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Class 10 Mathematics Chapter 3 Test 2 | Matrices and Determinants | Unit Wise Class Test
Class 10 Math Chapter 3 Test 2
Class 10 Mathematics Chapter 3 Test 2 Exercise 3.1 to Exercise 3.3 is a useful practice test based on Matrices and Determinants for Punjab Board students. This Class 10 Math Test 2 includes objective and subjective-type questions covering important concepts of matrices, order of matrices, diagonal and symmetric matrices, matrix equality, transpose, matrix operations, and multiplication. Students can use this Class 10 Mathematics Chapter 3 test to prepare according to the board examination pattern and improve their practice for the Punjab Board Class 10 Mathematics 2027 examination.
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Exercise 3.1 to Exercise 3.3
T-2
Unit 3: Matrices and Determinants
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer.
(1 × 5 = 5)
(i) Order of the matrix
[
]
is:
| 5 | 2 | 1 |
| 0 | −1 | 2 |
Answer: (b) 2-by-3
- The order of a matrix is written as (number of rows) × (number of columns).
- Count the rows: there are 2 rows (5 2 1 and 0 −1 2).
- Count the columns: there are 3 columns.
- So the order is 2 × 3, read as “2-by-3”. Option (b).
- (Option (a) 6 is the total number of entries, not the order.)
(ii) If
[
]
=
[
]
then x =
| x − 1 | 3 |
| 2 | x + 1 |
| −2 | 3 |
| 2 | 0 |
Answer: (b) −1
- Two matrices are equal when their corresponding entries are equal.
- Compare the (1,1) entries: x − 1 = −2 ⟹ x = −2 + 1 = −1.
- Compare the (2,2) entries: x + 1 = 0 ⟹ x = −1.
- Both equations give the same value, so the matrices are equal for x = −1. (The entries 3 and 2 already match.)
- Hence the correct option is (b).
(iii) Which of the following is diagonal matrix?
Answer: (a)
- A diagonal matrix is a square matrix in which every entry outside the main diagonal is zero and at least one diagonal entry is non-zero.
- The main diagonal runs from the top-left to the bottom-right corner.
- (a): diagonal entries are 3, 5, 2 and all other entries are 0 ⟹ diagonal matrix ✔
- (b): the non-zero entries 3, 5, 2 lie on the other diagonal (top-right to bottom-left), while the main diagonal has 0, 5, 0 and non-zero entries lie outside it ⟹ not diagonal ✘
- (c): every entry is 0, so it is the zero (null) matrix, not a diagonal matrix ✘
- (d) is wrong because (b) is not diagonal. So the answer is (a).
(iv) Which of the following is symmetric matrix?
Answer: (c)
- A matrix A is symmetric if At = A, that is, the entry in row i, column j equals the entry in row j, column i.
- For a 2 × 2 matrix [a b; c d], this means b = c.
- (a): b = 4 but c = −4, so b ≠ c ⟹ not symmetric ✘
- (b): b = 4 but c = −4, so b ≠ c ⟹ not symmetric (it is skew-symmetric) ✘
- (c): b = 4 and c = 4, so b = c ⟹ symmetric ✔
- (d) is wrong, since (a) and (b) are not symmetric. Answer: (c).
(v) If
A = [
]
and
B = [
]
then BA =
| 5 | 3 |
| −2 |
| 1 |
Answer: (c)
- Order of A = 1 × 2 and order of B = 2 × 1.
- For BA: columns of B (1) = rows of A (1), so BA is possible and its order is 2 × 2.
- Multiply each entry of B (column) with each entry of A (row):
Row 1: (−2)(5) = −10, (−2)(3) = −6
Row 2: (1)(5) = 5, (1)(3) = 3 - BA = [
], which is option (c).−10 −6 5 3 - (Option (a) [−7] is AB, not BA.)
Unit Wise Class Test
T-2
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) Write the order of matrix.
E = [
]
| 0 | 1 |
| 3 | 4 |
| 5 | 6 |
Answer: Order of E = 3 × 2
- Order = (number of rows) × (number of columns).
- Rows of E: (0 1), (3 4), (5 6) ⟹ 3 rows.
- Columns of E: first column (0, 3, 5) and second column (1, 4, 6) ⟹ 2 columns.
- So the order of E is 3 × 2 (“3-by-2”).
(ii) If
[
]
=
[
]
then find the values of x and y.
| 2x + 1 | 4 |
| 7 | 5 |
| 9 | 4 |
| 7 | y |
Answer: x = 4, y = 5
- Equal matrices have equal corresponding entries.
- Finding x (compare the (1,1) entries): 2x + 1 = 9
2x = 9 − 1 = 8
x = 8 ÷ 2 = 4 - Finding y (compare the (2,2) entries): y = 5
- The remaining entries 4 and 7 already match.
- Check: 2(4) + 1 = 9 ✔ and y = 5 ✔
(iii) Define Diagonal Matrix with example.
Answer
- Definition: A square matrix in which all the entries outside the main diagonal are zero (and at least one entry on the main diagonal is non-zero) is called a diagonal matrix.
- Example: D = [
]3 0 0 5 - Here the diagonal entries are 3 and 5, and the non-diagonal entries are both 0. Another example is a 3 × 3 matrix with diagonal 3, 5, 2 and zeros elsewhere.
(iv) If
B = [
]
then verify that (Bt)t = B
| 7 | 6 |
| 5 | 8 |
Verified: (Bt)t = B
- The transpose Bt is obtained by writing the rows of B as columns.
- Row 1 of B is (7, 6) and row 2 is (5, 8).
- Bt = [
]7 5 6 8 - Now take the transpose again (rows of Bt become columns): Row 1 of Bt is (7, 5) and row 2 is (6, 8).
- (Bt)t = [
]7 6 5 8 - This is the same as B. Hence (Bt)t = B ✔
(v) If
X = [
],
Y = [
]
and
Z = [
],
then find the following: X − 4Y + Z
| 1 | −1 |
| −2 | 2 |
| 1 | 2 |
| 3 | 4 |
| 3 | 0 |
| 0 | −2 |
Answer: X − 4Y + Z = [0 −9; −14 −16]
- Step 1: find 4Y (multiply every entry of Y by 4):
4Y = [
]4 8 12 16 - Step 2: find X − 4Y (subtract corresponding entries):
(1,1): 1 − 4 = −3 (1,2): −1 − 8 = −9
(2,1): −2 − 12 = −14 (2,2): 2 − 16 = −14
X − 4Y = [
]−3 −9 −14 −14 - Step 3: add Z (add corresponding entries):
(1,1): −3 + 3 = 0 (1,2): −9 + 0 = −9
(2,1): −14 + 0 = −14 (2,2): −14 + (−2) = −16 - X − 4Y + Z = [
]0 −9 −14 −16
(vi) If
A = [
],
B = [
]
and
C = [
],
then verify the following: 3(A + B) = 3A + 3B
| 2 | 3 |
| −3 | 2 |
| 3 | 4 |
| 5 | 6 |
| 1 | −2 |
| 0 | 5 |
Verified: 3(A + B) = 3A + 3B
- (C is not needed for this statement.)
- L.H.S. = 3(A + B)
A + B: (1,1): 2 + 3 = 5, (1,2): 3 + 4 = 7, (2,1): −3 + 5 = 2, (2,2): 2 + 6 = 8
A + B = [
]5 7 2 8 - 3(A + B) = [
] …(1)15 21 6 24 - R.H.S. = 3A + 3B
3A = [
] 3B = [6 9 −9 6
]9 12 15 18 - 3A + 3B: (1,1): 6 + 9 = 15, (1,2): 9 + 12 = 21, (2,1): −9 + 15 = 6, (2,2): 6 + 18 = 24
3A + 3B = [
] …(2)15 21 6 24 - From (1) and (2), L.H.S. = R.H.S. Hence 3(A + B) = 3A + 3B ✔
Note: Answer the following Questions.
(4 + 4 = 8)
3. If
A = [
]
and
B = [
],
then find AB and its order.
| 5 | 3 |
| −2 |
| 1 |
Answer: AB = [−7], order 1 × 1
- Order of A = 1 × 2 and order of B = 2 × 1.
- Check if AB exists: columns of A (2) = rows of B (2), so AB is possible.
- Order of AB: (rows of A) × (columns of B) = 1 × 1.
- Multiply (row of A × column of B):
AB = [(5)(−2) + (3)(1)]
= [−10 + 3]
= [−7] - So AB = [−7] and its order is 1 × 1.
4. If
A = [
]
and
B = [
],
then verify that: (A − B)t = At − Bt
| 6 | −2 |
| 0 | 3 |
| 8 | −1 |
| 3 | 0 |
Verified: (A − B)t = At − Bt
- L.H.S. = (A − B)t
A − B: (1,1): 6 − 8 = −2, (1,2): −2 − (−1) = −2 + 1 = −1, (2,1): 0 − 3 = −3, (2,2): 3 − 0 = 3
A − B = [
]−2 −1 −3 3 - Take the transpose (rows become columns):
(A − B)t = [
] …(1)−2 −3 −1 3 - R.H.S. = At − Bt
At = [
] Bt = [6 0 −2 3
]8 3 −1 0 - At − Bt: (1,1): 6 − 8 = −2, (1,2): 0 − 3 = −3, (2,1): −2 − (−1) = −1, (2,2): 3 − 0 = 3
At − Bt = [
] …(2)−2 −3 −1 3 - From (1) and (2), L.H.S. = R.H.S. Hence (A − B)t = At − Bt ✔
