What is Class 12 Mathematics Chapter 3 (Matrices) about?

This chapter introduces matrices as rectangular arrays, their order and types such as row, column, square, diagonal, scalar, identity and zero matrices, and when two matrices are equal. It covers addition, scalar multiplication and multiplication of matrices with their properties, the transpose, symmetric and skew symmetric matrices, and invertible matrices with the uniqueness of the inverse.

Are these solutions based on the new NCERT syllabus?

Yes. The answers follow the current NCERT textbook for Class 12 Mathematics and the latest CBSE syllabus, so the question numbers and wording match the book you use in class.

How should I use these solutions to prepare for exams?

Attempt each question on your own first, then compare your working with the answer here. Note the steps or keywords you missed, and revise the chapter notes before taking a chapter-wise MCQ test to check your understanding.

NCERT Solutions for Class 12 Mathematics Chapter 3 Matrices

Q: What do these NCERT solutions for Class 12 Mathematics Chapter 3 (Matrices) cover?

All 23 questions from the latest NCERT textbook for this chapter are answered step by step, organised as Exercise 3.1, Exercise 3.2, Exercise 3.3, Exercise 3.4, Miscellaneous Exercise on Chapter 3 and Important Extra Questions.

Q: Are the NCERT solutions for Class 12 Mathematics Chapter 3 (Matrices) free?

Yes. All Circle of Learning NCERT solutions and chapter notes are free to read online, with no sign-up needed.

Notation: every matrix is shown as a small table (rows of the table = rows of the matrix). A′ is the transpose of A, I the identity matrix and O the zero matrix.

Exercise 3.1

Q1. In the matrix A =

2	5	19	–7
35	–2	5/2	12
√3	1	–5	17

write: (i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

Ans. (i) A has 3 rows and 4 columns → order 3 × 4.

(ii) Number of elements = 3 × 4 = 12.

(iii) a₁₃ = 19, a₂₁ = 35, a₃₃ = –5, a₂₄ = 12, a₂₃ = 5/2.

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

Answer

An m × n matrix has mn elements, so we need all natural-number pairs with mn = 24.

Possible orders: 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4.

13 is prime, so the only orders are 1 × 13 and 13 × 1.

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

Answer

mn = 18
⇒ possible orders: 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3.

5 is prime
⇒ possible orders: 1 × 5 and 5 × 1.

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

Answer

(i) a₁₁ = 4/2 = 2, a₁₂ = 9/2, a₂₁ = 9/2, a₂₂ = 16/2 = 8. So A =

2	9/2
9/2	8

(ii) a₁₁ = 1/1, a₁₂ = 1/2, a₂₁ = 2/1, a₂₂ = 2/2. So A =

1	1/2
2	1

(iii) a₁₁ = 9/2, a₁₂ = 25/2, a₂₁ = 16/2, a₂₂ = 36/2. So A =

9/2	25/2
8	18

Construct a 3 × 4 matrix, whose elements are given by: (i) aᵢⱼ = (1/2)|–3i + j| (ii) aᵢⱼ = 2i – j

Answer

(i) Row i uses (1/2)|–3i + j| for j = 1, 2, 3, 4: row 1
⇒ |–2|/2, |–1|/2, 0, |1|/2; row 2
⇒ |–5|/2, |–4|/2, |–3|/2, |–2|/2; row 3
⇒ |–8|/2, |–7|/2, |–6|/2, |–5|/2. So A =

1	1/2	0	1/2
5/2	2	3/2	1
4	7/2	3	5/2

(ii) aᵢⱼ = 2i – j: row 1
⇒ 2 – j, row 2
⇒ 4 – j, row 3
⇒ 6 – j. So A =

1	0	–1	–2
3	2	1	0
5	4	3	2

Q6. Find the values of x, y and z from the following equations

(i)

4	3
x	5

=

y	z
1	5

(ii)

x + y	2
5 + z	xy

=

6	2
5	8

(iii)

x + y + z

x + z

y + z

=

Ans. Equal matrices have equal corresponding entries.

(i) 4 = y, 3 = z, x = 1 → x = 1, y = 4, z = 3.

(ii) 5 + z = 5 → z = 0. x + y = 6 and xy = 8, so x, y are roots of t² – 6t + 8 = 0 → t = 2 or 4. Hence x = 2, y = 4, z = 0 or x = 4, y = 2, z = 0.

(iii) x + y + z = 9, x + z = 5, y + z = 7 → y = 9 – 5 = 4 → x = 9 – 7 = 2 → z = 5 – 2 = 3. So x = 2, y = 4, z = 3.

Q7. Find the value of a, b, c and d from the equation

a – b	2a + c
2a – b	3c + d

=

–1	5
0	13

Ans. a – b = –1 and 2a – b = 0 → subtracting, a = 1 → b = 2.

2a + c = 5 → c = 3 → 3c + d = 13 → d = 4. So a = 1, b = 2, c = 3, d = 4.

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

Answer

A square matrix has as many rows as columns. Correct answer: (C) m = n.

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

Answer

3x + 7	5
y + 1	2 – 3x

0	y – 2
8	4

(A) x = –1/3, y = 7 (B) Not possible to find (C) y = 7, x = –2/3 (D) x = –1/3, y = –2/3

Ans. 3x + 7 = 0 gives x = –7/3, while 2 – 3x = 4 gives x = –2/3. (y – 2 = 5 and y + 1 = 8 both give y = 7.)

The two conditions on x contradict each other, so the matrices can never be equal. Correct answer: (B) Not possible to find.

Q10

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

Answer

9 entries, each with 2 choices
⇒ 2⁹ = 512. Correct answer: (D).

Exercise 3.2

Q1. Let A =

2	4
3	2

B =

1	3
–2	5

C =

–2	5
3	4

Find each of the following: (i) A + B (ii) A – B (iii) 3A – C (iv) AB (v) BA

Ans. (i) A + B = add corresponding entries:

3	7
1	7

(ii) A – B =

1	1
5	–3

(iii) 3A =

6	12
9	6

3A – C =

8	7
6	2

(iv) AB: row 1 → 2(1) + 4(–2) = –6, 2(3) + 4(5) = 26; row 2 → 3(1) + 2(–2) = –1, 3(3) + 2(5) = 19. So AB =

–6	26
–1	19

(v) BA: row 1 → 1(2) + 3(3) = 11, 1(4) + 3(2) = 10; row 2 → –2(2) + 5(3) = 11, –2(4) + 5(2) = 2. So BA =

11	10
11	2

Q2. Compute the following

(i)

+

a	b
b	a

(ii)

a² + b²	b² + c²
a² + c²	a² + b²

+

2ab	2bc
–2ac	–2ab

(iii)

–1	4	–6
8	5	16
2	8	5

+

12	7	6
8	0	5
3	2	4

(iv)

cos²x	sin²x
sin²x	cos²x

+

sin²x	cos²x
cos²x	sin²x

Ans. (i) Sum =

2a	2b
0	2a

(ii) Sum =

a² + b² + 2ab	b² + c² + 2bc
a² + c² – 2ac	a² + b² – 2ab

(a + b)²	(b + c)²
(a – c)²	(a – b)²

(iii) Sum =

11	11	0
16	5	21
5	10	9

(iv) Every entry becomes cos²x + sin²x = 1, so the sum =

1	1
1	1

Compute the indicated products.

Answer

(i)

×

a	–b
b	a

(ii)

×

(iii)

1	–2
2	3

×

1	2	3
2	3	1

(iv)

2	3	4
3	4	5
4	5	6

×

1	–3	5
0	2	4
3	0	5

(v)

2	1
3	2
–1	1

×

1	0	1
–1	2	1

(vi)

3	–1	3
–1	0	2

×

2	–3
1	0
3	1

Ans. (i) Product: row 1 → a·a + b·b, a(–b) + b·a; row 2 → –b·a + a·b, (–b)(–b) + a·a. So product =

a² + b²	0
0	a² + b²

(ii) (3 × 1)(1 × 3) gives a 3 × 3 matrix; entry (i, j) = (row entry i) × (column entry j):

2	3	4
4	6	8
6	9	12

(iii) Row 1: 1 – 4, 2 – 6, 3 – 2; Row 2: 2 + 6, 4 + 9, 6 + 3. Product =

–3	–4	1
8	13	9

(iv) Row 1: 2 + 0 + 12, –6 + 6 + 0, 10 + 12 + 20; Row 2: 3 + 0 + 15, –9 + 8 + 0, 15 + 16 + 25; Row 3: 4 + 0 + 18, –12 + 10 + 0, 20 + 20 + 30. Product =

14	0	42
18	–1	56
22	–2	70

(v) Row 1: 2 – 1, 0 + 2, 2 + 1; Row 2: 3 – 2, 0 + 4, 3 + 2; Row 3: –1 – 1, 0 + 2, –1 + 1. Product =

1	2	3
1	4	5
–2	2	0

(vi) Row 1: 6 – 1 + 9, –9 + 0 + 3; Row 2: –2 + 0 + 6, 3 + 0 + 2. Product =

14	–6
4	5

Q4. If A =

1	2	–3
5	0	2
1	–1	1

B =

3	–1	2
4	2	5
2	0	3

and C =

4	1	2
0	3	2
1	–2	3

then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.

Ans. A + B =

4	1	–1
9	2	7
3	–1	4

B – C =

–1	–2	0
4	–1	3
1	2	0

A + (B – C) =

0	0	–3
9	–1	5
2	1	1

(A + B) – C =

0	0	–3
9	–1	5
2	1	1

Both results are the same matrix, so A + (B – C) = (A + B) – C. Verified.

Q5. If A =

2/3	1	5/3
1/3	2/3	4/3
7/3	2	2/3

and B =

2/5	3/5	1
1/5	2/5	4/5
7/5	6/5	2/5

then compute 3A – 5B.

Ans. 3A =

2	3	5
1	2	4
7	6	2

5B =

2	3	5
1	2	4
7	6	2

3A – 5B =

0	0	0
0	0	0
0	0	0

Since 3A = 5B, the result is the zero matrix O.

Q6. Simplify cos θ ×

cos θ	sin θ
–sin θ	cos θ

+ sin θ ×

sin θ	–cos θ
cos θ	sin θ

Ans. Sum =

cos²θ + sin²θ	sin θ cos θ – sin θ cos θ
–sin θ cos θ + sin θ cos θ	cos²θ + sin²θ

= I (identity matrix of order 2).

Q7. Find X and Y, if

(i) X + Y =

7	0
2	5

and X – Y =

3	0
0	3

(ii) 2X + 3Y =

2	3
4	0

and 3X + 2Y =

2	–2
–1	5

Ans. (i) Adding: 2X =

10	0
2	8

→ X =

5	0
1	4

Subtracting: 2Y =

4	0
2	2

→ Y =

2	0
1	1

(ii) Multiply the first by 2 and the second by 3: 4X + 6Y and 9X + 6Y. Subtracting, 5X = 3(3X + 2Y) – 2(2X + 3Y) =

2	–12
–11	15

→ X =

2/5	–12/5
–11/5	3

Similarly 5Y = 3(2X + 3Y) – 2(3X + 2Y) =

2	13
14	–10

→ Y =

2/5	13/5
14/5	–2

Q8. Find X, if Y =

3	2
1	4

and 2X + Y =

1	0
–3	2

Ans. 2X = (2X + Y) – Y =

–2	–2
–4	–2

→ X =

–1	–1
–2	–1

Q9. Find x and y, if 2 ×

1	3
0	x

+

y	0
1	2

=

5	6
1	8

Ans. Left side =

2 + y	6
1	2x + 2

Equating: 2 + y = 5 → y = 3; 2x + 2 = 8 → x = 3.

Q10. Solve the equation for x, y, z and t, if 2 ×

x	z
y	t

+ 3 ×

1	–1
0	2

= 3 ×

3	5
4	6

Ans. Left side =

2x + 3	2z – 3
2y	2t + 6

Right side =

9	15
12	18

2x + 3 = 9 → x = 3; 2y = 12 → y = 6; 2z – 3 = 15 → z = 9; 2t + 6 = 18 → t = 6.

Q11. If x ×

+ y ×

–1

=

find the values of x and y.

Ans. 2x – y = 10 and 3x + y = 5. Adding: 5x = 15 → x = 3 → y = 2(3) – 10 = –4. So x = 3, y = –4.

Q12. Given 3 ×

x	y
z	w

=

x	6
–1	2w

+

4	x + y
z + w	3

find the values of x, y, z and w.

Ans. Right side =

x + 4	6 + x + y
z + w – 1	2w + 3

3x = x + 4 → x = 2. 3w = 2w + 3 → w = 3. 3y = 6 + x + y → 2y = 8 → y = 4. 3z = z + w – 1 → 2z = 2 → z = 1.

So x = 2, y = 4, z = 1, w = 3.

Q13. If F(x) =

cos x	–sin x	0
sin x	cos x	0
0	0	1

show that F(x) F(y) = F(x + y).

Ans. F(x) F(y) =

cos x cos y – sin x sin y	–cos x sin y – sin x cos y	0
sin x cos y + cos x sin y	–sin x sin y + cos x cos y	0
0	0	1

cos(x + y)	–sin(x + y)	0
sin(x + y)	cos(x + y)	0
0	0	1

= F(x + y), using the cosine and sine addition formulas. Hence proved.

Q14. Show that

(i)

×

≠

×

(ii)

1	2	3
0	1	0
1	1	0

×

–1	1	0
0	–1	1
2	3	4

≠

–1	1	0
0	–1	1
2	3	4

×

1	2	3
0	1	0
1	1	0

Ans. (i) Left side: row 1 → 10 – 3, 5 – 4; row 2 → 12 + 21, 6 + 28 =

7	1
33	34

Right side: row 1 → 10 + 6, –2 + 7; row 2 → 15 + 24, –3 + 28 =

16	5
39	25

The two products differ, so the inequality holds.

(ii) Left side =

5	8	14
0	–1	1
–1	0	1

Right side =

–1	–1	–3
1	0	0
6	11	6

The products differ, so the inequality holds (matrix multiplication is not commutative).

Q15. Find A² – 5A + 6I, if A =

2	0	1
2	1	3
1	–1	0

Ans. A² = A × A =

5	–1	2
9	–2	5
0	–1	–2

5A =

10	0	5
10	5	15
5	–5	0

A² – 5A + 6I =

1	–1	–3
–1	–1	–10
–5	4	4

Q16. If A =

1	0	2
0	2	1
2	0	3

prove that A³ – 6A² + 7A + 2I = 0

Ans. A² =

5	0	8
2	4	5
8	0	13

A³ = A² × A =

21	0	34
12	8	23
34	0	55

A³ – 6A² + 7A + 2I: first entry = 21 – 6(5) + 7(1) + 2 = 0, and in the same way every entry is 0:

0	0	0
0	0	0
0	0	0

Hence A³ – 6A² + 7A + 2I = O. Proved.

Q17. If A =

3	–2
4	–2

and I =

find k so that A² = kA – 2I

Ans. A² =

1	–2
4	–4

kA – 2I =

3k – 2	–2k
4k	–2k – 2

Equating: 3k – 2 = 1 → k = 1 (check: –2k = –2, 4k = 4, –2k – 2 = –4 all hold). So k = 1.

Q18. 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) ×

Ans. Let t = tan(α/2). Then cos α = (1 – t²)/(1 + t²) and sin α = 2t/(1 + t²).

I + A =

I – A =

1	t
–t	1

(I – A) × (rotation matrix): entry (1,1) = cos α + t sin α = (1 – t² + 2t²)/(1 + t²) = 1; entry (1,2) = –sin α + t cos α = (–2t + t – t³)/(1 + t²) = –t(1 + t²)/(1 + t²) = –t; entry (2,1) = –t cos α + sin α = (–t + t³ + 2t)/(1 + t²) = t; entry (2,2) = t sin α + cos α = 1. So the product =

= I + A. Hence proved.

Q19

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

Answer

Let ₹x go into the first bond and ₹(30000 – x) into the second. Annual interest = (row matrix of investments) × (column matrix of rates):

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 ₹15,000 in the first bond and ₹15,000 in the second.

(b) 2100 – 0.02x = 2000
⇒ 0.02x = 100
⇒ x = 5000. Invest ₹5,000 in the first bond and ₹25,000 in the second.

Q20

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.

Answer

Number of books: 10 dozen = 120, 8 dozen = 96, 10 dozen = 120. Total amount = (row matrix of quantities) × (column matrix of prices):

120

= [120 × 80 + 96 × 60 + 120 × 40] = [9600 + 5760 + 4800] = [20160]. The bookshop receives ₹20,160.

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.

Q21

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

Answer

PY: (p × k)(3 × k) needs k = 3, giving order p × k. WY: (n × 3)(3 × k) is defined, giving n × k. For the sum, p × k = n × k
⇒ p = n. Correct answer: (A).

Q22

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

Answer

X is 2 × n and Z is 2 × p = 2 × n, so 7X – 5Z is 2 × n. Correct answer: (B).

Exercise 3.3

Q1. Find the transpose of each of the following matrices

(i)

1/2

–1

(ii)

1	–1
2	3

(iii)

–1	5	6
√3	5	6
2	3	–1

Ans. Interchange rows and columns. (i) Transpose =

1/2

–1

(ii) Transpose =

1	2
–1	3

(iii) Transpose =

–1	√3	2
5	5	3
6	6	–1

Q2. If A =

–1	2	3
5	7	9
–2	1	1

and B =

–4	1	–5
1	2	0
1	3	1

then verify that (i) (A + B)′ = A′ + B′, (ii) (A – B)′ = A′ – B′

Ans. (i) A + B =

–5	3	–2
6	9	9
–1	4	2

(A + B)′ =

–5	6	–1
3	9	4
–2	9	2

A′ =

–1	5	–2
2	7	1
3	9	1

B′ =

–4	1	1
1	2	3
–5	0	1

A′ + B′ =

–5	6	–1
3	9	4
–2	9	2

Both are equal. Verified.

(ii) A – B =

3	1	8
4	5	9
–3	–2	0

(A – B)′ =

3	4	–3
1	5	–2
8	9	0

A′ – B′ =

3	4	–3
1	5	–2
8	9	0

Both are equal. Verified.

Q3. If A′ =

3	4
–1	2
0	1

and B =

–1	2	1
1	2	3

then verify that (i) (A + B)′ = A′ + B′ (ii) (A – B)′ = A′ – B′

Ans. A = (A′)′ =

3	–1	0
4	2	1

(i) A + B =

2	1	1
5	4	4

(A + B)′ =

2	5
1	4
1	4

B′ =

–1	1
2	2
1	3

A′ + B′ =

2	5
1	4
1	4

Equal. Verified.

(ii) A – B =

4	–3	–1
3	0	–2

(A – B)′ =

4	3
–3	0
–1	–2

A′ – B′ =

4	3
–3	0
–1	–2

Equal. Verified.

Q4. If A′ =

–2	3
1	2

and B =

–1	0
1	2

then find (A + 2B)′

Ans. A = (A′)′ =

–2	1
3	2

A + 2B =

–4	1
5	6

(A + 2B)′ =

–4	5
1	6

Q5. For the matrices A and B, verify that (AB)′ = B′A′, where

(i) A =

–4

B =

–1

(ii) A =

B =

Ans. (i) AB =

–1	2	1
4	–8	–4
–3	6	3

(AB)′ =

–1	4	–3
2	–8	6
1	–4	3

B′A′ =

–1	4	–3
2	–8	6
1	–4	3

Both are equal. Verified.

(ii) AB =

0	0	0
1	5	7
2	10	14

(AB)′ =

0	1	2
0	5	10
0	7	14

B′A′ =

0	1	2
0	5	10
0	7	14

Both are equal. Verified.

Q6. (i) If A =

then verify that A′A = I

(ii) If A =

then verify that A′A = I

Ans. (i) A′A =

cos²α + sin²α	cos α sin α – sin α cos α
sin α cos α – cos α sin α	sin²α + cos²α

= I. Verified.

(ii) A′A =

sin α	–cos α
cos α	sin α

sin²α + cos²α	sin α cos α – cos α sin α
cos α sin α – sin α cos α	cos²α + sin²α

= I. Verified.

Q7. (i) Show that the matrix A =

1	–1	5
–1	2	1
5	1	3

is a symmetric matrix.

(ii) Show that the matrix A =

0	1	–1
–1	0	1
1	–1	0

is a skew symmetric matrix.

Ans. (i) A′ =

1	–1	5
–1	2	1
5	1	3

= A, so A is symmetric.

(ii) A′ =

0	–1	1
1	0	–1
–1	1	0

= –A (each entry is the negative of the matching entry of A), so A is skew symmetric.

Q8. For the matrix A =

1	5
6	7

verify that (i) (A + A′) is a symmetric matrix (ii) (A – A′) is a skew symmetric matrix

Ans. A′ =

1	6
5	7

(i) A + A′ =

2	11
11	14

Its transpose is the same matrix, so A + A′ is symmetric.

(ii) A – A′ =

0	–1
1	0

Its transpose is

0	1
–1	0

= –(A – A′), so A – A′ is skew symmetric.

Q9. Find (1/2)(A + A′) and (1/2)(A – A′), when A =

0	a	b
–a	0	c
–b	–c	0

Ans. A′ =

0	–a	–b
a	0	–c
b	c	0

= –A. So A + A′ = O and (1/2)(A + A′) =

0	0	0
0	0	0
0	0	0

A – A′ = 2A, so (1/2)(A – A′) = A =

0	a	b
–a	0	c
–b	–c	0

Q10

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

Answer

(i)

3	5
1	–1

(ii)

6	–2	2
–2	3	–1
2	–1	3

(iii)

3	3	–1
–2	–2	1
–4	–5	2

(iv)

1	5
–1	2

Ans. Use A = P + Q, where P = (1/2)(A + A′) is symmetric and Q = (1/2)(A – A′) is skew symmetric.

(i) A′ =

3	1
5	–1

P = (1/2)(A + A′) =

3	3
3	–1

(P′ = P, symmetric) and Q = (1/2)(A – A′) =

0	2
–2	0

(Q′ = –Q, skew symmetric). Then P + Q = A.

(ii) A′ =

6	–2	2
–2	3	–1
2	–1	3

P = (1/2)(A + A′) =

6	–2	2
–2	3	–1
2	–1	3

(P′ = P, symmetric) and Q = (1/2)(A – A′) =

0	0	0
0	0	0
0	0	0

(Q′ = –Q, skew symmetric). Then P + Q = A.

(iii) A′ =

3	–2	–4
3	–2	–5
–1	1	2

P = (1/2)(A + A′) =

3	1/2	–5/2
1/2	–2	–2
–5/2	–2	2

(P′ = P, symmetric) and Q = (1/2)(A – A′) =

0	5/2	3/2
–5/2	0	3
–3/2	–3	0

(Q′ = –Q, skew symmetric). Then P + Q = A.

(iv) A′ =

1	–1
5	2

P = (1/2)(A + A′) =

1	2
2	2

(P′ = P, symmetric) and Q = (1/2)(A – A′) =

0	3
–3	0

(Q′ = –Q, skew symmetric). Then P + Q = A.

Note for (ii): A′ = A, so A is already symmetric and Q = O.

Q11

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

Answer

(AB – BA)′ = B′A′ – A′B′ = BA – AB = –(AB – BA). So AB – BA is skew symmetric. Correct answer: (A).

Q12. If A =

and A + A′ = I, then the value of α is (A) π/6 (B) π/3 (C) π (D) 3π/2

Ans. A + A′ =

2cos α	0
0	2cos α

= I → 2cos α = 1 → cos α = 1/2 → α = π/3. Correct answer: (B).

Exercise 3.4

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

Answer

By definition, B is the inverse of A when AB = BA = I. Correct answer: (D).

Miscellaneous Exercise on Chapter 3

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

Answer

A′ = A and B′ = B. (AB – BA)′ = (AB)′ – (BA)′ = B′A′ – A′B′ = BA – AB = –(AB – BA). Hence AB – BA is skew symmetric.

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

Answer

(B′AB)′ = B′A′(B′)′ = B′A′B.

If A is symmetric (A′ = A): (B′AB)′ = B′AB
⇒ B′AB is symmetric.

If A is skew symmetric (A′ = –A): (B′AB)′ = B′(–A)B = –B′AB
⇒ B′AB is skew symmetric.

Q3. Find the values of x, y, z if the matrix A =

0	2y	z
x	y	–z
x	–y	z

satisfy the equation A′A = I.

Ans. A′A =

0	x	x
2y	y	–y
z	–z	z

0	2y	z
x	y	–z
x	–y	z

2x²	0	0
0	6y²	0
0	0	3z²

Equating with I: 2x² = 1, 6y² = 1, 3z² = 1 → x = ±1/√2, y = ±1/√6, z = ±1/√3.

Q4. For what values of x

×

1	2	0
2	0	1
1	0	2

×

= O?

Ans. First product:

1	2	0
2	0	1
1	0	2

Then

= [0 + 4 + 4x] = [4 + 4x] (a 1 × 1 matrix). Setting 4 + 4x = 0 → x = –2.

Q5. If A =

3	1
–1	2

show that A² – 5A + 7I = 0.

Ans. A² =

8	5
–5	3

5A =

15	5
–5	10

A² – 5A + 7I =

0	0
0	0

= O. Proved.

Q6. Find x, if

–5

–1

×

1	0	2
0	2	1
2	0	3

×

= O

Ans. First product =

x + 0 – 2

0 – 10 + 0

2x – 5 – 3

Then

= [x² – 2x – 40 + 2x – 8] = [x² – 48].

x² – 48 = 0 → x = ±√48 = ±4√3.

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; Market II: 6,000, 20,000, 8,000 (units of x, y, z). (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.

Answer

(a) Revenue =

2.50

1.50

1.00

25000 + 3000 + 18000

15000 + 30000 + 8000

46000

53000

Revenue: Market I ₹46,000; Market II ₹53,000.

(b) Cost =

2.00

1.00

0.50

20000 + 2000 + 9000

12000 + 20000 + 4000

31000

36000

Profit = Revenue – Cost: Market I ₹46,000 – ₹31,000 = ₹15,000; Market II ₹53,000 – ₹36,000 = ₹17,000. Gross profit = ₹15,000 + ₹17,000 = ₹32,000.

Q8. Find the matrix X so that X ×

1	2	3
4	5	6

=

–7	–8	–9
2	4	6

Ans. The given matrix is 2 × 3 and the product is 2 × 3, so X must be 2 × 2. Let X =

a	b
c	d

Then X × (given) =

a + 4b	2a + 5b	3a + 6b
c + 4d	2c + 5d	3c + 6d

a + 4b = –7, 2a + 5b = –8 → a = 1, b = –2 (check: 3a + 6b = –9 ✓).

c + 4d = 2, 2c + 5d = 4 → c = 2, d = 0 (check: 3c + 6d = 6 ✓). So X =

1	–2
2	0

Choose the correct answer in the following questions

Q9. If A =

α	β
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

Ans. A² =

α² + βγ	αβ – αβ
αγ – αγ	βγ + α²

α² + βγ	0
0	α² + βγ

= I → α² + βγ = 1 → 1 – α² – βγ = 0. Correct answer: (C).

Q10

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

Answer

A′ = A and A′ = –A
⇒ A = –A
⇒ 2A = O
⇒ A = O. Correct answer: (B).

Q11

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

Answer

A² = A gives A³ = A² · A = A · A = A. (I + A)³ = I + 3A + 3A² + A³ (I and A commute) = I + 3A + 3A + A = I + 7A.

So (I + A)³ – 7A = I. Correct answer: (C).

Important Extra Questions

If a matrix has 8 elements, list all its possible orders.

Answer

mn = 8 → 1 × 8, 8 × 1, 2 × 4, 4 × 2.

Q2. If A =

1	2
3	4

find A² – 5A.

Ans. A² =

7	10
15	22

A² – 5A =

2	0
0	2

(This equals 2I, which shows A² – 5A – 2I = O.)

If A is a square matrix, show that A + A′ is symmetric and A – A′ is skew symmetric.

Answer

(A + A′)′ = A′ + (A′)′ = A′ + A
⇒ symmetric. (A – A′)′ = A′ – A = –(A – A′)
⇒ skew symmetric.

What are the diagonal elements of a skew symmetric matrix?

Answer

aᵢᵢ = –aᵢᵢ
⇒ 2aᵢᵢ = 0
⇒ every diagonal element is 0.

Q5. Show that B is the inverse of A, where A =

2	3
1	2

and B =

2	–3
–1	2

Ans. AB =

and BA =

Both equal I, so B = A⁻¹.