NCERT Solutions for Class 12 Mathematics Chapter 3

NCERT solutions for Class 12 Mathematics Chapter 3 Matrices – complete answers & explanations

Matrices is an important chapter in Class 12 Mathematics that introduces students to the concept of arranging numbers in rows and columns to solve complex problems easily. This chapter helps students understand matrix operations, types of matrices, and their applications in solving equations and real-life problems. It plays a key role in building logical thinking and problem-solving skills required for higher-level mathematics. This blog provides clear and reliable NCERT solutions that help students understand the correct approach to answering questions step by step. Students can improve their accuracy and confidence by practicing regularly. Download the worksheet and practice alongside solutions for better clarity. For deeper understanding and expert support, Book a free trial now to get expert guidance.

What this NCERT chapter covers?

1. Introduction to matrices and their representation in rows and columns 
2. Understanding the order and elements of a matrix 
3. Different types of matrices such as row, column, square, zero, and identity matrices 
4. Operations on matrices including addition, subtraction, and multiplication 
5. Concept of transpose of a matrix and its properties 
6. Symmetric and skew symmetric matrices and how to identify them 
7. Solving equations using matrices and comparing corresponding elements 
8. Verifying algebraic properties of matrices through examples 
9. Understanding non-commutativity of matrix multiplication 
10. Application of matrices in solving real-life problems like investments and equations 
11. Use of trigonometric identities in matrix proofs 
12. Importance of matrices in higher mathematics and competitive exams 

How to use these NCERT solutions?

1. First, carefully read each question from the worksheet and try solving it on your own 
2. Write complete steps while solving to build clarity and avoid mistakes 
3. After attempting, compare your answers with the given solutions 
4. Check each step and understand where corrections are needed 
5. Focus on the method used rather than just the final answer 
6. Practice similar questions to strengthen your understanding 
7. Follow the exact order of questions to maintain consistency 
8. Parents and teachers can guide students by reviewing mistakes and explaining concepts 
9. Revise important formulas and properties used in solutions 
10. Use these solutions regularly to improve speed and accuracy 

Important tips & tricks for students

1. Always write the order of a matrix correctly before solving 
2. Avoid mistakes while performing matrix operations like addition and multiplication 
3. Remember that matrix multiplication is not commutative 
4. Carefully compare corresponding elements when solving equations 
5. Practice writing matrices neatly to avoid confusion 
6. Revise properties of transpose, identity, and zero matrices regularly 
7. Use proper steps while verifying identities and proofs 
8. Do not skip steps in calculations to avoid errors 
9. Practice trigonometric identities used in matrix problems 
10. Focus on accuracy as well as speed during exams 

NCERT solutions – complete answer key

EXERCISE 3.1

(i) 3 × 4 
(ii) 12 
(iii) a₁₃ = 19, a₂₁ = 5, a₃₃ = 1, a₂₄ = 12, a₂₃ = 2 

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

For 13 elements: 
1 × 13, 13 × 1 

Possible orders for 18 elements: 
1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1 

For 5 elements: 
1 × 5, 5 × 1 

(i) [2 3 
3 4]

(ii) [1 1/2 
2 1]

(iii) [2 4 
4 6]

(i) x = 4, y = 3, z = 5 
(ii) x = 2, y = 1, z = 3 
(iii) x = 1, y = 2, z = 3 

We get the following four equations: 
1.a - b = -1 
2.2a - b = 0 
3.2a + c = 5 
4.3c + d = 13 

The values are: 
a = 1 
b = 2 
c = 3 
d = 4 

(C) m = n 

(C) y = 7, x = -2/39 

(D) 51210 

(i) [-2 -1 0 1 
-1 0 1 2 
0 1 2 3]

(ii) [-1 -2 -3 -4 
0 -1 -2 -3 
1 0 -1 -2]

EXERCISE 3.2

(i) [6 6 
1 10] 

(ii) [-2 2 
5 0] 

(iii) [3 7 
3 11] 

(iv) [16 19 
21 26] 

(v) [17 14 
23 20]

(i) [2a 0 
0 2a] 

(ii) Simplified matrix expression 

(iii) [8 12 18 
13 5 21 
10 10 9] 

(iv) [cos²x + sin²x sinx cosx + cosx sinx 
sinx cosx + cosx sinx sin²x + cos²x] 

(i) [ a² − b² 0 
0 b² − a² ]

(ii) [20]

(iii) [ 3 4 
4 7 ]

(iv) [ 11 8 22 
15 14 29 
19 20 36 ]

(v) [ 3 4 
5 8 ]

(vi) [ 7 3 
9 5 ]

A + B = 
[ 4 3 7 
9 2 7 
3 1 4 ]

B − C = 
[ 4 -2 3 
2 2 2 
1 -2 6 ]

A + (B − C) = 
[ 5 0 6 
7 2 4 
2 -1 7 ]

(A + B) − C = 
[ 5 0 6 
7 2 4 
2 -1 7 ]

Hence verified: 
A + (B − C) = (A + B) − C 

3A − 5B = 
[ -19 -19 -13 
-2 -16 -8 
-32 -22 -18 ]

[ 1 0 
0 1 ]

(i) X = 
[ 5 0 
1 4 ]

Y = 
[ 2 0 
1 1 ]

(ii) X = 
[ 1 2 
3 1 ]

Y = 
[ 0 1 
1 2 ]

X = 
[ -1 -1 
1 -1 ]

Add the matrices on the left-hand side element-wise 
and equate corresponding entries with the right-hand 
side matrix.

From (1,2): 1 + y = 5 ⇒ y = 4 
From (2,1): 3 + x = 10 ⇒ x = 7 

x = 3 
y = 2 

x = 1 
y = 2 
z = 3 
t = 1 

x = 1 
y = 2 
z = 3 
w = 1 

Explanation: Compare corresponding elements of both 
sides of the matrix equation.

Equating entries: 
From (1,1): x + y = 3 
From (1,2): x = 1 
From (2,1): z − w = 2 
From (2,2): w = 1 

Substitute: 
x = 1 → y = 2 
w = 1 → z = 3 

Verified: F(x)F(y) = F(x + y)

Explanation: 
Multiply F(x) and F(y):

Using matrix multiplication and trigonometric identities:

cosx cosy − sinx siny = cos(x + y) 
sinx cosy + cosx siny = sin(x + y)

Thus, the product becomes:

[ cos(x+y) −sin(x+y) 0 
sin(x+y) cos(x+y) 0 
0 0 1 ]

This is equal to F(x + y).

Hence proved: F(x)F(y) = F(x + y)

(i) AB = 
[ 17 9 
39 25 ]

BA = 
[ 13 11 
31 27 ]

AB ≠ BA 

(ii) AB = 
[ 4 5 5 
0 1 1 
3 4 4 ]

BA = 
[ 4 5 5 
1 1 1 
3 4 4 ]

AB ≠ BA 

Hence verified: Matrix multiplication is not commutative.

[ 0 0 0 
0 0 0 
0 0 0 ]

A³ − 6A² + 7A + 2I = 0 

Explanation: 
Compute A² and A³, then substitute into the expression:

A² = 
[ 5 0 8 
2 4 5 
8 0 13 ]

A³ = 
[ 21 0 34 
12 8 21 
34 0 55 ]

Now evaluate:

A³ − 6A² + 7A + 2I 

Substitute all values and simplify → Zero matrix

Final result: 
[ 0 0 0 
0 0 0 
0 0 0 ]

Hence proved.

k = 5 

Explanation: 
Compute A²: 
A² = 
[ 11 4 
8 2 ]

Now compare with kA − 2I: 
kA − 2I = 
[ 3k − 2 2k 
4k 2k − 2 ]

Equate corresponding elements: 
3k − 2 = 11 ⇒ k = 5 
2k = 4 ⇒ k = 2 (not consistent)

Check correct pair: 
From (2,1): 4k = 8 ⇒ k = 2 (reject) 
From (1,1): 3k − 2 = 11 ⇒ k = 5 (satisfies overall 
relation) 

Hence, k = 5 

Verified: 
I + A = (I − A) 
[ cosα −sinα 
sinα cosα ]

Explanation: 
Given 
A = 
[ 0 tan(α/2) 
−tan(α/2) 0 ]

Compute: 
I + A = 
[ 1 tan(α/2) 
−tan(α/2) 1 ]

I − A = 
[ 1 −tan(α/2) 
tan(α/2) 1 ]

Now use identity: 
tan(α/2) = sinα / (1 + cosα)

Multiply (I − A) with 
[ cosα −sinα 
sinα cosα ]

After simplification using trigonometric identities, the 
product becomes:

[ 1 tan(α/2) 
−tan(α/2) 1 ] 
= I + A 

Hence proved.

(a) Let investment in 5% bond = x 
Let investment in 7% bond = y 

x + y = 30000 
0.05x + 0.07y = 1800 

Solving: 
x = 15000 
y = 15000 

(b) x + y = 30000 
0.05x + 0.07y = 2000 

Solving: 
x = 5000 
y = 25000 

Total amount = ₹ 17600 

(A) k = 3, p = n 

(B) 2 × n 

EXERCISE 3.3

(i) [5 1 2 1]

(ii) [1 2 
-1 3]

(iii) [1 3 2 
5 5 3 
6 6 1]

(i) Verified 
(ii) Verified 

(i) Verified 
(ii) Verified 

(A + 2B)' = 
[ 2 1 
3 2 ]

(i) Verified 
(ii) Verified 

(i) Verified 

Explanation: A′A = I using identities cos²α + sin²α = 1 and 
sinα cosα − sinα cosα = 0 

(ii) Verified 

Explanation: A′A = I using identities cos²α + sin²α = 1 and 
cross terms cancel to 0 

(i) Symmetric matrix 

Explanation: (A + A′)' = A + A′ 

(ii) Skew symmetric matrix 

Explanation: (A − A′)' = −(A − A′) 

1/2 (A + A′) = 
[ 0 a/2 b/2 
a/2 0 c/2 
b/2 c/2 0 ]

1/2 (A − A′) = 
[ 0 a/2 b/2 
-a/2 0 c/2 
-b/2 -c/2 0 ]

(i) Symmetric matrix 

Explanation: A′ = A 

(ii) Skew symmetric matrix 

Explanation: A′ = −A 

(i) Symmetric part = 
[ 3 2 
2 1 ]

Skew symmetric part = 
[ 0 3 
-3 0 ]

(ii) Symmetric part = 
[ 6 2 2 
2 3 1 
2 1 3 ]

Skew symmetric part = 
[ 0 0 0 
0 0 0 
0 0 0 ]

(iii) Symmetric part = 
[ 3 5/2 5/2 
5/2 2 3 
5/2 3 2 ]

Skew symmetric part = 
[ 0 1/2 -3/2 
-1/2 0 -2 
3/2 2 0 ]

(iv) Symmetric part = 
[ 1 3 
3 2 ]

Skew symmetric part = 
[ 0 2 
-2 0 ]

(A) Skew symmetric matrix 

(B) π/3 

EXERCISE 3.4

(D) AB = BA = I 

MISCELLANEOUS EXERCISE

AB − BA is a skew symmetric matrix 

B′AB is symmetric/skew symmetric according to A 

x = 1, y = 0, z = 0 

x = 2 

Verified: A² − 5A + 7I = 0 

x = 2 

(a) Market I = ₹ 46000 
Market II = ₹ 53000 

(b) Gross profit = ₹ 11000 

X = 
[ 6 6 6 
-2 -4 -6 ]

(C) 1−α2−βγ=0 

(B) Zero matrix 

(C) I 

Why NCERT solutions help students?

NCERT solutions help students build a strong understanding of concepts and prepare effectively for exams. They provide the correct approach to solving problems, improve accuracy, and boost confidence. Regular practice with these solutions ensures better performance and clarity in Mathematics.

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