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🔢 Unit 2: Algebra

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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 1 CBSE 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 that A = [aij]2×2 where aij = i + j
A = [a₁₁ a₁₂; a₂₁ a₂₂] = [(1+1) (1+2); (2+1) (2+2)] = [2 3; 3 4]
Question 2 CBSE 2022
If a matrix such that A² = A, then (I + A)² - 3A is equal to:
  • A I
  • B 2A
  • C 3I
  • D A
✓ Correct Answer: (A) I
Solution:
Given that A² = A
Consider: (I + A)² - 3A
= (I + A)(I + A) - 3A
= I² + IA + AI + A² - 3A
= I + A + A + A - 3A
= I + 3A - 3A = I
Question 3 CBSE 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 0; -2 1] × [-5 10; -10 -5]
= [(1×(-5) + 0×(-10)) (1×10 + 0×(-5)); ((-2)×(-5) + 1×(-10)) ((-2)×10 + 1×(-5))]
= [-5 10; 0 -25]
Question 4 CBSE 2023
If [x+y x+2; 2x-y 16] = [8 5; 1 3y+1], then the values of x and y are:
  • A x = 3, y = 5
  • B x = 5, y = 3
  • C x = 2, y = 7
  • D x = 7, y = 2
✓ Correct Answer: (A) x = 3, y = 5
Solution:
From equality of matrices:
x + y = 8 and x + 2 = 5
From second equation: x = 3
Substituting in first: 3 + y = 8
Therefore: y = 5
Question 5 CBSE 2023-Comptt
If A and B are two matrices such that AB = A and BA = B, then B² is equal to:
  • A B
  • B A
  • C I
  • D O
✓ Correct Answer: (A) B
Solution:
Given: B = BA
Therefore: B² = B · B = B · (BA) = (B · B) · A = B² · A
Also: B² = (BA) · B = B · (AB) = B · A = BA = B
Hence: B² = B
Question 6 CBSE 2023-Comptt
If A = [5 x; y 0] is symmetric, then:
  • A x = 0, y = 5
  • B x = 5, y = 0
  • C x = y
  • D x + y = 0
✓ Correct Answer: (C) x = y
Solution:
For A to be symmetric: A = A'
[5 x; y 0] = [5 y; x 0]
Comparing corresponding elements: x = y
Question 7 CBSE 2024-Comptt
If A = [4 1; 3 2] and I = [1 0; 0 1], then (A² - 6A) equals:
  • A 3I
  • B -5I
  • C 5I
  • D -3I
✓ Correct Answer: (B) -5I
Solution:
A² = [4 1; 3 2] × [4 1; 3 2] = [19 6; 18 7]
6A = 6 × [4 1; 3 2] = [24 6; 18 12]
A² - 6A = [19 6; 18 7] - [24 6; 18 12]
= [-5 0; 0 -5] = -5I
Question 8 CBSE 2022
If A is a square matrix of order 3×3 such that |A| = 4, then |3A| is equal to:
  • A 27
  • B 81
  • C 108
  • D 256
✓ Correct Answer: (C) 108
Solution:
For a square matrix of order n: |kA| = kn|A|
Here, n = 3 and k = 3
Therefore: |3A| = 3³ × |A| = 27 × 4 = 108
Question 9 CBSE 2022
If the points (1,3), (x,5) and (2,7) are collinear, then the value of x is:
  • A 2
  • B 3/2
  • C 1
  • D 3/4
✓ Correct Answer: (B) 3/2
Solution:
For collinear points, area of triangle = 0
½|1 3 1; x 5 1; 2 7 1| = 0
1(5-7) - 3(x-2) + 1(7x-10) = 0
-2 - 3x + 6 + 7x - 10 = 0
4x - 6 = 0
x = 3/2
Question 10 CBSE 2025
If |2x 5; 4 x| = |3 5; 4 6|, then the value of x is:
  • A 3/2
  • B 6
  • C 3
  • D ±3
✓ Correct Answer: (D) ±3
Solution:
|2x 5; 4 x| = |3 5; 4 6|
2x² - 20 = 18 - 20
2x² - 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 1 CBSE 2024
If [x-y 2x+z; 2x-y 3z+w] = [-1 5; 0 13], find the values of x, y, z and w.
Solution:
From equality of matrices, we get four equations:
x - y = -1 ...(i)
2x - y = 0 ...(ii)
2x + z = 5 ...(iii)
3z + w = 13 ...(iv)
Solving (i) and (ii): From (ii), 2x = y
Substituting in (i): x - 2x = -1 ⇒ x = 1
Therefore: y = 2x = 2
From (iii): 2(1) + z = 5 ⇒ z = 3
From (iv): 3(3) + w = 13 ⇒ w = 4
Final Answer: x = 1, y = 2, z = 3, w = 4
Question 2 CBSE 2024
If A = [1 0; -1 7], find the value of k such that A² - 8A + kI = 0.
Solution:
Given: A = [1 0; -1 7]
Calculate A²: A² = [1 0; -1 7] × [1 0; -1 7] = [1 0; -8 49]
Calculate 8A: 8A = 8[1 0; -1 7] = [8 0; -8 56]
Given equation: A² - 8A + kI = 0
[1 0; -8 49] - [8 0; -8 56] + k[1 0; 0 1] = [0 0; 0 0]
[1-8+k 0; -8+8 49-56+k] = [0 0; 0 0]
From first element: 1 - 8 + k = 0 ⇒ k = 7
Verify from fourth element: 49 - 56 + 7 = 0 ✓
Final Answer: k = 7
Question 3 CBSE 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:
BA = [1 1 -5; -5 1 -5; 1 -2 4] × [2 0 1; 3 4 5; 0 2 3]
First row: [1×2+1×3-5×0 1×0+1×4-5×2 1×1+1×5-5×3]
= [2+3-0 0+4-10 1+5-15] = [5 -6 -9]
Second row: [-5×2+1×3-5×0 -5×0+1×4-5×2 -5×1+1×5-5×3]
= [-10+3-0 0+4-10 -5+5-15] = [-7 -6 -15]
Third row: [1×2-2×3+4×0 1×0-2×4+4×2 1×1-2×5+4×3]
= [2-6+0 0-8+8 1-10+12] = [-4 0 3]
Final Answer: BA = [5 -6 -9; -7 -6 -15; -4 0 3]
Question 4 CBSE 2023
Solve the following system of equations by Cramer's rule: 2x - y = 17, 3x + 5y = 6
Solution:
The system can be written as: AX = B
Where A = [2 -1; 3 5], X = [x; y], B = [17; 6]
Calculate D = |A| = |2 -1; 3 5| = 10 - (-3) = 13 ≠ 0
Calculate Dx = |17 -1; 6 5| = 85 - (-6) = 91
Calculate Dy = |2 17; 3 6| = 12 - 51 = -39
By Cramer's rule: x = Dx/D = 91/13 = 7
y = Dy/D = -39/13 = -3
Final Answer: x = 7, y = -3
Question 5 CBSE 2023-Comptt
Write the matrix A = [7 -3 -3; -1 1 0; -1 0 1] as a sum of a symmetric and skew-symmetric matrix.
Solution:
Let A = [7 -3 -3; -1 1 0; -1 0 1]
Find A': A' = [7 -1 -1; -3 1 0; -3 0 1]
Symmetric part P = ½(A + A')
A + A' = [14 -4 -4; -4 2 0; -4 0 2]
P = ½[14 -4 -4; -4 2 0; -4 0 2] = [7 -2 -2; -2 1 0; -2 0 1]
Verify: P' = [7 -2 -2; -2 1 0; -2 0 1] = P ✓ (symmetric)
Skew-symmetric part Q = ½(A - A')
A - A' = [0 -2 -2; 2 0 0; 2 0 0]
Q = ½[0 -2 -2; 2 0 0; 2 0 0] = [0 -1 -1; 1 0 0; 1 0 0]
Verify: Q' = [0 1 1; -1 0 0; -1 0 0] = -Q ✓ (skew-symmetric)
Final Answer: A = P + Q where
P = [7 -2 -2; -2 1 0; -2 0 1] (symmetric)
Q = [0 -1 -1; 1 0 0; 1 0 0] (skew-symmetric)

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Case Studies

Real-world application based questions

Case Study 1 CBSE 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 that can be formulated from the above described situation.
Solution:
Total students: x + y + z = 10
Double first group + second group = 13: 2x + y = 13
First + second = 4 × third: x + y = 4z or x + y - 4z = 0
System of equations:
x + y + z = 10
2x + y = 13
x + y - 4z = 0
(b) Write the coefficient matrix, say A
Solution:
The system can be written as: AX = B
Where X = [x; y; z] and B = [10; 13; 0]
Coefficient Matrix: A = [1 1 1; 2 1 0; 1 1 -4]
(c)(i) Write the matrix of cofactors of every element of matrix A
Solution:
A₁₁ = |-4 - 0| = -4
A₁₂ = -|(-8) - 0| = 8
A₁₃ = |2 - 1| = 1
A₂₁ = -|(-4) - 1| = 5
A₂₂ = |(-4) - 1| = -5
A₂₃ = -|1 - 1| = 0
A₃₁ = |0 - 1| = -1
A₃₂ = -|0 - 2| = 2
A₃₃ = |1 - 2| = -1
Cofactor Matrix = [-4 8 1; 5 -5 0; -1 2 -1]
(c)(ii) OR: Determine the number of students of each group
Solution:
|A| = 1(-4-0) - 1(-8-0) + 1(2-1) = -4 + 8 + 1 = 5 ≠ 0
Transpose of cofactor matrix gives adj(A):
adj(A) = [-4 5 -1; 8 -5 2; 1 0 -1]
A⁻¹ = (1/|A|) × adj(A) = (1/5)[-4 5 -1; 8 -5 2; 1 0 -1]
X = A⁻¹B = (1/5)[-4 5 -1; 8 -5 2; 1 0 -1][10; 13; 0]
= (1/5)[-40+65-0; 80-65+0; 10+0-0] = (1/5)[25; 15; 10]
= [5; 3; 2]
Number of students:
First group (hard workers): x = 5
Second group (honest): y = 3
Third group (vigilant): z = 2
Case Study 2 CBSE 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 in the orphanage home be x and the amount to be donated to each child be Rs y.
(i) Write the system of linear equations in x and y formed from the given situation
Solution:
Let total amount = xy (constant)
Condition 1: 8 children less, each gets Rs 10 more
(x - 8)(y + 10) = xy
xy + 10x - 8y - 80 = xy
10x - 8y = 80
Dividing by 2: 5x - 4y = 40 ...(i)
Condition 2: 16 children more, each gets Rs 10 less
(x + 16)(y - 10) = xy
xy - 10x + 16y - 160 = xy
-10x + 16y = 160
Dividing by -2: 5x - 8y = -80 ...(ii)
System of equations:
5x - 4y = 40
5x - 8y = -80
(ii) Write the system of linear equations in matrix form AX = B
Solution:
The equations can be written as AX = B
Where X = [x; y]
A = [5 -4; 5 -8], X = [x; y], B = [40; -80]
(iii)(a) Find the inverse of matrix A
Solution:
|A| = 5(-8) - 5(-4) = -40 + 20 = -20 ≠ 0
adj(A) = [-8 4; -5 5]
A⁻¹ = (1/|A|) × adj(A)
= (1/-20)[-8 4; -5 5]
= (1/20)[8 -4; 5 -5]
A⁻¹ = (1/20)[8 -4; 5 -5]
(iii)(b) OR: Determine the values of x and y
Solution:
X = A⁻¹B
= (1/20)[8 -4; 5 -5][40; -80]
= (1/20)[8×40 + (-4)×(-80); 5×40 + (-5)×(-80)]
= (1/20)[320 + 320; 200 + 400]
= (1/20)[640; 600]
= [32; 30]
Number of children: x = 32
Amount per child: y = Rs 30

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