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Class 10 Math Chapter 3 Test 3 | Matrices and Determinants | Punjab Board Model Papers 2027
Class 10 Math Chapter 3 Test 3 Model Papers 2027
Class 10 Mathematics Chapter 3 Test 3 on Matrices and Determinants is designed according to the latest board-style examination pattern for Punjab Board students. This Class 10 Math Chapter 3 Test provides practice questions from Matrices and Determinants and helps students prepare effectively for the Class 10 Mathematics Board Exam. Students can use this Mathematics Chapter 3 Test 3 for revision, self-assessment, and exam preparation.
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Unit 03:
Total Time: 35 min
Chapter Wise Class Test
(Test # 3)
Mathematics
Total Marks: 25
1ABCD
2ABCD
3ABCD
4ABCD
5ABCD
6ABCD
7ABCD
8ABCD
NOTE: Four possible answers A, B, C and D to each question are given.
The choice which you think is correct, fill that circle in front of that
question with Marker or Pen ink. Cutting or filling two or more circles
will result in zero mark in that question.
Q.1: (8)
1. If
=
,
then what is the value of a?
| a + 2 |
| 3 |
| 7 |
| 3 |
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.
- Second entries: 3 = 3 ✔ (already equal).
- Check: 5 + 2 = 7 ✔. So the answer is (B).
2. If AT = −A, then A is ______ matrix.
Answer: (D) skew-symmetric
- If AT = A, the matrix is symmetric.
- If AT = −A (the transpose equals the negative of the matrix), the matrix is called skew-symmetric.
- Example: A = [0 4; −4 0] has AT = [0 −4; 4 0] = −A.
- So the answer is (D).
3. If A = [3 4], B = [7 8], then B − A =
Answer: (C) [4 4]
- Both matrices have the same order (1 × 2), so subtraction is possible.
- Subtract corresponding entries (B − A, B first):
First entry: 7 − 3 = 4
Second entry: 8 − 4 = 4 - B − A = [4 4], which is option (C).
- (Option (D) would be A − B, the wrong order.)
4. If Q =
then −4Q is:
| 2 | −3 |
| 0 | −4 |
Answer: (A)
- To find −4Q, multiply every entry of Q by −4.
- (1,1): (−4)(2) = −8 (1,2): (−4)(−3) = 12
(2,1): (−4)(0) = 0 (2,2): (−4)(−4) = 16 - −4Q =
, which is option (A).−8 12 0 16
5. What is product of
[1 2]
?
| 5 |
| 0 |
Answer: (C) [5]
- The first matrix is 1 × 2 and the second is 2 × 1.
- Columns of first (2) = rows of second (2), so the product exists and has order 1 × 1.
- Multiply the row by the column:
[(1)(5) + (2)(0)] = [5 + 0] = [5] - So the answer is option (C).
6. The associative law of multiplication of matrices is:
Answer: (D) (AB)C = A(BC)
- The associative law says that the way the matrices are grouped does not change the product: (AB)C = A(BC).
- (A) AB = BA is the commutative law, which is not true in general for matrices.
- (B) AB = I only holds when B is the inverse of A.
- (C) is a (mis-written) form of the distributive law A(B + C) = AB + AC.
- Hence the answer is (D).
7. If B =
,
then |B| will be:
| a | b |
| c | d |
Answer: (B) ad − bc
- The determinant of a 2 × 2 matrix is the product of the main diagonal minus the product of the other diagonal.
- Main diagonal: a and d ⟹ ad. Other diagonal: b and c ⟹ bc.
- |B| = ad − bc, which is option (B).
- Example: for [3 1; 2 5], |B| = (3)(5) − (1)(2) = 13.
8. If A =
,
then order of AT is:
| 3 | 1 | 3 |
| 2 | 0 | 2 |
Answer: (A) 3-by-2
- Order of A = 2 rows × 3 columns = 2 × 3.
- The transpose AT turns rows into columns, so the numbers of rows and columns are swapped.
- AT =
which has 3 rows and 2 columns.3 2 1 0 3 2 - Order of AT = 3-by-2, option (A).
(PART – I)
Q.2 Write short answers to following questions.
(6×2=12)
(i) Write the order of the matrix A = [1 3 7]
Answer: 1 × 3
- Order = (number of rows) × (number of columns).
- A has 1 row and 3 columns (1, 3, 7).
- Order of A = 1 × 3 (“1-by-3”). It is a row matrix.
(ii) Define a “Column matrix”.
Answer
- Definition: A matrix that has only one column (and any number of rows) is called a column matrix. Its order is m × 1.
- Example: B =
has order 3 × 1.2 −1 4
(iii) Find the transpose of matrix A =
| 1 | 2 |
| 3 | 4 |
Answer: AT = [1 3; 2 4]
- The transpose is obtained by writing the rows of A as columns.
- Row 1 of A = (1, 2) becomes column 1. Row 2 of A = (3, 4) becomes column 2.
- AT =
1 3 2 4
(iv) If Z = [−3 0 15], then find −7ZT.
Answer: −7ZT = [21; 0; −105]
- Step 1: find ZT (the row becomes a column):
ZT =−3 0 15 - Step 2: multiply every entry by −7:
(−7)(−3) = 21, (−7)(0) = 0, (−7)(15) = −105 - −7ZT =
21 0 −105
(v) Find the value of
| −5 | 8 |
| −3 | −7 |
Answer: 59
- For a 2 × 2 determinant, value = ad − bc.
- Here a = −5, b = 8, c = −3, d = −7.
- ad = (−5)(−7) = 35 and bc = (8)(−3) = −24.
- Value = 35 − (−24) = 35 + 24 = 59.
(vi) Find negative of the matrix. A =
| −3 | 0 |
| 5 | 6 |
Answer: −A = [3 0; −5 −6]
- The negative of a matrix is found by changing the sign of every entry (multiplying by −1).
- (1,1): −(−3) = 3 (1,2): −(0) = 0
(2,1): −(5) = −5 (2,2): −(6) = −6 - −A =
3 0 −5 −6 - Check: A + (−A) = zero matrix ✔
(PART – II)
Note: Attempt the following question.
(5)
Q.3:
If A =
,
then find multiplicative inverse of A and verify that
AA−1 = A−1A = I
| 8 | 5 |
| 4 | 3 |
Answer: A−1 = (1/4)[3 −5; −4 8] = [3/4 −5/4; −1 2]
- Step 1: Determinant. |A| = (8)(3) − (5)(4) = 24 − 20 = 4. Since |A| ≠ 0, A is non-singular and its inverse exists.
- Step 2: Adjoint. Swap the diagonal entries 8 and 3, and change the signs of 5 and 4:
Adj A =3 −5 −4 8 - Step 3: Inverse. A−1 = (1/|A|) Adj A = (1/4)
=3 −5 −4 8 3/4 −5/4 −1 2 - Step 4: Verify AA−1 = I. First multiply A by Adj A:
(1,1): (8)(3) + (5)(−4) = 24 − 20 = 4
(1,2): (8)(−5) + (5)(8) = −40 + 40 = 0
(2,1): (4)(3) + (3)(−4) = 12 − 12 = 0
(2,2): (4)(−5) + (3)(8) = −20 + 24 = 4
So A · Adj A =
and AA−1 = (1/4) × that =4 0 0 4
= I ✔1 0 0 1 - Step 5: Verify A−1A = I. Multiply Adj A by A:
(1,1): (3)(8) + (−5)(4) = 24 − 20 = 4
(1,2): (3)(5) + (−5)(3) = 15 − 15 = 0
(2,1): (−4)(8) + (8)(4) = −32 + 32 = 0
(2,2): (−4)(5) + (8)(3) = −20 + 24 = 4
So Adj A · A =
and A−1A = (1/4) × that = I ✔4 0 0 4 - Hence AA−1 = A−1A = I. Verified.
