Unit 2: Algebra - Free Study Resources | Boundless Maths

🔢 Unit 2: Algebra

Matrices, Determinants, Inverse & System of Linear Equations

Weightage: 10 Marks in Board Exams
Home › Class 12 Applied Maths › Unit 2: Algebra
Topics Covered MCQs (10) Short Answers (5) Case Studies (2)
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Topics Covered in Unit 2

Master these 6 important topics

1. Matrices

Types of matrices, matrix operations, transpose, symmetric matrices

2. Matrix Operations

Addition, subtraction, multiplication, scalar multiplication, properties

3. Symmetric & Skew-Symmetric

Properties, decomposition, expressing matrices as sum

4. Determinants

Calculation methods, properties, cofactors, minors

5. Inverse of a Matrix

Adjoint method, properties of inverses

6. System of Linear Equations

Matrix method, Cramer's rule, applications

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Practice MCQs with Answers

Click "Show Answer" to reveal explanations

Note: Each Matrix row separated by ';'

Question 1CBSE 2022
If matrix is given by A = [aij]2×2, where aij = i + j, then A is equal to:
  • A[1 2; 3 4]
  • B[2 3; 3 4]
  • C[1 1; 2 2]
  • D[1 2; 1 2]
✓ Correct Answer: (B) [2 3; 3 4]
Solution:
Given aij = i + j
A = [(1+1) (1+2); (2+1) (2+2)] = [2 3; 3 4]
Question 2CBSE 2022
If a matrix such that A² = A, then (I + A)² - 3A is equal to:
  • AI
  • B2A
  • C3I
  • DA
✓ Correct Answer: (A) I
Solution:
(I + A)² - 3A = I² + IA + AI + A² - 3A
= I + A + A + A - 3A = I + 3A - 3A = I
Question 3CBSE 2023
If A = [1 0; -2 1] and B = [-5 10; -10 -5], then AB is:
  • A[-5 10; 0 -5]
  • B[0 -5; 25 10]
  • C[10 -25; -5 0]
  • D[-5 10; 0 -25]
✓ Correct Answer: (D) [-5 10; 0 -25]
Solution:
AB = [1×(-5)+0×(-10) 1×10+0×(-5); (-2)×(-5)+1×(-10) (-2)×10+1×(-5)]
= [-5 10; 0 -25]
Question 4CBSE 2023
If [x+y x+2; 2x-y 16] = [8 5; 1 3y+1], then the values of x and y are:
  • Ax = 3, y = 5
  • Bx = 5, y = 3
  • Cx = 2, y = 7
  • Dx = 7, y = 2
✓ Correct Answer: (A) x = 3, y = 5
Solution:
x + 2 = 5 → x = 3; x + y = 8 → y = 5
Question 5CBSE 2023-Comptt
If A and B are two matrices such that AB = A and BA = B, then B² is equal to:
  • AB
  • BA
  • CI
  • DO
✓ Correct Answer: (A) B
Solution:
B² = (BA)·B = B·(AB) = B·A = BA = B
Question 6CBSE 2023-Comptt
If A = [5 x; y 0] is symmetric, then:
  • Ax = 0, y = 5
  • Bx = 5, y = 0
  • Cx = y
  • Dx + y = 0
✓ Correct Answer: (C) x = y
Solution:
For symmetric matrix A = A'. Comparing: x = y
Question 7CBSE 2024-Comptt
If A = [4 1; 3 2] and I = [1 0; 0 1], then (A² - 6A) equals:
  • A3I
  • B-5I
  • C5I
  • D-3I
✓ Correct Answer: (B) -5I
Solution:
A² = [19 6; 18 7], 6A = [24 6; 18 12]
A² - 6A = [-5 0; 0 -5] = -5I
Question 8CBSE 2022
If A is a square matrix of order 3×3 such that |A| = 4, then |3A| is equal to:
  • A27
  • B81
  • C108
  • D256
✓ Correct Answer: (C) 108
Solution:
|kA| = kn|A| → |3A| = 3³ × 4 = 27 × 4 = 108
Question 9CBSE 2022
If the points (1,3), (x,5) and (2,7) are collinear, then the value of x is:
  • A2
  • B3/2
  • C1
  • D3/4
✓ Correct Answer: (B) 3/2
Solution:
Area = 0: 1(5-7) - 3(x-2) + 1(7x-10) = 0
-2 - 3x + 6 + 7x - 10 = 0 → 4x = 6 → x = 3/2
Question 10CBSE 2025
If |2x 5; 4 x| = |3 5; 4 6|, then the value of x is:
  • A3/2
  • B6
  • C3
  • D±3
✓ Correct Answer: (D) ±3
Solution:
2x² - 20 = 18 - 20 = -2 → 2x² = 18 → x² = 9 → x = ±3

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Short Answer Questions with Step-by-Step Solutions

Practice 2-mark and 3-mark questions

Question 1CBSE 2024
If [x-y 2x+z; 2x-y 3z+w] = [-1 5; 0 13], find the values of x, y, z and w.
Solution:
x - y = -1 ...(i)   2x - y = 0 ...(ii)   2x + z = 5 ...(iii)   3z + w = 13 ...(iv)
From (ii): y = 2x. Substitute in (i): x - 2x = -1 → x = 1, y = 2
From (iii): 2 + z = 5 → z = 3. From (iv): 9 + w = 13 → w = 4
Final Answer: x = 1, y = 2, z = 3, w = 4
Question 2CBSE 2024
If A = [1 0; -1 7], find the value of k such that A² - 8A + kI = 0.
Solution:
A² = [1 0; -8 49]   8A = [8 0; -8 56]
A² - 8A = [-7 0; 0 -7]. So A² - 8A + kI = 0 → k = 7
Final Answer: k = 7
Question 3CBSE 2025
Given A = [2 0 1; 3 4 5; 0 2 3] and B = [1 1 -5; -5 1 -5; 1 -2 4], find BA.
Solution:
Row 1: [1×2+1×3-5×0   1×0+1×4-5×2   1×1+1×5-5×3] = [5 -6 -9]
Row 2: [-5×2+1×3-5×0   -5×0+1×4-5×2   -5×1+1×5-5×3] = [-7 -6 -15]
Row 3: [1×2-2×3+4×0   1×0-2×4+4×2   1×1-2×5+4×3] = [-4 0 3]
Final Answer: BA = [5 -6 -9; -7 -6 -15; -4 0 3]
Question 4CBSE 2023
Solve by Cramer's rule: 2x - y = 17, 3x + 5y = 6
Solution:
D = |2 -1; 3 5| = 10 + 3 = 13
Dx = |17 -1; 6 5| = 85 + 6 = 91   Dy = |2 17; 3 6| = 12 - 51 = -39
x = 91/13 = 7   y = -39/13 = -3
Final Answer: x = 7, y = -3
Question 5CBSE 2023-Comptt
Write A = [7 -3 -3; -1 1 0; -1 0 1] as a sum of a symmetric and skew-symmetric matrix.
Solution:
A' = [7 -1 -1; -3 1 0; -3 0 1]
P = ½(A+A') = [7 -2 -2; -2 1 0; -2 0 1] (symmetric ✓)
Q = ½(A-A') = [0 -1 -1; 1 0 0; 1 0 0] (skew-symmetric ✓)
Final Answer: A = P + Q where P = [7 -2 -2; -2 1 0; -2 0 1], Q = [0 -1 -1; 1 0 0; 1 0 0]

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Case Studies

Real-world application based questions

Case Study 1CBSE 2023-Comptt
School Awards Distribution
10 students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers, the second group has honest and law abiding students and the third group contains vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 13, while the combined strength of first and second group is four times that of the third group. Assume that x, y and z denote the number of students in first, second and third group respectively.
(a) Write the system of linear equations
Solution:
x + y + z = 10
2x + y = 13
x + y - 4z = 0
System: x+y+z=10; 2x+y=13; x+y-4z=0
(b) Write the coefficient matrix A
Solution:
Coefficient Matrix: A = [1 1 1; 2 1 0; 1 1 -4]
(c)(i) Write the matrix of cofactors of every element of A
Solution:
A₁₁ = -4, A₁₂ = 8, A₁₃ = 1
A₂₁ = 5, A₂₂ = -5, A₂₃ = 0
A₃₁ = -1, A₃₂ = 2, A₃₃ = -1
Cofactor Matrix = [-4 8 1; 5 -5 0; -1 2 -1]
(c)(ii) OR: Determine the number of students in each group
Solution:
|A| = -4 + 8 + 1 = 5 ≠ 0
A⁻¹ = (1/5)[-4 5 -1; 8 -5 2; 1 0 -1]
X = A⁻¹B = (1/5)[25; 15; 10] = [5; 3; 2]
x = 5 (hard workers), y = 3 (honest), z = 2 (vigilant)
Case Study 2CBSE 2024
Orphanage Donation
On her birthday, Prema decides to donate some money to children of an orphanage home. If there are 8 children less, everyone gets Rs 10 more. However, if there are 16 children more, everyone gets Rs 10 less. Let the number of children be x and the amount per child be Rs y.
(i) Write the system of linear equations
Solution:
(x-8)(y+10) = xy → 5x - 4y = 40 ...(i)
(x+16)(y-10) = xy → 5x - 8y = -80 ...(ii)
5x - 4y = 40   and   5x - 8y = -80
(ii) Write the system in matrix form AX = B
Solution:
A = [5 -4; 5 -8], X = [x; y], B = [40; -80]
(iii)(a) Find the inverse of matrix A
Solution:
|A| = -40 + 20 = -20   adj(A) = [-8 4; -5 5]
A⁻¹ = (1/20)[8 -4; 5 -5]
(iii)(b) OR: Determine x and y
Solution:
X = A⁻¹B = (1/20)[8×40+(-4)×(-80); 5×40+(-5)×(-80)] = (1/20)[640; 600]
Number of children: x = 32   Amount per child: y = ₹30

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