Class 12 Maths Chapter 3 Matrices Worksheet with Answers, free PDF (CBSE and JEE Main; based on NCERT)

A 20-mark worksheet on Matrices for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

A matrix has 12 elements and more rows than columns. The number of different orders it can have is

(a) 6(b) 2(c) 4(d) 3

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Answer: (d) 3

An m × n matrix has mn elements, so mn = 12 with m > n. The pairs are 12 × 1, 6 × 2 and 4 × 3, so there are 3 orders.

Question 2

If A = [[2, −1, 3], [0, 4, 1]] and B = [[1, 2], [5, −3], [−2, 0]], the element in the second row and first column of AB is

(a) −9(b) −12(c) 18(d) 22

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Answer: (c) 18

The (2, 1) element of AB takes row 2 of A and column 1 of B: (0)(1) + (4)(5) + (1)(−2) = 0 + 20 − 2 = 18.

Question 3

A = [a_ij] is a 2 × 3 matrix with a_ij = ij − 2. The sum of the elements in the second row of A is

(a) 12(b) 2(c) 0(d) 6

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Answer: (d) 6

a₂₁ = 2 − 2 = 0, a₂₂ = 4 − 2 = 2, a₂₃ = 6 − 2 = 4, so A = [[−1, 0, 1], [0, 2, 4]] and the second row sums to 0 + 2 + 4 = 6.

Question 4

The matrix [[2x − 1, y − 3], [0, 5]] is a scalar matrix. Then x + y equals

(a) 6(b) 3(c) 9(d) 0

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Answer: (a) 6

A scalar matrix is diagonal, so y − 3 = 0, y = 3; its diagonal elements are equal, so 2x − 1 = 5, x = 3. Hence x + y = 6.

Question 5

Which of the following statements is true?

(a) Every scalar matrix is an identity matrix.(b) Every diagonal matrix is a scalar matrix.(c) Every zero matrix is a square matrix.(d) Every identity matrix is a scalar matrix.

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Answer: (d) Every identity matrix is a scalar matrix.

An identity matrix is a diagonal matrix whose diagonal elements are all equal to 1, so it is a scalar matrix with k = 1. A scalar matrix with k = 3 is not an identity matrix; diag(1, 2) is diagonal but not scalar; [0, 0] is a zero matrix of order 1 × 2, not square.

Question 6 · 2 marks

Find p, q, r and s if [[p + q, q − r], [2r + s, p − s]] = [[6, 1], [9, 2]].

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Equating corresponding elements: p + q = 6, q − r = 1, 2r + s = 9, p − s = 2. Write p = s + 2, q = 6 − p = 4 − s, r = q − 1 = 3 − s. Then 2(3 − s) + s = 9 gives 6 − s = 9, s = −3. Hence p = −1, q = 7, r = 6, s = −3.

four equations from equality of matrices (1)
p = −1, q = 7, r = 6, s = −3 (1)

Question 7 · 2 marks

Find x, y and z if the matrix [[0, 2x − 1, 4], [3, 0, y + 2], [z, −5, 0]] is skew symmetric.

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For a skew symmetric matrix a_ji = −a_ij. From a₂₁ = 3: a₁₂ = 2x − 1 = −3, so x = −1. From a₃₂ = −5: a₂₃ = y + 2 = 5, so y = 3. From a₁₃ = 4: a₃₁ = z = −4. The diagonal elements are already 0.

uses a_ji = −a_ij (1)
x = −1, y = 3, z = −4 (1)

Question 8 · 2 marks

Construct the 2 × 3 matrix A = [a_ij] whose elements are given by a_ij = |2i − 3j|.

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a₁₁ = |2 − 3| = 1, a₁₂ = |2 − 6| = 4, a₁₃ = |2 − 9| = 7; a₂₁ = |4 − 3| = 1, a₂₂ = |4 − 6| = 2, a₂₃ = |4 − 9| = 5. So A = [[1, 4, 7], [1, 2, 5]].

first row 1, 4, 7 (1)
second row 1, 2, 5, written as a 2 × 3 matrix (1)

Question 9 · 3 marks

Express C = [[1, 4, −2], [0, 3, 5], [6, −1, 2]] as the sum of a symmetric and a skew symmetric matrix.

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C′ = [[1, 0, 6], [4, 3, −1], [−2, 5, 2]]. P = (1/2)(C + C′) = (1/2)[[2, 4, 4], [4, 6, 4], [4, 4, 4]] = [[1, 2, 2], [2, 3, 2], [2, 2, 2]], and P′ = P. Q = (1/2)(C − C′) = (1/2)[[0, 4, −8], [−4, 0, 6], [8, −6, 0]] = [[0, 2, −4], [−2, 0, 3], [4, −3, 0]], and Q′ = −Q. P + Q = C.

C′ and P = (1/2)(C + C′), symmetric (1)
Q = (1/2)(C − C′), skew symmetric (1)
P + Q = C stated (1)

Question 10 · 3 marks

For a real number t, let G(t) = [[1, t], [0, 1]]. (a) Show that G(s)G(t) = G(s + t). (b) Hence show that G(t) is invertible with G(t)⁻¹ = G(−t), and find [G(2)]³.

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(a) G(s)G(t) = [[1·1 + s·0, 1·t + s·1], [0, 1]] = [[1, s + t], [0, 1]] = G(s + t). (b) G(t)G(−t) = G(0) = I and G(−t)G(t) = G(0) = I, so G(−t) is the inverse of G(t). [G(2)]³ = G(2)G(2)G(2) = G(4)G(2) = G(6) = [[1, 6], [0, 1]].

product G(s)G(t) = G(s + t) (1)
G(t)G(−t) = G(−t)G(t) = I, so G(t)⁻¹ = G(−t) (1)
[G(2)]³ = [[1, 6], [0, 1]] (1)

Question 11 · 3 marks

Find matrices X and Y if 2X + Y = [[5, 0], [−3, 10]] and X − 2Y = [[0, 5], [−4, 0]].

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Let A = [[5, 0], [−3, 10]], B = [[0, 5], [−4, 0]]. Multiplying the first equation by 2 and adding the second: 5X = 2A + B = [[10, 5], [−10, 20]], so X = [[2, 1], [−2, 4]]. Then Y = A − 2X = [[5, 0], [−3, 10]] − [[4, 2], [−4, 8]] = [[1, −2], [1, 2]]. Check: X − 2Y = [[0, 5], [−4, 0]].

eliminates Y: 5X = 2A + B (1)
X = [[2, 1], [−2, 4]] (1)
Y = [[1, −2], [1, 2]] (1)

Class 12 Maths Chapter 3: Matrices worksheet with answer key