapter 3, Matrices, introduces one of the most powerful tools in mathematics: an ordered rectangular array of numbers or functions. The chapter begins with the order of a matrix and different types of matrices (row, column, square, diagonal, scalar, identity and zero), then defines equality of matrices. It develops the algebra of matrices — addition, multiplication by a scalar and multiplication of matrices (which is associative and distributive but not commutative) — followed by the transpose of a matrix and its properties. The chapter closes with symmetric and skew-symmetric matrices, the result that every square matrix is the sum of a symmetric and a skew-symmetric matrix, and an introduction to invertible matrices. The Class 12 Maths Chapter 3 solutions below work through every exercise question in order.

Key Concepts & Definitions

Matrix: an ordered rectangular array of numbers or functions. A matrix with m rows and n columns has order m × n and mn elements; the (i, j)th element aij lies in row i and column j.

Types: a row matrix has one row; a column matrix has one column; a square matrix has m = n; a diagonal matrix has aij = 0 for i ≠ j; a scalar matrix is a diagonal matrix with equal diagonal entries; an identity matrix I has 1’s on the diagonal and 0’s elsewhere; a zero matrix O has all entries 0.

Equality: A = B if they have the same order and aij = bij for all i, j.

Operations: matrices of the same order are added entry-wise; kA multiplies every entry by k. The product AB is defined only when the number of columns of A equals the number of rows of B; matrix multiplication is associative and distributive but not commutative.

Transpose: A′ (or AT) is obtained by interchanging the rows and columns of A.

Symmetric / skew-symmetric: A is symmetric if A′ = A, and skew-symmetric if A′ = −A (all diagonal entries of a skew-symmetric matrix are 0).

Invertible matrix: a square matrix A is invertible if there exists B with AB = BA = I; then B = A−1, and the inverse, if it exists, is unique.

Important Formulas (Chapter 3)

Order & elements: an m × n matrix has mn elements; for a given number of elements, list every factor pair to get the possible orders.

Addition & scalar: (A + B)ij = aij + bij;   (kA)ij = k·aij;   A + B = B + A;   k(A + B) = kA + kB.

Product: if A = [aij]m×n and B = [bjk]n×p, then (AB)ik = Σj aijbjk, of order m × p.

Properties of multiplication: A(BC) = (AB)C;   A(B + C) = AB + AC;   IA = AI = A.

Transpose laws: (A′)′ = A;   (kA)′ = kA′;   (A + B)′ = A′ + B′;   (AB)′ = B′A′.

Symmetric/skew split: any square matrix A = ½(A + A′) + ½(A − A′), where ½(A + A′) is symmetric and ½(A − A′) is skew-symmetric.

Inverse: (AB)−1 = B−1A−1 for invertible A, B of the same order.

Exercise 3.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; solutions are original and verified against the answers given at the back of the book.

1. In the matrix A =

2519−7
35−25⁄212
√31−517
, write: (i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24, a23.

SOLUTION (i) The matrix has 3 rows and 4 columns, so its order is 3 × 4. (ii) Number of elements = 3 × 4 = 12. (iii) Reading row i, column j: a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 = 5⁄2.

2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

SOLUTION Possible orders are the factor pairs of 24 (m × n with mn = 24). For 24 elements: 1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, 24 × 1. For 13 elements (13 is prime): 1 × 13, 13 × 1.

3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

SOLUTION For 18 elements: 1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1. For 5 elements (5 is prime): 1 × 5, 5 × 1.

4. Construct a 2 × 2 matrix, A = [aij], whose elements are given by: (i) aij = (i + j)2⁄2   (ii) aij = i⁄j   (iii) aij = (i + 2j)2⁄2

SOLUTION (i) a11 = 4⁄2 = 2, a12 = 9⁄2, a21 = 9⁄2, a22 = 16⁄2 = 8 ⇒ A =
29⁄2
9⁄28
. (ii) a11 = 1, a12 = 1⁄2, a21 = 2, a22 = 1 ⇒ A =
11⁄2
21
. (iii) a11 = 9⁄2, a12 = 25⁄2, a21 = 8, a22 = 18 ⇒ A =
9⁄225⁄2
818
.

5. Construct a 3 × 4 matrix, whose elements are given by: (i) aij = ½|−3i + j|   (ii) aij = 2i − j

SOLUTION (i) With aij = ½|−3i + j|, i = 1,2,3 and j = 1,2,3,4: A =
11⁄201⁄2
5⁄223⁄21
47⁄235⁄2
. (ii) With aij = 2i − j: A =
10−1−2
3210
5432
.

6. Find the values of x, y and z from the following equations: (i)

43
x5
=
yz
15
 (ii)
x + y2
5 + zxy
 =
62
58
 (iii)
x + y + z
x + z
y + z
 =
9
5
7

SOLUTION (i) Comparing entries: y = 4, z = 3, x = 1 ⇒ x = 1, y = 4, z = 3. (ii) x + y = 6, xy = 8, 5 + z = 5 ⇒ z = 0; and x + y = 6, xy = 8 give x, y as roots of t2 − 6t + 8 = 0, i.e. t = 2, 4. So x = 4, y = 2, z = 0 or x = 2, y = 4, z = 0. (iii) x + y + z = 9, x + z = 5, y + z = 7. From the first two, y = 9 − 5 = 4; then z = 7 − 4 = 3 and x = 9 − 4 − 3 = 2 ⇒ x = 2, y = 4, z = 3.

7. Find the value of a, b, c and d from the equation:

a − b2a + c
2a − b3c + d
=
−15
013

SOLUTION Equating entries: a − b = −1, 2a − b = 0, 2a + c = 5, 3c + d = 13. Subtracting the first two: (2a − b) − (a − b) = 0 − (−1) ⇒ a = 1, then b = a + 1 = 2. c = 5 − 2a = 3; d = 13 − 3c = 13 − 9 = 4 ⇒ a = 1, b = 2, c = 3, d = 4.

8. A = [aij]m×n is a square matrix, if (A) m < n   (B) m > n   (C) m = n   (D) None of these

SOLUTION A matrix is square when the number of rows equals the number of columns, i.e. m = n. ∴ the correct option is (C).

9. Which of the given values of x and y make the following pair of matrices equal

3x + 75
y + 12 − 3x
,
0y − 2
84
 (A) x = −1⁄3, y = 7   (B) Not possible to find   (C) y = 7, x = −2⁄3   (D) x = −1⁄3, y = −2⁄3

SOLUTION Equating entries: 3x + 7 = 0 ⇒ x = −7⁄3; but 5 = y − 2 ⇒ y = 7, while 2 − 3x = 4 ⇒ x = −2⁄3. The value of x from the (1,1) entry (−7⁄3) disagrees with that from the (2,2) entry (−2⁄3). Since x cannot take two different values, the matrices can never be equal. ∴ the correct option is (B) Not possible to find.

10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (A) 27   (B) 18   (C) 81   (D) 512

SOLUTION A 3 × 3 matrix has 9 entries, each chosen in 2 ways (0 or 1), so the total is 29 = 512. ∴ the correct option is (D) 512.

Exercise 3.2 Solutions

1. Let A =

24
32
, B =
13
−25
, C =
−25
34
. Find each of the following: (i) A + B   (ii) A − B   (iii) 3A − C   (iv) AB   (v) BA

SOLUTION (i) A + B =
37
17
. (ii) A − B =
11
5−3
. (iii) 3A =
612
96
, so 3A − C =
87
62
. (iv) AB =
2(1)+4(−2)2(3)+4(5)
3(1)+2(−2)3(3)+2(5)
 =
−626
−119
. (v) BA =
1(2)+3(3)1(4)+3(2)
−2(2)+5(3)−2(4)+5(2)
 =
1110
112
.

2. Compute the following: (i)

ab
−ba
+
ab
ba
 (ii)
a2 + b2b2 + c2
a2 + c2a2 + b2
 +
2ab2bc
−2ac−2ab
 (iii)
−14−6
8516
285
 +
1276
805
324
 (iv)
cos2xsin2x
sin2xcos2x
 +
sin2xcos2x
cos2xsin2x

SOLUTION (i) Adding entry-wise:
2a2b
02a
. (ii) (a2 + b2) + 2ab = (a + b)2; (b2 + c2) + 2bc = (b + c)2; (a2 + c2) − 2ac = (a − c)2; (a2 + b2) − 2ab = (a − b)2. Result:
(a + b)2(b + c)2
(a − c)2(a − b)2
. (iii) Adding entry-wise:
11110
16521
5109
. (iv) Each entry becomes cos2x + sin2x = 1, giving
11
11
.

3. Compute the indicated products. (i)

ab
−ba
a−b
ba
 (ii)
1
2
3
[2 3 4] (iii)
1−2
23
123
231
 (iv)
234
345
456
1−35
024
305
 (v)
21
32
−11
101
−121
 (vi)
3−13
−102
2−3
10
31

SOLUTION (i)
a·a + b·b−ab + ab
−ab + abb2 + a2
 =
a2 + b20
0a2 + b2
. (ii) (3 × 1)(1 × 3) gives a 3 × 3 matrix:
234
468
6912
. (iii)
1−42−63−2
2+64+96+3
 =
−3−41
8139
. (iv) Computing row × column:
14042
18−156
22−270
. (v)
2−10+22+1
3−20+43+2
−1−10+2−1+1
 =
123
145
−220
. (vi)
6−1+9−9+0+3
−2+0+63+0+2
 =
14−6
45
.

4. If A =

12−3
502
1−11
, B =
3−12
425
203
 and C =
412
032
1−23
, then compute (A + B) and (B − C). Also, verify that A + (B − C) = (A + B) − C.

SOLUTION A + B =
41−1
927
3−14
. B − C =
−1−20
4−13
120
. A + (B − C) =
00−3
9−15
211
, and (A + B) − C gives the same matrix, so the identity is verified.

5. If A =

2⁄315⁄3
1⁄32⁄34⁄3
7⁄322⁄3
and B =
2⁄53⁄51
1⁄52⁄54⁄5
7⁄56⁄52⁄5
, then compute 3A − 5B.

SOLUTION 3A =
235
124
762
 and 5B =
235
124
762
. ∴ 3A − 5B =
000
000
000
 (the zero matrix).

6. Simplify cosθ

cosθsinθ
−sinθcosθ
+ sinθ
sinθ−cosθ
cosθsinθ

SOLUTION First product:
cos2θsinθcosθ
−sinθcosθcos2θ
; second:
sin2θ−sinθcosθ
sinθcosθsin2θ
. Adding, each diagonal entry is cos2θ + sin2θ = 1 and each off-diagonal entry is 0 ⇒
10
01
.

7. Find X and Y, if (i) X + Y =

70
25
and X − Y =
30
03
 (ii) 2X + 3Y =
23
40
 and 3X + 2Y =
2−2
−15

SOLUTION (i) Adding the two equations: 2X =
100
28
 ⇒ X =
50
14
. Subtracting: 2Y =
40
22
 ⇒ Y =
20
11
. (ii) Multiply the first by 3 and the second by 2 and subtract to eliminate X (or use 2×Eq1 − 3×Eq2 for Y). Solving gives X =
2⁄5−12⁄5
−11⁄53
 and Y =
2⁄513⁄5
14⁄5−2
.

8. Find X, if Y =

32
14
and 2X + Y =
10
−32

SOLUTION 2X =
10
−32
 −
32
14
 =
−2−2
−4−2
. ∴ X =
−1−1
−2−1
.

9. Find x and y, if 2

13
0x
+
y0
12
 =
56
18

SOLUTION LHS =
2 + y6
12x + 2
. Equating entries: 2 + y = 5 ⇒ y = 3; 2x + 2 = 8 ⇒ x = 3. ∴ x = 3, y = 3.

10. Solve the equation for x, y, z and t, if 2

xz
yt
+ 3
1−1
02
 = 3
35
46

SOLUTION LHS =
2x + 32z − 3
2y2t + 6
; RHS =
915
1218
. 2x + 3 = 9 ⇒ x = 3; 2z − 3 = 15 ⇒ z = 9; 2y = 12 ⇒ y = 6; 2t + 6 = 18 ⇒ t = 6. ∴ x = 3, y = 6, z = 9, t = 6.

11. If x

2
3
+ y
−1
1
 =
10
5
, find the values of x and y.

SOLUTION 2x − y = 10 and 3x + y = 5. Adding: 5x = 15 ⇒ x = 3; then y = 2(3) − 10 = −4. ∴ x = 3, y = −4.

12. Given 3

xy
zw
=
x6
−12w
 +
4x + y
z + w3
, find the values of x, y, z and w.

SOLUTION Equating entries: 3x = x + 4 ⇒ x = 2; 3w = 2w + 3 ⇒ w = 3; 3z = −1 + z + w ⇒ 2z = −1 + 3 = 2 ⇒ z = 1; 3y = 6 + x + y ⇒ 2y = 6 + 2 = 8 ⇒ y = 4. ∴ x = 2, y = 4, z = 1, w = 3.

13. If F(x) =

cos x−sin x0
sin xcos x0
001
, show that F(x) F(y) = F(x + y).

SOLUTION Multiplying the top-left 2 × 2 blocks: (1,1) = cos x cos y − sin x sin y = cos(x + y); (1,2) = −cos x sin y − sin x cos y = −sin(x + y); (2,1) = sin x cos y + cos x sin y = sin(x + y); (2,2) = −sin x sin y + cos x cos y = cos(x + y). The third row/column give 0’s and a 1, so F(x)F(y) =
cos(x+y)−sin(x+y)0
sin(x+y)cos(x+y)0
001
 = F(x + y). Hence proved.

14. Show that (i)

5−1
67
21
34
 ≠
21
34
5−1
67
 (ii)
123
010
110
−110
0−11
234
 ≠
−110
0−11
234
123
010
110

SOLUTION (i) LHS =
10−35−4
12+216+28
 =
71
3334
; RHS =
10+6−2+7
15+24−3+28
 =
165
3925
. LHS ≠ RHS. (ii) Computing both products gives LHS =
5814
0−11
−101
, RHS =
−1−1−3
100
6116
. Since LHS ≠ RHS, matrix multiplication is not commutative.

15. Find A2 − 5A + 6I, if A =

201
213
1−10

SOLUTION A2 = A·A =
5−12
9−25
0−1−2
; 5A =
1005
10515
5−50
; 6I =
600
060
006
. A2 − 5A + 6I =
1−1−3
−1−1−10
−544
.

16. If A =

102
021
203
, prove that A3 − 6A2 + 7A + 2I = 0.

SOLUTION A2 =
508
245
8013
, A3 = A2·A =
21034
12823
34055
. Then A3 − 6A2 + 7A + 2I: each entry, e.g. (1,1) = 21 − 30 + 7 + 2 = 0, (1,3) = 34 − 48 + 14 + 0 = 0, (3,3) = 55 − 78 + 21 + 2 = 0, etc. All entries vanish, so A3 − 6A2 + 7A + 2I = O. Hence proved.

17. If A =

3−2
4−2
and I =
10
01
, find k so that A2 = kA − 2I.

SOLUTION A2 =
9−8−6+4
12−8−8+4
 =
1−2
4−4
. kA − 2I =
3k − 2−2k
4k−2k − 2
. Comparing the (1,2) entry: −2k = −2 ⇒ k = 1 (and all other entries agree). ∴ k = 1.

18. If A =

0−tan(α⁄2)
tan(α⁄2)0
and I is the identity matrix of order 2, show that I + A = (I − A)
cosα−sinα
sinαcosα

SOLUTION Let t = tan(α⁄2). Then I + A =
1−t
t1
 and I − A =
1t
−t1
. Using cosα = (1 − t2)⁄(1 + t2) and sinα = 2t⁄(1 + t2), the product (I − A)
cosα−sinα
sinαcosα
 has (1,1) entry cosα + t sinα = (1 − t2 + 2t2)⁄(1 + t2) = 1. Likewise (1,2) = −sinα + t cosα = (−2t + t − t3)⁄(1 + t2) = −t; (2,1) = −t cosα + sinα = t; (2,2) = t sinα + cosα = 1. This equals I + A. Hence proved.

19. A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) ₹ 1800   (b) ₹ 2000

SOLUTION Let ₹ x be invested in the 5% bond and ₹ (30000 − x) in the 7% bond. The interest is [x   30000 − x]
0.05
0.07
 = 0.05x + 0.07(30000 − x) = 2100 − 0.02x. (a) 2100 − 0.02x = 1800 ⇒ 0.02x = 300 ⇒ x = 15000. Invest ₹ 15000 in each bond. (b) 2100 − 0.02x = 2000 ⇒ 0.02x = 100 ⇒ x = 5000. Invest ₹ 5000 at 5% and ₹ 25000 at 7%.

20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

SOLUTION In dozens: chemistry 10, physics 8, economics 10 ⇒ 120, 96, 120 books respectively. Total = [120   96   120]
80
60
40
. = 120(80) + 96(60) + 120(40) = 9600 + 5760 + 4800 = ₹ 20160.

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

21. The restriction on n, k and p so that PY + WY will be defined are: (A) k = 3, p = n   (B) k is arbitrary, p = 2   (C) p is arbitrary, k = 3   (D) k = 2, p = 3

SOLUTION PY: P is p × k, Y is 3 × k. For PY to be defined we need k = 3, and the product is p × k. WY: W is n × 3, Y is 3 × k, giving an n × k matrix. For PY + WY to be defined, both must have the same order, so p = n. ∴ the correct option is (A) k = 3, p = n.

22. If n = p, then the order of the matrix 7X − 5Z is: (A) p × 2   (B) 2 × n   (C) n × 3   (D) p × n

SOLUTION X is 2 × n and Z is 2 × p. With n = p they have the same order 2 × n, so 7X − 5Z is also 2 × n. ∴ the correct option is (B) 2 × n.

Exercise 3.3 Solutions

1. Find the transpose of each of the following matrices: (i)

5
1⁄2
−1
  (ii)
1−1
23
   (iii)
−156
√356
23−1

SOLUTION (i) [5   1⁄2   −1] (a 1 × 3 row matrix). (ii)
12
−13
. (iii)
−1√32
553
66−1
.

2. If A =

−123
579
−211
and B =
−41−5
120
131
, then verify that (i) (A + B)′ = A′ + B′,   (ii) (A − B)′ = A′ − B′

SOLUTION (i) A + B =
−53−2
699
−142
, so (A + B)′ =
−56−1
394
−292
; adding A′ and B′ gives the same matrix. (ii) A − B =
318
459
−3−20
, so (A − B)′ =
34−3
15−2
890
 = A′ − B′. Both identities verified.

3. If A′ =

34
−12
01
and B =
−121
123
, then verify that (i) (A + B)′ = A′ + B′   (ii) (A − B)′ = A′ − B′

SOLUTION Here A = (A′)′ =
3−10
421
. (i) A + B =
211
544
, so (A + B)′ =
25
14
14
 = A′ + B′. (ii) A − B =
4−3−1
30−2
, so (A − B)′ =
43
−30
−1−2
 = A′ − B′. Verified.

4. If A′ =

−23
12
and B =
−10
12
, then find (A + 2B)′.

SOLUTION A = (A′)′ =
−21
32
; 2B =
−20
24
. A + 2B =
−41
56
, so (A + 2B)′ =
−45
16
.

5. For the matrices A and B, verify that (AB)′ = B′A′, where (i) A =

1
−4
3
, B = [−1 2 1] (ii) A =
0
1
2
, B = [1 5 7]

SOLUTION (i) AB =
−121
4−8−4
−363
, so (AB)′ =
−14−3
2−86
1−43
. B′A′ gives the same matrix. (ii) AB =
000
157
21014
, so (AB)′ =
012
0510
0714
 = B′A′. Verified.

6. If (i) A =

cosαsinα
−sinαcosα
, then verify that A′A = I. (ii) If A =
sinαcosα
−cosαsinα
, then verify that A′A = I.

SOLUTION (i) A′ =
cosα−sinα
sinαcosα
. A′A diagonal entries = cos2α + sin2α = 1, off-diagonal = 0 ⇒ A′A =
10
01
 = I. (ii) Similarly A′A diagonal entries = sin2α + cos2α = 1, off-diagonal = 0 ⇒ A′A = I. Verified.

7. (i) Show that the matrix A =

1−15
−121
513
is a symmetric matrix. (ii) Show that the matrix A =
01−1
−101
1−10
 is a skew symmetric matrix.

SOLUTION (i) Interchanging rows and columns leaves A unchanged because aij = aji (e.g. a12 = a21 = −1, a13 = a31 = 5). So A′ = A and A is symmetric. (ii) Here aji = −aij and the diagonal entries are 0, so A′ = −A and A is skew symmetric.

8. For the matrix A =

15
67
, verify that (i) (A + A′) is a symmetric matrix (ii) (A − A′) is a skew symmetric matrix

SOLUTION A′ =
16
57
. (i) A + A′ =
211
1114
; its transpose is itself, so it is symmetric. (ii) A − A′ =
0−1
10
; its transpose is its negative, so it is skew symmetric.

9. Find ½(A + A′) and ½(A − A′), when A =

0ab
−a0c
−b−c0

SOLUTION A′ =
0−a−b
a0−c
bc0
. ½(A + A′) =
000
000
000
 (the zero matrix, since A is already skew symmetric). ½(A − A′) =
0ab
−a0c
−b−c0
 = A.

10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix: (i)

35
1−1
(ii)
6−22
−23−1
2−13
 (iii)
33−1
−2−21
−4−52
 (iv)
15
−12

SOLUTION Use A = ½(A + A′) [symmetric] + ½(A − A′) [skew symmetric]. (i) A =
33
3−1
 +
02
−20
. (ii) A is already symmetric, so A =
6−22
−23−1
2−13
 +
000
000
000
. (iii) A =
31⁄2−5⁄2
1⁄2−2−2
−5⁄2−22
 +
05⁄23⁄2
−5⁄203
−3⁄2−30
. (iv) A =
12
22
 +
03
−30
.

Choose the correct answer in the Exercises 11 and 12.

11. If A, B are symmetric matrices of same order, then AB − BA is a (A) Skew symmetric matrix   (B) Symmetric matrix   (C) Zero matrix   (D) Identity matrix

SOLUTION (AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′ = BA − AB = −(AB − BA), so AB − BA is skew symmetric. ∴ the correct option is (A).

12. If A =

cosα−sinα
sinαcosα
, and A + A′ = I, then the value of α is (A) π⁄6   (B) π⁄3   (C) π   (D) 3π⁄2

SOLUTION A + A′ =
2cosα0
02cosα
. Setting this equal to I gives 2cosα = 1 ⇒ cosα = ½ ⇒ α = π⁄3. ∴ the correct option is (B) π⁄3.

Exercise 3.4 Solutions

1. Matrices A and B will be inverse of each other only if (A) AB = BA   (B) AB = BA = 0   (C) AB = 0, BA = I   (D) AB = BA = I

SOLUTION By definition, B is the inverse of A when AB = BA = I (and then A and B are square matrices of the same order). ∴ the correct option is (D) AB = BA = I.

Miscellaneous Exercise Solutions

1. If A and B are symmetric matrices, prove that AB − BA is a skew symmetric matrix.

SOLUTION Since A and B are symmetric, A′ = A and B′ = B. (AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′ = BA − AB = −(AB − BA). Since the transpose equals the negative, AB − BA is skew symmetric. Hence proved.

2. Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

SOLUTION (B′AB)′ = B′A′(B′)′ = B′A′B. If A is symmetric (A′ = A): (B′AB)′ = B′AB, so B′AB is symmetric. If A is skew symmetric (A′ = −A): (B′AB)′ = B′(−A)B = −B′AB, so B′AB is skew symmetric. Hence proved.

3. Find the values of x, y, z if the matrix A =

02yz
xy−z
x−yz
satisfy the equation A′A = I.

SOLUTION A′A = I means the columns of A are mutually orthogonal unit vectors. Column sums of squares: column 1: x2 + x2 = 2x2 = 1; column 2: 4y2 + y2 + y2 = 6y2 = 1; column 3: z2 + z2 + z2 = 3z2 = 1. ∴ x = ±1⁄√2, y = ±1⁄√6, z = ±1⁄√3.

4. For what values of x : [1 2 1]

120
201
102
0
2
x
 = O?

SOLUTION [1 2 1]
120
201
102
 = [1+4+1   2+0+0   0+2+2] = [6   2   4]. [6 2 4]
0
2
x
 = 0 + 4 + 4x = 4 + 4x. Setting 4 + 4x = 0 gives x = −1.

5. If A =

31
−12
, show that A2 − 5A + 7I = 0.

SOLUTION A2 =
9−13+2
−3−2−1+4
 =
85
−53
; 5A =
155
−510
; 7I =
70
07
. A2 − 5A + 7I =
8−15+75−5+0
−5+5+03−10+7
 =
00
00
 = O. Hence proved.

6. Find x, if [x −5 −1]

102
021
203
x
4
1
 = O

SOLUTION [x −5 −1]
102
021
203
 = [x − 2   −10   2x − 8]. Multiplying by
x
4
1
: (x − 2)x − 40 + (2x − 8) = x2 − 2x − 40 + 2x − 8 = x2 − 48. Setting x2 − 48 = 0 gives x = ±4√3.

7. A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below: Market I: 10,000   2,000   18,000   (products x, y, z) Market 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. (b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.

SOLUTION (a) Revenue = sales × prices:
10000200018000
6000200008000
2.50
1.50
1.00
. Market I: 25000 + 3000 + 18000 = ₹ 46000; Market II: 15000 + 30000 + 8000 = ₹ 53000. (b) Total cost = sales × costs: Market I: 20000 + 2000 + 9000 = ₹ 31000; Market II: 12000 + 20000 + 4000 = ₹ 36000. Gross profit = revenue − cost: Market I = 46000 − 31000 = ₹ 15000; Market II = 53000 − 36000 = ₹ 17000.

8. Find the matrix X so that X

123
456
=
−7−8−9
246

SOLUTION The RHS is 2 × 3 and the known matrix is 2 × 3, so X must be 2 × 2. Let X =
ab
cd
. Row 1: a + 4b = −7, 2a + 5b = −8, 3a + 6b = −9 ⇒ a = 1, b = −2. Row 2: c + 4d = 2, 2c + 5d = 4, 3c + 6d = 6 ⇒ c = 2, d = 0. ∴ X =
1−2
20
.

9. If A =

αβ
γ−α
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 A2 =
α2 + βγαβ − αβ
γα − αγβγ + α2
 =
α2 + βγ0
0α2 + βγ
. Setting A2 = I gives α2 + βγ = 1, i.e. 1 − α2 − βγ = 0. ∴ the correct option is (C).

10. 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 Symmetric means A′ = A; skew symmetric means A′ = −A. Together A = −A ⇒ 2A = O ⇒ A = O. ∴ the correct option is (B) A is a zero matrix.

11. If A is square matrix such that A2 = A, then (I + A)3 − 7A is equal to (A) A   (B) I − A   (C) I   (D) 3A

SOLUTION (I + A)3 = I + 3A + 3A2 + A3. Since A2 = A, also A3 = A2·A = A·A = A. So (I + A)3 = I + 3A + 3A + A = I + 7A, and (I + A)3 − 7A = I. ∴ the correct option is (C) I.

Common Mistakes to Avoid

Watch out for these

  • Treating matrix multiplication as commutative — in general AB ≠ BA, and sometimes only one of them is even defined.
  • Multiplying matrices when the orders don’t conform: AB needs (columns of A) = (rows of B).
  • Forgetting the order reversal in (AB)′ = B′A′ — not A′B′.
  • Assuming AB = O forces A = O or B = O — false for matrices.
  • In equality problems (like Ex 3.1 Q9), checking only one entry — every corresponding entry must match, and a clash means the matrices can’t be equal.
  • Mis-stating definitions: a scalar matrix is a diagonal matrix with equal diagonal entries; a skew-symmetric matrix always has zero diagonal.

Practice MCQs & Assertion–Reason

1. A matrix with 15 elements can have order:

(a) 2 × 7    (b) 4 × 4    (c) 3 × 5    (d) 2 × 8

2. The number of all possible matrices of order 2 × 2 with each entry 0, 1 or 2 is:

(a) 16    (b) 64    (c) 81    (d) 512

3. If A is a 3 × 4 matrix and B is a 4 × 5 matrix, the order of AB is:

(a) 3 × 5    (b) 4 × 4    (c) 5 × 3    (d) not defined

4. If A and B are matrices such that both AB and BA are defined and A is m × n, B is n × m, then AB has order:

(a) n × n    (b) m × m    (c) m × n    (d) n × m

5. The diagonal entries of any skew-symmetric matrix are:

(a) all equal    (b) all 1    (c) all 0    (d) all negative

6. For any matrices A and B of suitable order, (AB)′ equals:

(a) A′B′    (b) B′A′    (c) AB    (d) BA

7. A square matrix A such that A′ = A is called:

(a) skew symmetric    (b) symmetric    (c) diagonal    (d) scalar

8. If A is invertible with inverse B, then:

(a) AB = O    (b) AB = BA = I    (c) AB = I, BA = O    (d) A + B = I

9. For invertible matrices A and B of the same order, (AB)−1 equals:

(a) A−1B−1    (b) B−1A−1    (c) AB    (d) (BA)−1

10. A scalar matrix is a diagonal matrix in which the diagonal entries are:

(a) all 1    (b) all 0    (c) all equal    (d) all distinct

Answer key: 1-(c), 2-(c), 3-(a), 4-(b), 5-(c), 6-(b), 7-(b), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: For two matrices A and B, AB may not equal BA.

Reason: Matrix multiplication is not commutative in general.

A-R 2. Assertion: If A, B are symmetric matrices of the same order, then AB − BA is skew symmetric.

Reason: (AB − BA)′ = BA − AB = −(AB − BA).

A-R 3. Assertion: A matrix that is both symmetric and skew symmetric must be the zero matrix.

Reason: A′ = A and A′ = −A together give 2A = O.

A-R 4. Assertion: If AB = O for matrices A and B, then at least one of A or B is the zero matrix.

Reason: The product of two non-zero matrices is always non-zero.

A-R 5. Assertion: The inverse of a square matrix, if it exists, is unique.

Reason: If B and C are both inverses of A, then B = BI = B(AC) = (BA)C = IC = C.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • A matrix of order m × n has mn elements; for a square matrix m = n.
  • Matrices of the same order add entry-wise and obey A + B = B + A and associativity; kA scales every entry.
  • AB is defined only when (columns of A) = (rows of B); multiplication is associative and distributive but not commutative.
  • AB = O does not imply A = O or B = O.
  • Transpose laws: (A′)′ = A, (kA)′ = kA′, (A + B)′ = A′ + B′, (AB)′ = B′A′.
  • A is symmetric if A′ = A and skew symmetric if A′ = −A; every square matrix = ½(A + A′) + ½(A − A′).
  • B is the inverse of A if AB = BA = I; the inverse, if it exists, is unique, and (AB)−1 = B−1A−1.

How to score full marks in this chapter

Always check that two matrices are of conforming orders before adding or multiplying, and write out each product entry as a row-times-column sum so no partial mark is lost. For equality problems, equate every corresponding entry. In transpose and symmetric/skew-symmetric proofs, quote the law (AB)′ = B′A′ explicitly and remember the order reversal. For polynomial-in-A questions (A2 − 5A + 6I etc.), compute A2 first, keep the identity matrix scaled correctly, and add the matrices entry-by-entry to reach the zero matrix.

Frequently Asked Questions

What is Class 12 Maths Chapter 3 Matrices about?

Chapter 3, Matrices, covers the order and types of matrices, equality of matrices, addition and scalar multiplication, multiplication of matrices and its properties, the transpose of a matrix, symmetric and skew-symmetric matrices, and an introduction to invertible matrices.

How many exercises are there in Class 12 Maths Chapter 3?

There are four numbered exercises — Exercise 3.1, 3.2, 3.3 and 3.4 — plus a Miscellaneous Exercise on Chapter 3. Every question of all five is solved on this page.

Is matrix multiplication commutative?

No. In general AB ≠ BA, and sometimes only one of the products is even defined. Matrix multiplication is, however, associative and distributive over addition.

Are these Class 12 Maths Chapter 3 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics Part I textbook for the 2026–27 session, with every answer cross-checked against the book’s answer key.

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