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Class 10 Math Chapter 3 Full Unit Paper 2026 | Matrices and Determinants | Punjab Board New Syllabus 2027

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Class 10 Mathematics Chapter 3 Full Unit Paper 2026 is prepared for students following the Punjab Board New Syllabus 2027. This Mathematics Full Unit Paper covers important concepts from Matrices and Determinants and includes objective and subjective-type questions for exam preparation. Students can use this Class 10 Math Chapter 3 Test to practice matrices, determinants, inverse of matrices, matrix operations, and related questions according to the new Punjab Board paper pattern.

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Full Unit Test
Unit 3: Matrices and Determinants
New Board Paper Pattern 2027
Objective Type
Time Allowed: 20 Min. Max. Marks: 15
Note: Four possible answer A, B, C and D to each question are given. The choice which you think is correct, fill that circle front of that question with Marker or ink pen in the answer book. Cutting filling two or more circles will result in zero mark in that question.
(i) M = [
100
010
001
] is a/an __________ matrix.
(A) rectangular
(B) identity
(C) column
(D) row
Answer: (B) identity
  1. M has 3 rows and 3 columns, so it is a square matrix (not rectangular, column or row).
  2. All entries on the main diagonal are 1 and all other entries are 0.
  3. A square matrix with 1s on the main diagonal and 0s elsewhere is the identity matrix (I3).
  4. Hence the answer is (B).
(ii) If At = − A, then A is __________ matrix.
(A) symmetric
(B) row
(C) rectangular
(D) skew-symmetric
Answer: (D) skew-symmetric
  1. If At = A, the matrix is symmetric.
  2. If At = −A, the matrix is skew-symmetric.
  3. Example: A = [0 3; −3 0]. Then At = [0 −3; 3 0] = −A ✔
  4. So the answer is (D).
(iii) If [
a + 2
3
] = [
7
3
] , then a = :
(A) 3
(B) 5
(C) 6
(D) 7
Answer: (B) 5
  1. Equal matrices have equal corresponding entries.
  2. First entries: a + 2 = 7.
  3. a = 7 − 2 = 5.
  4. Second entries: 3 = 3 ✔. Check: 5 + 2 = 7 ✔. Answer (B).
(iv) If A = [
313
202
] , then order of At is:
(A) 3-by-2
(B) 2-by-3
(C) 3-by-3
(D) 2-by-2
Answer: (A) 3-by-2
  1. A has 2 rows and 3 columns, so its order is 2 × 3.
  2. The transpose changes rows into columns, so the order is reversed.
  3. At = [
    32
    10
    32
    ]     has 3 rows and 2 columns.
  4. Order of At = 3-by-2, option (A).
(v) If [
x − 13
2x + 1
] = [
−23
20
] then x =
(A) 0
(B) −1
(C) 1
(D) not possible
Answer: (B) −1
  1. Equate corresponding entries.
  2. (1,1): x − 1 = −2 ⟹ x = −2 + 1 = −1.
  3. (2,2): x + 1 = 0 ⟹ x = −1.
  4. Both entries give the same value, and 3 = 3, 2 = 2 already match. So x = −1, option (B).
(vi) Which of the following is diagonal matrix?
(A) [
300
050
002
]
(B) [
003
050
200
]
(C) [
000
000
000
]
(D) both (A) and (B)
Answer: (A)
  1. A diagonal matrix is a square matrix whose entries outside the main diagonal are all zero (with at least one non-zero diagonal entry).
  2. (A): diagonal entries 3, 5, 2; all others 0 ⟹ diagonal ✔
  3. (B): the non-zero entries 3, 5, 2 lie along the other diagonal; the entries 3 and 2 are outside the main diagonal ⟹ not diagonal ✘
  4. (C): all entries are zero, so it is the null (zero) matrix ✘
  5. (D) is wrong because (B) is not diagonal. Answer (A).
(vii) (At)t =
(A) −A
(B) −At
(C) A
(D) I
Answer: (C) A
  1. Taking the transpose swaps rows and columns.
  2. Taking the transpose a second time swaps them back, returning the original matrix.
  3. Example: A = [1 2; 3 4], At = [1 3; 2 4], (At)t = [1 2; 3 4] = A.
  4. So (At)t = A, option (C).
(viii) ______ matrix is a special case of a diagonal matrix.
(A) square
(B) null
(C) scalar
(D) all of these
Answer: (C) scalar
  1. A scalar matrix is a diagonal matrix in which all diagonal entries are equal. Example: [4 0; 0 4].
  2. So every scalar matrix is a diagonal matrix, but not every diagonal matrix is scalar (for example [3 0; 0 5]). This makes it a special case.
  3. “Square” is a wider category: a diagonal matrix is a special case of a square matrix, not the other way round.
  4. The null matrix is not the textbook example of a special diagonal matrix, and (D) is therefore wrong.
  5. Answer (C).
(ix) If A = [
53
] and B = [
−2
1
] then AB =
(A) [−7]
(B) [−13]
(C) [7]
(D) [13]
Answer: (A) [−7]
  1. A is 1 × 2 and B is 2 × 1. Columns of A (2) = rows of B (2), so AB exists with order 1 × 1.
  2. Multiply the row of A by the column of B:
    AB = [(5)(−2) + (3)(1)]
    = [−10 + 3] = [−7]
  3. Answer (A).
(x) If A = [
abc
] and B = [
d
e
f
] then order of BA is:
(A) 3-by-3
(B) 1-by-1
(C) 1-by-3
(D) 3-by-1
Answer: (A) 3-by-3
  1. Order of A = 1 × 3 and order of B = 3 × 1.
  2. For BA: columns of B (1) = rows of A (1), so BA exists.
  3. Order of BA = (rows of B) × (columns of A) = 3 × 3.
  4. (Note: AB would be 1 × 1, which is option (B), but the question asks for BA.)
  5. Answer (A).
(xi) If A is of order m-by-n and B is of order n-by-f then AB is of order:
(A) n-by-n
(B) f-by-m
(C) m-by-f
(D) none of these
Answer: (C) m-by-f
  1. AB is possible because columns of A (n) = rows of B (n).
  2. The two “inner” numbers (n and n) match and cancel.
  3. The order of the product is formed by the two “outer” numbers: (rows of A) × (columns of B) = m × f.
  4. Answer (C).
(xii) The property of matrices A + (B + C) = (A + B) + C is called ______ under addition.
(A) associative law
(B) commutative law
(C) distributive law
(D) none of these
Answer: (A) associative law
  1. In A + (B + C) = (A + B) + C, only the grouping of the matrices changes; the order stays the same.
  2. That property is the associative law of addition.
  3. (B) the commutative law would be A + B = B + A (order changes).
  4. (C) the distributive law involves both multiplication and addition, e.g. A(B + C) = AB + AC.
  5. Answer (A).
(xiii) If A and B are negative of each other then:
(A) A + B = O
(B) A − B = O
(C) A = B
(D) At = B
Answer: (A) A + B = O
  1. “Negative of each other” means B = −A.
  2. Then A + B = A + (−A) = O (the zero matrix).
  3. Example: A = [2 −1] and B = [−2 1]; A + B = [0 0] = O ✔
  4. (B) A − B = O would mean A = B, which is not the case. Answer (A).
(xiv) Inverse of the identity matrix [
10
01
] is:
(A) [
01
10
]
(B) [
−10
0−1
]
(C) [
0−1
−10
]
(D) [
10
01
]
Answer: (D) the identity matrix itself
  1. Let I = [1 0; 0 1]. Determinant |I| = (1)(1) − (0)(0) = 1.
  2. Adjoint: swap the diagonal entries (1 and 1) and change the signs of the zeros: Adj I = [1 0; 0 1] = I.
  3. I−1 = (1/|I|) Adj I = (1/1) [1 0; 0 1] = [1 0; 0 1].
  4. Check: I × I = I, so I is its own inverse.
  5. Answer (D).
(xv) (AB)−1 =
(A) A−1B−1
(B) AB
(C) BA
(D) B−1A−1
Answer: (D) B−1A−1
  1. The inverse of a product reverses the order of the factors (the “socks and shoes” rule).
  2. Proof: (AB)(B−1A−1) = A(BB−1)A−1 = A·I·A−1 = AA−1 = I.
  3. So B−1A−1 is the inverse of AB.
  4. (A) A−1B−1 is the wrong order in general. Answer (D).

Here is a Subjective Part of Paper. Solve it carefull.

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Subjective Type
Time Allowed:
2:10 hours
(Part-I)
Max. Marks: 60
2. Write short answers to any six (06) questions:
(i) If [
a + 2c − 3
b − 1d + 4
] = [
58
64
], then find the values of a, b, c and d.
Answer: a = 3, b = 7, c = 11, d = 0
  1. Equal matrices have equal corresponding entries.
  2. (1,1): a + 2 = 5 ⟹ a = 5 − 2 = 3
  3. (2,1): b − 1 = 6 ⟹ b = 6 + 1 = 7
  4. (1,2): c − 3 = 8 ⟹ c = 8 + 3 = 11
  5. (2,2): d + 4 = 4 ⟹ d = 4 − 4 = 0
  6. Check: 3 + 2 = 5 ✔, 7 − 1 = 6 ✔, 11 − 3 = 8 ✔, 0 + 4 = 4 ✔
(ii) If A = [
23
45
] and B = [
76
58
], then verify that
(At)t = A
Verified: (At)t = A
  1. (B is not needed for this statement.)
  2. Write the rows of A as columns: At = [2 4; 3 5].
  3. Now take the transpose of At (rows become columns again): (At)t = [2 3; 4 5].
  4. This is exactly A. Hence (At)t = A ✔
(iii) Show that L = [
234
3−25
450
] is a symmetric matrix.
Shown: Lt = L, so L is symmetric
  1. A matrix is symmetric if Lt = L.
  2. Rows of L: (2, 3, 4), (3, −2, 5), (4, 5, 0).
  3. Write these rows as columns: Lt = [2 3 4; 3 −2 5; 4 5 0].
  4. Lt is identical to L (the (1,2) and (2,1) entries are both 3, the (1,3) and (3,1) entries are both 4, and the (2,3) and (3,2) entries are both 5).
  5. Hence L is a symmetric matrix ✔
(iv) If [
p + q5
11p − 2q
] = [
65
110
], then find the values of p and q.
Answer: p = 4, q = 2
  1. Equate corresponding entries: p + q = 6 …(1) and p − 2q = 0 …(2).
  2. From (2): p = 2q.
  3. Put in (1): 2q + q = 6 ⟹ 3q = 6 ⟹ q = 2.
  4. Then p = 2(2) = 4.
  5. Check: 4 + 2 = 6 ✔ and 4 − 2(2) = 0 ✔
(v) If A = [
23
−32
], B = [
34
56
] and C = [
1−2
05
], then verify that A + B = B + A.
Verified: A + B = B + A
  1. (C is not needed here.)
  2. 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. B + A: (1,1): 3 + 2 = 5, (1,2): 4 + 3 = 7, (2,1): 5 + (−3) = 2, (2,2): 6 + 2 = 8 ⟹ B + A = [5 7; 2 8].
  4. Both results are equal, so A + B = B + A (commutative law of addition) ✔
(vi) If A = [
6−2
03
] and B = [
8−1
30
], then verify that:
(A + B)t = At + Bt.
Verified: (A + B)t = At + Bt
  1. L.H.S. A + B: (1,1): 6 + 8 = 14, (1,2): −2 + (−1) = −3, (2,1): 0 + 3 = 3, (2,2): 3 + 0 = 3 ⟹ A + B = [14 −3; 3 3].
  2. (A + B)t = [14 3; −3 3] …(1)
  3. R.H.S. At = [6 0; −2 3] and Bt = [8 3; −1 0].
  4. At + Bt: (1,1): 6 + 8 = 14, (1,2): 0 + 3 = 3, (2,1): −2 + (−1) = −3, (2,2): 3 + 0 = 3 ⟹ [14 3; −3 3] …(2)
  5. From (1) and (2), L.H.S. = R.H.S. ✔
(vii) If A = [
56
7−2
] and B = [
−5−6
−72
], then show that B is additive inverse of A and A is additive inverse of B.
Shown: A + B = B + A = O
  1. Two matrices are additive inverses if their sum is the zero matrix O.
  2. A + B: (1,1): 5 + (−5) = 0, (1,2): 6 + (−6) = 0, (2,1): 7 + (−7) = 0, (2,2): −2 + 2 = 0 ⟹ A + B = [0 0; 0 0] = O.
  3. B + A gives the same entries (addition is commutative) ⟹ B + A = O.
  4. Since A + B = O, B is the additive inverse of A; since B + A = O, A is the additive inverse of B ✔
(viii) If A = [
ab
cd
], B = [
lm
np
], then find: AB, if possible.
Answer: AB = [al + bn, am + bp; cl + dn, cm + dp]
  1. A is 2 × 2 and B is 2 × 2. Columns of A (2) = rows of B (2), so AB is possible and has order 2 × 2.
  2. (1,1): (row 1 of A)(column 1 of B) = a·l + b·n
  3. (1,2): (row 1 of A)(column 2 of B) = a·m + b·p
  4. (2,1): (row 2 of A)(column 1 of B) = c·l + d·n
  5. (2,2): (row 2 of A)(column 2 of B) = c·m + d·p
  6. AB = [
    al + bnam + bp
    cl + dncm + dp
    ]
(ix) If A = [
53
] and B = [
−2
1
], then find AB and its order.
Answer: AB = [−7], order 1 × 1
  1. A is 1 × 2 and B is 2 × 1. Columns of A (2) = rows of B (2), so AB exists.
  2. Order of AB = (rows of A) × (columns of B) = 1 × 1.
  3. AB = [(5)(−2) + (3)(1)] = [−10 + 3] = [−7].
3. Write short answers to any six (06) questions:
(i) Find AB and BA, if possible. A = [
1−2
], B = [
3
−4
].
Answer: AB = [11], BA = [3 −6; −4 8]
  1. A is 1 × 2 and B is 2 × 1.
  2. AB: columns of A (2) = rows of B (2), so AB exists with order 1 × 1.
    AB = [(1)(3) + (−2)(−4)] = [3 + 8] = [11]
  3. BA: columns of B (1) = rows of A (1), so BA exists with order 2 × 2.
    Row 1: (3)(1) = 3, (3)(−2) = −6
    Row 2: (−4)(1) = −4, (−4)(−2) = 8
    BA = [3 −6; −4 8]
(ii) Verify the statement AB ≠ BA, using A = [
51
−14
], B = [
2−1
03
] and C = [
51
14
].
Verified: AB ≠ BA
  1. Only A and B are needed (C is not used).
  2. AB: (1,1): (5)(2) + (1)(0) = 10; (1,2): (5)(−1) + (1)(3) = −2; (2,1): (−1)(2) + (4)(0) = −2; (2,2): (−1)(−1) + (4)(3) = 13 ⟹ AB = [10 −2; −2 13].
  3. BA: (1,1): (2)(5) + (−1)(−1) = 11; (1,2): (2)(1) + (−1)(4) = −2; (2,1): (0)(5) + (3)(−1) = −3; (2,2): (0)(1) + (3)(4) = 12 ⟹ BA = [11 −2; −3 12].
  4. The (1,1), (2,1) and (2,2) entries differ (10 ≠ 11, −2 ≠ −3, 13 ≠ 12). Hence AB ≠ BA ✔
(iii) Define identity matrix.
Answer
  1. A square matrix in which every entry on the main diagonal is 1 and every other entry is 0 is called an identity matrix. It is denoted by I.
  2. Example: I = [1 0; 0 1] (order 2 × 2). For any matrix A of suitable order, AI = IA = A.
(iv) Solve the following matrix inversion method:
2x + 3y = 13
4x − 5y = −7
Answer: x = 2, y = 3
  1. Matrix form AX = B: A = [2 3; 4 −5], X = [x; y], B = [13; −7].
  2. Determinant: |A| = (2)(−5) − (3)(4) = −10 − 12 = −22 (not zero, so A−1 exists).
  3. Adjoint: swap 2 and −5, change signs of 3 and 4: Adj A = [−5 −3; −4 2].
  4. A−1 = (1/−22) Adj A = (−1/22) [−5 −3; −4 2].
  5. X = A−1B. First Adj A × B:
    Row 1: (−5)(13) + (−3)(−7) = −65 + 21 = −44
    Row 2: (−4)(13) + (2)(−7) = −52 − 14 = −66
  6. X = (−1/22) [−44; −66] = [2; 3]. So x = 2, y = 3.
  7. Check: 2(2) + 3(3) = 13 ✔ and 4(2) − 5(3) = −7 ✔
(v) Which condition is required for multiplication of two matrices.
Answer
  1. The product AB of two matrices is possible only if the number of columns of the first matrix A equals the number of rows of the second matrix B.
  2. If A is m × n and B is n × p, then AB exists and has order m × p.
  3. Example: a 2 × 3 matrix can be multiplied by a 3 × 2 matrix, but not by a 2 × 2 matrix.
(vi) Define matrix with example.
Answer
  1. Definition: A matrix is a rectangular arrangement of numbers (entries) in rows and columns, enclosed in brackets [ ].
  2. Example: A = [1 2 3; 4 5 6] has 2 rows and 3 columns, so its order is 2 × 3.
(vii) If [
4a
b3
] [
6
6
] = [
6
3
], then find the values of a and b.
Answer: a = −3, b = −5/2
  1. (The second matrix was unclear in the original print, so it is taken as the column [6; 6].)
  2. Multiply: Row 1: (4)(6) + (a)(6) = 24 + 6a. Row 2: (b)(6) + (3)(6) = 6b + 18.
  3. So [24 + 6a; 6b + 18] = [6; 3].
  4. Finding a: 24 + 6a = 6 ⟹ 6a = −18 ⟹ a = −3.
  5. Finding b: 6b + 18 = 3 ⟹ 6b = −15 ⟹ b = −5/2.
  6. Check: 24 + 6(−3) = 6 ✔ and 6(−5/2) + 18 = 3 ✔
(viii) If [
x1
y2
] [
10
3−1
] = [
7−1
4−2
], then find the values of x and y.
Answer: x = 4, y = −2
  1. Multiply the left side:
    (1,1): (x)(1) + (1)(3) = x + 3; (1,2): (x)(0) + (1)(−1) = −1
    (2,1): (y)(1) + (2)(3) = y + 6; (2,2): (y)(0) + (2)(−1) = −2
  2. So [x + 3, −1; y + 6, −2] = [7, −1; 4, −2].
  3. x + 3 = 7 ⟹ x = 4.
  4. y + 6 = 4 ⟹ y = −2.
  5. The entries −1 and −2 already match ✔
(ix) Define Singular Matrix with example.
Answer
  1. Definition: A square matrix A is called singular if its determinant is zero, |A| = 0. A singular matrix has no multiplicative inverse.
  2. Example: A = [2 4; 1 2]. |A| = (2)(2) − (4)(1) = 4 − 4 = 0, so A is singular.
4. Write short answers to any six (06) questions:
(i) Take any 2 by 2 matrix and check whether it is singular or non-singular. Also find its adjoint.
Answer (example): A = [3 2; 1 4] is non-singular, Adj A = [4 −2; −1 3]
  1. Take A = [3 2; 1 4].
  2. |A| = (3)(4) − (2)(1) = 12 − 2 = 10. Since |A| ≠ 0, A is non-singular.
  3. Adjoint: swap the diagonal entries 3 and 4, and change the signs of 2 and 1: Adj A = [4 −2; −1 3].
(ii) Find the values of each of the determinant.
105
46
Answer: 40
  1. Value = ad − bc with a = 10, b = 5, c = 4, d = 6.
  2. = (10)(6) − (5)(4) = 60 − 20 = 40.
(iii) Find whether the matrix A = [
53
32
] is singular or non-singular.
Answer: non-singular
  1. |A| = (5)(2) − (3)(3) = 10 − 9 = 1.
  2. Since |A| ≠ 0, A is non-singular.
(iv) Find the value of x when A = [
x6
515
] is a singular matrix.
Answer: x = 2
  1. A is singular when |A| = 0.
  2. |A| = (x)(15) − (6)(5) = 15x − 30.
  3. 15x − 30 = 0 ⟹ 15x = 30 ⟹ x = 2.
  4. Check: (2)(15) − (6)(5) = 0 ✔
(v) Find multiplicative inverse of the matrix [
50
05
]
Answer: [1/5 0; 0 1/5]
  1. Let A = [5 0; 0 5]. |A| = (5)(5) − (0)(0) = 25 ≠ 0, so the inverse exists.
  2. Adj A: swap the diagonal entries (5 and 5), change the signs of 0 and 0 ⟹ Adj A = [5 0; 0 5].
  3. A−1 = (1/25) [5 0; 0 5] = [1/5 0; 0 1/5].
  4. Check: A·A−1 = [5·(1/5), 0; 0, 5·(1/5)] = [1 0; 0 1] = I ✔
(vi) Find the adjoint of the matrix C = [
−35
−32
]
Answer: Adj C = [2 −5; 3 −3]
  1. For [a b; c d], Adj = [d −b; −c a] (swap the diagonal, change the signs of the other two).
  2. Here a = −3, b = 5, c = −3, d = 2.
  3. Diagonal swapped: 2 and −3. Signs changed: −b = −5, −c = 3.
  4. Adj C = [2 −5; 3 −3].
(vii) Solve the following by matrix inversion method:
2x + 3y = 13
4x − 5y = −7
Answer: x = 2, y = 3
  1. AX = B with A = [2 3; 4 −5], X = [x; y], B = [13; −7].
  2. |A| = (2)(−5) − (3)(4) = −22 ≠ 0.
  3. Adj A = [−5 −3; −4 2], so A−1 = (−1/22) [−5 −3; −4 2].
  4. X = A−1B = (−1/22) [(−5)(13) + (−3)(−7); (−4)(13) + (2)(−7)] = (−1/22) [−44; −66] = [2; 3].
  5. x = 2, y = 3. Check: 4 + 9 = 13 ✔, 8 − 15 = −7 ✔
(viii) Use Cramer’s rule to solve the system of equations:
2x + 3y = 13
4x − 5y = −7
Answer: x = 2, y = 3
  1. D (coefficients) = |2 3; 4 −5| = (2)(−5) − (3)(4) = −22.
  2. Dx (replace the x-column by 13, −7) = |13 3; −7 −5| = (13)(−5) − (3)(−7) = −65 + 21 = −44.
  3. Dy (replace the y-column by 13, −7) = |2 13; 4 −7| = (2)(−7) − (13)(4) = −14 − 52 = −66.
  4. x = Dx / D = −44 / −22 = 2.
  5. y = Dy / D = −66 / −22 = 3.
(ix) Define Symmetric Matrix with example.
Answer
  1. Definition: A square matrix A is called symmetric if At = A, that is, the entry in row i, column j equals the entry in row j, column i.
  2. Example: A = [1 2; 2 3]. Then At = [1 2; 2 3] = A, so A is symmetric.
(Part-II)
Note: Attempt any two (02) questions.
5:(a) Show that the following matrices are multiplicative inverse of each other.
[
2−1
−32
], [
21
32
]
Shown: AB = BA = I
  1. Let A = [2 −1; −3 2] and B = [2 1; 3 2]. They are inverses of each other if AB = BA = I.
  2. AB: (1,1): (2)(2) + (−1)(3) = 4 − 3 = 1; (1,2): (2)(1) + (−1)(2) = 0; (2,1): (−3)(2) + (2)(3) = 0; (2,2): (−3)(1) + (2)(2) = 1 ⟹ AB = [1 0; 0 1] = I.
  3. BA: (1,1): (2)(2) + (1)(−3) = 1; (1,2): (2)(−1) + (1)(2) = 0; (2,1): (3)(2) + (2)(−3) = 0; (2,2): (3)(−1) + (2)(2) = 1 ⟹ BA = [1 0; 0 1] = I.
  4. Since AB = BA = I, A and B are multiplicative inverses of each other ✔
(b) Take any two linear equations in two variables and solve them by matrix inversion method.
Answer (example): x + y = 5, x − y = 1 gives x = 3, y = 2
  1. Take the equations x + y = 5 and x − y = 1.
  2. AX = B with A = [1 1; 1 −1], X = [x; y], B = [5; 1].
  3. |A| = (1)(−1) − (1)(1) = −2 ≠ 0.
  4. Adj A = [−1 −1; −1 1], so A−1 = (−1/2) [−1 −1; −1 1].
  5. X = A−1B = (−1/2) [(−1)(5) + (−1)(1); (−1)(5) + (1)(1)] = (−1/2) [−6; −4] = [3; 2].
  6. x = 3, y = 2. Check: 3 + 2 = 5 ✔, 3 − 2 = 1 ✔
6:(a) If A = [
85
43
], then find multiplicative inverse of A and verify that AA−1 = A−1A = I.
Answer: A−1 = (1/4)[3 −5; −4 8]
  1. |A| = (8)(3) − (5)(4) = 24 − 20 = 4 ≠ 0.
  2. Adj A: swap 8 and 3, change the signs of 5 and 4 ⟹ Adj A = [3 −5; −4 8].
  3. A−1 = (1/4) [3 −5; −4 8] = [3/4 −5/4; −1 2].
  4. AA−1: A · Adj A: (1,1): 24 − 20 = 4; (1,2): −40 + 40 = 0; (2,1): 12 − 12 = 0; (2,2): −20 + 24 = 4 ⟹ [4 0; 0 4]. Multiply by 1/4 ⟹ [1 0; 0 1] = I ✔
  5. A−1A: Adj A · A: (1,1): (3)(8) + (−5)(4) = 4; (1,2): (3)(5) + (−5)(3) = 0; (2,1): (−4)(8) + (8)(4) = 0; (2,2): (−4)(5) + (8)(3) = 4 ⟹ [4 0; 0 4]. Multiply by 1/4 ⟹ I ✔
  6. Hence AA−1 = A−1A = I.
(b) Two cyclists are 44 km apart and start out at the same time. If they go towards one another they meet in 2 hours, but if they go in the same direction the faster overtakes the slower in 71⁄2 hours. Find their speeds by using matrices.
Answer: faster = 209/15 ≈ 13.93 km/h, slower = 121/15 ≈ 8.07 km/h
  1. Let the faster speed = x km/h and the slower speed = y km/h.
  2. Towards each other: they cover 44 km together in 2 hours ⟹ 2(x + y) = 44 ⟹ x + y = 22.
  3. Same direction: the faster gains 44 km in 7½ hours ⟹ 7.5(x − y) = 44 ⟹ x − y = 44/7.5 = 88/15.
  4. AX = B with A = [1 1; 1 −1], X = [x; y], B = [22; 88/15].
  5. |A| = (1)(−1) − (1)(1) = −2. Adj A = [−1 −1; −1 1].
  6. Adj A · B: Row 1: −22 − 88/15 = −418/15; Row 2: −22 + 88/15 = −242/15.
  7. X = (−1/2) [−418/15; −242/15] = [209/15; 121/15].
  8. So the faster cyclist goes at 209/15 ≈ 13.93 km/h and the slower at 121/15 ≈ 8.07 km/h.
  9. Check: 209/15 + 121/15 = 330/15 = 22 ✔ and (209 − 121)/15 = 88/15, and (7.5)(88/15) = 44 ✔
7:(a) Two years ago a man was 5 times as old as his son was. After 6 years he will be 3 times as old as his son. Find their present ages by using matrices.
Answer: man = 42 years, son = 10 years
  1. Let the man’s present age = x and the son’s present age = y.
  2. Two years ago: x − 2 = 5(y − 2) ⟹ x − 5y = −8.
  3. After 6 years: x + 6 = 3(y + 6) ⟹ x − 3y = 12.
  4. AX = B with A = [1 −5; 1 −3], X = [x; y], B = [−8; 12].
  5. |A| = (1)(−3) − (−5)(1) = −3 + 5 = 2. Adj A = [−3 5; −1 1]. A−1 = (1/2) [−3 5; −1 1].
  6. Adj A · B: Row 1: (−3)(−8) + (5)(12) = 24 + 60 = 84; Row 2: (−1)(−8) + (1)(12) = 20.
  7. X = (1/2) [84; 20] = [42; 10].
  8. The man is 42 years old and the son is 10 years old.
  9. Check: 40 = 5 × 8 ✔ and 48 = 3 × 16 ✔
(b) If A = [
21
32
], B = [
34
24
], then verify (AB)−1 = B−1A−1. Can we say it is also a Reversal Law of Multiplicative Inverse?
Verified: (AB)−1 = B−1A−1 = [5 −3; −13/4 2]
  1. L.H.S. AB: (1,1): (2)(3) + (1)(2) = 8; (1,2): (2)(4) + (1)(4) = 12; (2,1): (3)(3) + (2)(2) = 13; (2,2): (3)(4) + (2)(4) = 20 ⟹ AB = [8 12; 13 20].
  2. |AB| = (8)(20) − (12)(13) = 160 − 156 = 4. Adj(AB) = [20 −12; −13 8].
  3. (AB)−1 = (1/4) [20 −12; −13 8] = [5 −3; −13/4 2] …(1)
  4. R.H.S. |A| = (2)(2) − (1)(3) = 1; Adj A = [2 −1; −3 2]; A−1 = [2 −1; −3 2].
  5. |B| = (3)(4) − (4)(2) = 4; Adj B = [4 −4; −2 3]; B−1 = (1/4) [4 −4; −2 3] = [1 −1; −1/2 3/4].
  6. B−1A−1 = [1 −1; −1/2 3/4] [2 −1; −3 2]:
    (1,1): (1)(2) + (−1)(−3) = 5; (1,2): (1)(−1) + (−1)(2) = −3
    (2,1): (−1/2)(2) + (3/4)(−3) = −1 − 9/4 = −13/4; (2,2): (−1/2)(−1) + (3/4)(2) = 1/2 + 3/2 = 2
    B−1A−1 = [5 −3; −13/4 2] …(2)
  7. From (1) and (2), L.H.S. = R.H.S. ✔
  8. Yes, this is the reversal law of the multiplicative inverse: the inverse of a product is the product of the inverses taken in reverse order.
(Part-III)
Note: Attempt any one (01) question.
8:(a) Eight years ago Huria’s was 34 of Jannat’s age.
After four years Huria’s age will be 67 of Jannat’s age.
Find their present ages by using matrices.
Answer: Huria = 20 years, Jannat = 24 years
  1. Let Huria’s present age = x and Jannat’s present age = y.
  2. Eight years ago: x − 8 = (3/4)(y − 8) ⟹ 4x − 32 = 3y − 24 ⟹ 4x − 3y = 8.
  3. After four years: x + 4 = (6/7)(y + 4) ⟹ 7x + 28 = 6y + 24 ⟹ 7x − 6y = −4.
  4. AX = B with A = [4 −3; 7 −6], X = [x; y], B = [8; −4].
  5. |A| = (4)(−6) − (−3)(7) = −24 + 21 = −3. Adj A = [−6 3; −7 4]. A−1 = (−1/3) [−6 3; −7 4].
  6. Adj A · B: Row 1: (−6)(8) + (3)(−4) = −48 − 12 = −60; Row 2: (−7)(8) + (4)(−4) = −56 − 16 = −72.
  7. X = (−1/3) [−60; −72] = [20; 24].
  8. Huria is 20 years old and Jannat is 24 years old.
  9. Check: 8 years ago 12 and 16, and 12 = (3/4)(16) ✔. After 4 years 24 and 28, and 24 = (6/7)(28) ✔
(b) Prove that A = [
4−2
−64
] and B = [
112
321
] are multiplicative inverse of each other.
Proved: AB = BA = I
  1. A = [4 −2; −6 4] and B = [1 1/2; 3/2 1].
  2. AB:
    (1,1): (4)(1) + (−2)(3/2) = 4 − 3 = 1
    (1,2): (4)(1/2) + (−2)(1) = 2 − 2 = 0
    (2,1): (−6)(1) + (4)(3/2) = −6 + 6 = 0
    (2,2): (−6)(1/2) + (4)(1) = −3 + 4 = 1
    AB = [1 0; 0 1] = I
  3. BA:
    (1,1): (1)(4) + (1/2)(−6) = 4 − 3 = 1
    (1,2): (1)(−2) + (1/2)(4) = −2 + 2 = 0
    (2,1): (3/2)(4) + (1)(−6) = 6 − 6 = 0
    (2,2): (3/2)(−2) + (1)(4) = −3 + 4 = 1
    BA = [1 0; 0 1] = I
  4. Since AB = BA = I, A and B are multiplicative inverses of each other ✔
9:(a) Solve by matrix inversion method, if possible.
2x + 5y = 19
4x − 3y = −1
Answer: x = 2, y = 3
  1. AX = B with A = [2 5; 4 −3], X = [x; y], B = [19; −1].
  2. |A| = (2)(−3) − (5)(4) = −6 − 20 = −26 ≠ 0, so the inverse exists and the system can be solved.
  3. Adj A: swap 2 and −3, change the signs of 5 and 4 ⟹ [−3 −5; −4 2]. A−1 = (−1/26) [−3 −5; −4 2].
  4. Adj A · B: Row 1: (−3)(19) + (−5)(−1) = −57 + 5 = −52; Row 2: (−4)(19) + (2)(−1) = −76 − 2 = −78.
  5. X = (−1/26) [−52; −78] = [2; 3]. So x = 2, y = 3.
  6. Check: 2(2) + 5(3) = 19 ✔ and 4(2) − 3(3) = −1 ✔
(b) Three forces act on a particle and must be in equilibrium i.e. F1 + F2 + F3 = 0, where F1 = [
8
x
], F2 = [
−2
−7
], F3 = [
y
−1
]. Find the value of x and y.
Answer: x = 8, y = −6
  1. Add the three column matrices (entry by entry): F1 + F2 + F3 = [8 + (−2) + y; x + (−7) + (−1)] = [6 + y; x − 8].
  2. For equilibrium this must equal the zero matrix [0; 0].
  3. Top entry: 6 + y = 0 ⟹ y = −6.
  4. Bottom entry: x − 8 = 0 ⟹ x = 8.
  5. Check: 8 − 2 − 6 = 0 ✔ and 8 − 7 − 1 = 0 ✔
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