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Class 10 Mathematics Chapter 3 Test 1 | Matrices and Determinants | Punjab Board paper pattern 2027
Class 10 Math Chapter 3 Test 1
Class 10 Mathematics Chapter 3 Test 1 Exercise No. 3.4 and Review Exercise on Matrices and Determinants is designed according to the Punjab Board examination pattern. This Class 10 Math Test 1 covers important concepts of Matrices and Determinants and provides useful practice for students preparing for the Class 10 Mathematics Board Exam 2027. Students can use this Chapter 3 Mathematics Test for revision, self-assessment, and board exam preparation.
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Exercise No. 3.4 and Review Exercise Test
Test No. 1
Unit 3: Matrices and Determinants
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer.
(1 × 5 = 5)
(i) Adjoint of a matrix
A =
[
1 −1
2 0
]
is:
Answer: (d)
- For a 2 × 2 matrix A = [a b; c d], the adjoint is found by swapping the diagonal entries (a and d) and changing the signs of the other two (b and c).
- Here a = 1, b = −1, c = 2, d = 0.
- Swap a and d: the diagonal becomes 0 and 1.
- Change signs of b and c: −b = 1 and −c = −2.
- So Adj A = [0 1−2 1], which is option (d).
(ii) A matrix A is called singular if:
Answer: (a) |A| = 0
- A matrix is singular when its determinant is zero, because then its inverse (A⁻¹ = Adj A / |A|) does not exist.
- If |A| ≠ 0, the matrix is called non-singular.
- Hence the correct option is |A| = 0.
(iii) Which of the following is a linear equation?
Answer: (d) both (a) and (b)
- A linear equation has variables with power 1 only, and no variables multiplied together.
- (a) 3x + 4y = 1: x and y both have power 1, so it is linear ✔
- (b) x − ½y + 25z = 3: x, y and z all have power 1, so it is linear ✔
- (c) xy = 1: two variables are multiplied, so the term xy has degree 2. Not linear ✘
- So both (a) and (b) are linear: option (d).
(iv) If
[
a + 2
2 + 1
]
=
[7
3
]
then a =
Answer: (b) 5
- Two matrices are equal when their corresponding entries are equal.
- First entries: a + 2 = 7.
- Subtract 2 from both sides: a = 7 − 2 = 5.
- Check the second entries: 2 + 1 = 3 ✔ (it matches the given 3, so the matrices are consistent).
- a = 5 only, so option (b).
(v) Determinant of the matrix
A =
[
3 0
0 0
]
is:
Answer: (a) 0
- For [a b; c d], |A| = ad − bc.
- Here a = 3, b = 0, c = 0, d = 0.
- |A| = (3)(0) − (0)(0) = 0 − 0 = 0.
- The matrix is singular, and its determinant is still defined (it equals 0).
Exercise No. 3.4 and Review Exercise Test
Test No. 1
Unit 3: Matrices and Determinants
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions:
(6 × 2 = 12)
(i) Find AB and BA, if possible.
A =
[
1 −2
],
B =
[3
−4
]
Answer: AB = [11], BA = [3 −6; −4 8]
- Order of A = 1 × 2 and order of B = 2 × 1.
- AB: columns of A (2) = rows of B (2), so AB is possible and has order 1 × 1.
- AB = [(1)(3) + (−2)(−4)] = [3 + 8] = [11].
- BA: columns of B (1) = rows of A (1), so BA is also possible and has order 2 × 2.
- Multiply each entry of B by each entry of A:
Row 1: (3)(1) = 3, (3)(−2) = −6
Row 2: (−4)(1) = −4, (−4)(−2) = 8 - BA = [3 −6−4 8]
- Both products exist, but they are different (even their orders differ).
(ii) Verify each statement, using
A =
[
5 1
−1 4
],
B =
[2 −1
0 3
]
and
C =
[5 1
1 4
].
AB ≠ BA
Verified: AB ≠ BA
- Only A and B are needed for this statement (C is used in other parts of the exercise).
- Find AB (row × column):
(1,1): (5)(2) + (1)(0) = 10
(1,2): (5)(−1) + (1)(3) = −5 + 3 = −2
(2,1): (−1)(2) + (4)(0) = −2
(2,2): (−1)(−1) + (4)(3) = 1 + 12 = 13 - AB = [10 −2−2 13]
- Find BA:
(1,1): (2)(5) + (−1)(−1) = 10 + 1 = 11
(1,2): (2)(1) + (−1)(4) = 2 − 4 = −2
(2,1): (0)(5) + (3)(−1) = −3
(2,2): (0)(1) + (3)(4) = 12 - BA = [11 −2−3 12]
- Compare: entries (1,1), (2,1) and (2,2) differ (10 ≠ 11, −2 ≠ −3, 13 ≠ 12). Hence AB ≠ BA. ✔
(iii) Find the values of each of the determinant.
[
3 8
0 2
]
Answer: 6
- For [a b; c d], determinant = ad − bc.
- Here a = 3, b = 8, c = 0, d = 2.
- Determinant = (3)(2) − (8)(0) = 6 − 0 = 6.
(iv) Find the value of x when
A =
[
x 6
5 15
]
is a singular matrix.
Answer: x = 2
- A matrix is singular when |A| = 0.
- |A| = (x)(15) − (6)(5) = 15x − 30.
- Put |A| = 0: 15x − 30 = 0.
- 15x = 30.
- x = 30 ÷ 15 = 2.
- Check: (2)(15) − (6)(5) = 30 − 30 = 0 ✔
(v) Find multiplicative inverse of the following matrices:
[
40 8
5 2
]
Answer: A⁻¹ = (1/40) [2 −8; −5 40]
- Let A = [40 8; 5 2]. Formula: A⁻¹ = (1/|A|) × Adj A.
- Determinant: |A| = (40)(2) − (8)(5) = 80 − 40 = 40. Since |A| ≠ 0, the inverse exists.
- Adjoint: swap 40 and 2, change signs of 8 and 5:
Adj A = [2 −8−5 40]
- Inverse: A⁻¹ = (1/40) [2 −8−5 40] = [1/20 −1/5−1/8 1]
- Check the first entry of A·A⁻¹: (40)(1/20) + (8)(−1/8) = 2 − 1 = 1 ✔
(vi) Solve by matrix inversion method, if possible.
3x + 2y = 2
x − 2y = −2
x − 2y = −2
Answer: x = 0, y = 1
- Write in matrix form AX = B:
[3 21 −2] [xy] = [2−2]
- Determinant: |A| = (3)(−2) − (2)(1) = −6 − 2 = −8. Since |A| ≠ 0, A⁻¹ exists, so the system can be solved.
- Adjoint: swap 3 and −2, change signs of 2 and 1:
Adj A = [−2 −2−1 3]
- A⁻¹ = (1/−8) Adj A = (−1/8) [−2 −2−1 3]
- Solution: X = A⁻¹B = (−1/8) [−2 −2−1 3] [2−2]
- Multiply: first row = (−2)(2) + (−2)(−2) = −4 + 4 = 0; second row = (−1)(2) + (3)(−2) = −2 − 6 = −8.
- X = (−1/8) [0−8] = [01]
- So x = 0, y = 1.
- Check: 3(0) + 2(1) = 2 ✔ and 0 − 2(1) = −2 ✔
Note: Answer the following Questions.
(4 + 4 = 8)
3. If
[
4 a
b 3
]
[6
6
]
=
[6
3
],
then find the values of a and b.
Answer: a = −3, b = −5/2
- Multiply the 2 × 2 matrix by the 2 × 1 matrix (row × column):
Row 1: (4)(6) + (a)(6) = 24 + 6a
Row 2: (b)(6) + (3)(6) = 6b + 18 - So the left side becomes [24 + 6a6b + 18] = [63].
- Equate corresponding entries.
- Finding a: 24 + 6a = 6 ⟹ 6a = 6 − 24 = −18 ⟹ a = −18 ÷ 6 = −3.
- Finding b: 6b + 18 = 3 ⟹ 6b = 3 − 18 = −15 ⟹ b = −15 ÷ 6 = −5/2 (= −2.5).
- Check: 24 + 6(−3) = 6 ✔ and 6(−5/2) + 18 = −15 + 18 = 3 ✔
4. An electrical engineer wants to determine the current
in two branches A and B of a simple electrical circuit.
The system of the equations is:
x + y = 7
2x − y = 2
where x is the current in branch A and y is the current in
branch B. Find x and y by using matrices.
2x − y = 2
Answer: x = 3, y = 4 (current in A = 3, current in B = 4)
- Write in matrix form AX = B:
[1 12 −1] [xy] = [72]
- Determinant: |A| = (1)(−1) − (1)(2) = −1 − 2 = −3 (not zero, so A⁻¹ exists).
- Adjoint: swap 1 and −1, change signs of 1 and 2:
Adj A = [−1 −1−2 1]
- A⁻¹ = (−1/3) [−1 −1−2 1]
- Solution: X = A⁻¹B. First multiply Adj A by B:
Row 1: (−1)(7) + (−1)(2) = −7 − 2 = −9
Row 2: (−2)(7) + (1)(2) = −14 + 2 = −12 - X = (−1/3) [−9−12] = [34]
- So x = 3 and y = 4.
- Check: 3 + 4 = 7 ✔ and 2(3) − 4 = 2 ✔
