Notation: a matrix is shown as a small table; a determinant is shown as a table after a label such as |A| = or Δ =. Mᵢⱼ is the minor and Aᵢⱼ the cofactor of aᵢⱼ, adj A is the adjoint of A, and Rᵢ, Cⱼ denote rows and columns.
Exercise 4.1
Evaluate the determinants in Exercises 1 and 2.
2 | 4 |
–5 | –1 |
Ans. Δ = 2(–1) – 4(–5) = –2 + 20 = 18.
Q2. (i) Δ =
cos θ | –sin θ |
sin θ | cos θ |
(ii) Δ =
x² – x + 1 | x – 1 |
x + 1 | x + 1 |
Ans. (i) Δ = cos θ · cos θ – (–sin θ)(sin θ) = cos²θ + sin²θ = 1.
(ii) Δ = (x² – x + 1)(x + 1) – (x – 1)(x + 1) = (x³ + 1) – (x² – 1) = x³ – x² + 2.
Q3. If A =
1 | 2 |
4 | 2 |
then show that |2A| = 4|A|
Ans. |A| = 1(2) – 2(4) = –6. Now 2A =
2 | 4 |
8 | 4 |
|2A| = 2(4) – 4(8) = 8 – 32 = –24 = 4 × (–6) = 4|A|. Hence shown.
Q4. If A =
1 | 0 | 1 |
0 | 1 | 2 |
0 | 0 | 4 |
then show that |3A| = 27|A|
Ans. Expanding along C₁: |A| = 1 × (1 × 4 – 2 × 0) = 4. Now 3A =
3 | 0 | 3 |
0 | 3 | 6 |
0 | 0 | 12 |
Expanding along C₁: |3A| = 3 × (3 × 12 – 6 × 0) = 108 = 27 × 4 = 27|A|. Hence shown.
Q5. Evaluate the determinants
(i)
3 | –1 | –2 |
0 | 0 | –1 |
3 | –5 | 0 |
(ii)
3 | –4 | 5 |
1 | 1 | –2 |
2 | 3 | 1 |
(iii)
0 | 1 | 2 |
–1 | 0 | –3 |
–2 | 3 | 0 |
(iv)
2 | –1 | –2 |
0 | 2 | –1 |
3 | –5 | 0 |
Ans. (i) Expand along R₂ (two zero entries): Δ = (–1) × A₂₃, where A₂₃ = (–1)²⁺³ × [3(–5) – (–1)(3)] = –(–12) = 12 → Δ = –12.
(ii) Δ = expanding along R₁: 3 × 7 – (–4) × 5 + 5 × 1 = 46.
(iii) Δ = expanding along R₁: 0 × 9 – 1 × (–6) + 2 × (–3) = 0.
(iv) Δ = expanding along R₁: 2 × (–5) – (–1) × 3 + (–2) × (–6) = 5.
Q6. If A =
1 | 1 | –2 |
2 | 1 | –3 |
5 | 4 | –9 |
find |A|
Ans. |A| = expanding along R₁: 1 × 3 – 1 × (–3) + (–2) × 3 = 0.
Q7. Find values of x, if
(i)
2 | 4 |
5 | 1 |
=
2x | 4 |
6 | x |
(ii)
2 | 3 |
4 | 5 |
=
x | 3 |
2x | 5 |
Ans. (i) 2 – 20 = 2x² – 24 → 2x² = 6 → x² = 3 → x = ±√3.
(ii) 10 – 12 = 5x – 6x → –2 = –x → x = 2.
Q8. If
x | 2 |
18 | x |
=
6 | 2 |
18 | 6 |
then x is equal to (A) 6 (B) ±6 (C) –6 (D) 0
Ans. x² – 36 = 36 – 36 = 0 → x² = 36 → x = ±6. Correct answer: (B).
Exercise 4.2
Find area of the triangle with vertices at the point given in each of the following: (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8)
Answer
Area = (1/2)|Δ|, where Δ is the determinant with rows (x₁, y₁, 1), (x₂, y₂, 1), (x₃, y₃, 1).
(i) Δ =
1 | 0 | 1 |
6 | 0 | 1 |
4 | 3 | 1 |
Δ = expanding along R₁: 1 × (–3) – 0 × 2 + 1 × 18 = 15
⇒ Area = (1/2) × |15| = 15/2 sq units.
(ii) Δ =
2 | 7 | 1 |
1 | 1 | 1 |
10 | 8 | 1 |
Δ = expanding along R₁: 2 × (–7) – 7 × (–9) + 1 × (–2) = 47
⇒ Area = (1/2) × |47| = 47/2 sq units.
(iii) Δ =
–2 | –3 | 1 |
3 | 2 | 1 |
–1 | –8 | 1 |
Δ = expanding along R₁: (–2) × 10 – (–3) × 4 + 1 × (–22) = –30
⇒ Area = (1/2) × |–30| = 15 sq units.
Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.
Answer
The points are collinear if the area of triangle ABC is 0, i.e. if Δ = 0, where Δ =
a | b + c | 1 |
b | c + a | 1 |
c | a + b | 1 |
Apply C₁
⇒ C₁ + C₂: every entry of C₁ becomes a + b + c. Taking (a + b + c) common from C₁, Δ = (a + b + c) ×
1 | b + c | 1 |
1 | c + a | 1 |
1 | a + b | 1 |
C₁ and C₃ are identical, so this determinant is 0. Hence Δ = 0, the area is zero and A, B, C are collinear.
Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)
Answer
(i) Δ =
k | 0 | 1 |
4 | 0 | 1 |
0 | 2 | 1 |
Δ = k(0 – 2) – 0 + 1(8 – 0) = 8 – 2k. (1/2)|8 – 2k| = 4
⇒ 8 – 2k = ±8
⇒ k = 0 or k = 8.
(ii) Δ =
–2 | 0 | 1 |
0 | 4 | 1 |
0 | k | 1 |
Δ = –2(4 – k) – 0 + 1(0 – 0) = 2k – 8. (1/2)|2k – 8| = 4
⇒ 2k – 8 = ±8
⇒ k = 8 or k = 0.
(i) Find equation of line joining (1, 2) and (3, 6) using determinants. (ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
Answer
Let P(x, y) be any point on the line. Then the area of the triangle formed by P and the two given points is 0.
(i) Δ =
x | y | 1 |
1 | 2 | 1 |
3 | 6 | 1 |
= 0
⇒ x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0
⇒ –4x + 2y = 0
⇒ y = 2x.
(ii) Δ =
x | y | 1 |
3 | 1 | 1 |
9 | 3 | 1 |
= 0
⇒ x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0
⇒ –2x + 6y = 0
⇒ x – 3y = 0.
If area of triangle is 35 sq units with vertices (2, –6), (5, 4) and (k, 4). Then k is (A) 12 (B) –2 (C) –12, –2 (D) 12, –2
Answer
Δ =
2 | –6 | 1 |
5 | 4 | 1 |
k | 4 | 1 |
Δ = 2(4 – 4) + 6(5 – k) + 1(20 – 4k) = 50 – 10k. (1/2)|50 – 10k| = 35
⇒ 50 – 10k = ±70
⇒ k = –2 or k = 12. Correct answer: (D).
Exercise 4.3
Write Minors and Cofactors of the elements of following determinants
Q1. (i)
2 | –4 |
0 | 3 |
(ii)
a | c |
b | d |
Ans. (i) M₁₁ = 3, M₁₂ = 0, M₂₁ = –4, M₂₂ = 2. Cofactors Aᵢⱼ = (–1)ⁱ⁺ʲ Mᵢⱼ: A₁₁ = 3, A₁₂ = 0, A₂₁ = 4, A₂₂ = 2.
(ii) M₁₁ = d, M₁₂ = b, M₂₁ = c, M₂₂ = a. A₁₁ = d, A₁₂ = –b, A₂₁ = –c, A₂₂ = a.
Q2. (i)
1 | 0 | 0 |
0 | 1 | 0 |
0 | 0 | 1 |
(ii)
1 | 0 | 4 |
3 | 5 | –1 |
0 | 1 | 2 |
Ans. (i) Each minor is a 2 × 2 determinant obtained by deleting row i and column j. Minors Mᵢⱼ (in position (i, j)):
1 | 0 | 0 |
0 | 1 | 0 |
0 | 0 | 1 |
Cofactors Aᵢⱼ = (–1)ⁱ⁺ʲ Mᵢⱼ:
1 | 0 | 0 |
0 | 1 | 0 |
0 | 0 | 1 |
So M₁₁ = M₂₂ = M₃₃ = 1 = A₁₁ = A₂₂ = A₃₃ and all other minors and cofactors are 0.
(ii) For example, M₁₁ = 5(2) – (–1)(1) = 11, M₁₂ = 3(2) – (–1)(0) = 6, M₁₃ = 3(1) – 5(0) = 3. All minors Mᵢⱼ:
11 | 6 | 3 |
–4 | 2 | 1 |
–20 | –13 | 5 |
Cofactors Aᵢⱼ = (–1)ⁱ⁺ʲ Mᵢⱼ:
11 | –6 | 3 |
4 | 2 | –1 |
–20 | 13 | 5 |
Q3. Using Cofactors of elements of second row, evaluate Δ =
5 | 3 | 8 |
2 | 0 | 1 |
1 | 2 | 3 |
Ans. A₂₁ = (–1)³ × (3 × 3 – 8 × 2) = –(–7) = 7; A₂₂ = (–1)⁴ × (5 × 3 – 8 × 1) = 7; A₂₃ = (–1)⁵ × (5 × 2 – 3 × 1) = –7.
Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 2(7) + 0(7) + 1(–7) = 14 + 0 – 7 = 7.
Q4. Using Cofactors of elements of third column, evaluate Δ =
1 | x | yz |
1 | y | zx |
1 | z | xy |
Ans. A₁₃ = (–1)⁴ × (z – y) = z – y; A₂₃ = (–1)⁵ × (z – x) = x – z; A₃₃ = (–1)⁶ × (y – x) = y – x.
Δ = yz(z – y) + zx(x – z) + xy(y – x) = yz² – y²z + x²z – xz² + xy² – x²y.
Factorising: Δ = (x – y)(y – z)(z – x).
Q5. If Δ =
a₁₁ | a₁₂ | a₁₃ |
a₂₁ | a₂₂ | a₂₃ |
a₃₁ | a₃₂ | a₃₃ |
and Aᵢⱼ is Cofactors of aᵢⱼ, then value of Δ is given by (A) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃ (B) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁ (C) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃ (D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Ans. Δ equals the sum of products of the elements of any row or column with their own cofactors. (D) is the expansion along C₁. Options (A) and (C) multiply elements of one row by cofactors of another row (sum = 0), and (B) mixes rows and columns. Correct answer: (D).
Exercise 4.4
Find adjoint of each of the matrices in Exercises 1 and 2.
1 | 2 |
3 | 4 |
Answer
A₁₁ = 4, A₁₂ = –3, A₂₁ = –2, A₂₂ = 1. adj A = transpose of the cofactor matrix =
4 | –2 |
–3 | 1 |
1 | –1 | 2 |
2 | 3 | 5 |
–2 | 0 | 1 |
Answer
Cofactors: A₁₁ = 3 – 0 = 3, A₁₂ = –(2 + 10) = –12, A₁₃ = 0 + 6 = 6, A₂₁ = –(–1 – 0) = 1, A₂₂ = 1 + 4 = 5, A₂₃ = –(0 – 2) = 2, A₃₁ = –5 – 6 = –11, A₃₂ = –(5 – 4) = –1, A₃₃ = 3 + 2 = 5.
adj A =
3 | 1 | –11 |
–12 | 5 | –1 |
6 | 2 | 5 |
Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4.
2 | 3 |
–4 | –6 |
Answer
|A| = 2 × (–6) – 3 × (–4) = 0. adj A =
–6 | –3 |
4 | 2 |
A (adj A) =
(adj A) A =
|A| I = 0 I =
All three are equal. Verified.
1 | –1 | 2 |
3 | 0 | –2 |
1 | 0 | 3 |
Answer
|A| = expanding along R₁: 1 × 0 – (–1) × 11 + 2 × 0 = 11. adj A =
0 | 3 | 2 |
–11 | 1 | 8 |
0 | –1 | 3 |
A (adj A) =
11 | 0 | 0 |
0 | 11 | 0 |
0 | 0 | 11 |
(adj A) A =
11 | 0 | 0 |
0 | 11 | 0 |
0 | 0 | 11 |
|A| I = 11 I =
11 | 0 | 0 |
0 | 11 | 0 |
0 | 0 | 11 |
All three are equal. Verified.
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
2 | –2 |
4 | 3 |
Answer
|A| = 2 × 3 – (–2) × 4 = 14.
adj A = transpose of the cofactor matrix =
A⁻¹ = (1/|A|) adj A = (1/14) ×
–1 | 5 |
–3 | 2 |
Answer
|A| = (–1) × 2 – 5 × (–3) = 13.
adj A = transpose of the cofactor matrix =
A⁻¹ = (1/|A|) adj A = (1/13) ×
1 | 2 | 3 |
0 | 2 | 4 |
0 | 0 | 5 |
Answer
|A| = expanding along R₁: 1 × 10 – 2 × 0 + 3 × 0 = 10.
Cofactors Aᵢⱼ (placed in position (i, j)):
10 | 0 | 0 |
–10 | 5 | 0 |
2 | –4 | 2 |
adj A = transpose of the cofactor matrix =
10 | –10 | 2 |
0 | 5 | –4 |
0 | 0 | 2 |
A⁻¹ = (1/|A|) adj A = (1/10) ×
10 | –10 | 2 |
0 | 5 | –4 |
0 | 0 | 2 |
1 | 0 | 0 |
3 | 3 | 0 |
5 | 2 | –1 |
Answer
|A| = expanding along R₁: 1 × (–3) – 0 × (–3) + 0 × (–9) = –3.
Cofactors Aᵢⱼ (placed in position (i, j)):
–3 | 3 | –9 |
0 | –1 | –2 |
0 | 0 | 3 |
adj A = transpose of the cofactor matrix =
–3 | 0 | 0 |
3 | –1 | 0 |
–9 | –2 | 3 |
A⁻¹ = (1/|A|) adj A = (–1/3) ×
–3 | 0 | 0 |
3 | –1 | 0 |
–9 | –2 | 3 |
2 | 1 | 3 |
4 | –1 | 0 |
–7 | 2 | 1 |
Answer
|A| = expanding along R₁: 2 × (–1) – 1 × 4 + 3 × 1 = –3.
Cofactors Aᵢⱼ (placed in position (i, j)):
–1 | –4 | 1 |
5 | 23 | –11 |
3 | 12 | –6 |
adj A = transpose of the cofactor matrix =
–1 | 5 | 3 |
–4 | 23 | 12 |
1 | –11 | –6 |
A⁻¹ = (1/|A|) adj A = (–1/3) ×
–1 | 5 | 3 |
–4 | 23 | 12 |
1 | –11 | –6 |
1 | –1 | 2 |
0 | 2 | –3 |
3 | –2 | 4 |
Answer
|A| = expanding along R₁: 1 × 2 – (–1) × 9 + 2 × (–6) = –1.
Cofactors Aᵢⱼ (placed in position (i, j)):
2 | –9 | –6 |
0 | –2 | –1 |
–1 | 3 | 2 |
adj A = transpose of the cofactor matrix =
2 | 0 | –1 |
–9 | –2 | 3 |
–6 | –1 | 2 |
A⁻¹ = (1/|A|) adj A = (–1) ×
2 | 0 | –1 |
–9 | –2 | 3 |
–6 | –1 | 2 |
1 | 0 | 0 |
0 | cos α | sin α |
0 | sin α | –cos α |
Answer
|A| = 1 × (–cos²α – sin²α) = –1 ≠ 0, so A⁻¹ exists.
Cofactors: A₁₁ = –cos²α – sin²α = –1, A₂₂ = –cos α, A₂₃ = –sin α, A₃₂ = –sin α, A₃₃ = cos α, all others 0. adj A =
–1 | 0 | 0 |
0 | –cos α | –sin α |
0 | –sin α | cos α |
A⁻¹ = (–1) × adj A =
1 | 0 | 0 |
0 | cos α | sin α |
0 | sin α | –cos α |
(So A⁻¹ = A.)
Q12. Let A =
3 | 7 |
2 | 5 |
and B =
6 | 8 |
7 | 9 |
Verify that (AB)⁻¹ = B⁻¹A⁻¹.
Ans. |A| = 15 – 14 = 1 → A⁻¹ =
5 | –7 |
–2 | 3 |
|B| = 54 – 56 = –2 → B⁻¹ = (–1/2) ×
9 | –8 |
–7 | 6 |
=
–9/2 | 4 |
7/2 | –3 |
AB =
67 | 87 |
47 | 61 |
|AB| = –2 → (AB)⁻¹ = (–1/2) adj(AB) = (–1/2) ×
61 | –87 |
–47 | 67 |
=
B⁻¹A⁻¹ =
Both are equal, so (AB)⁻¹ = B⁻¹A⁻¹. Verified.
Q13. If A =
3 | 1 |
–1 | 2 |
show that A² – 5A + 7I = O. Hence find A⁻¹.
Ans. A² =
8 | 5 |
–5 | 3 |
A² – 5A + 7I =
= O. Multiplying A² – 5A + 7I = O by A⁻¹: A – 5I + 7A⁻¹ = O → A⁻¹ = (1/7)(5I – A) = (1/7) ×
2 | –1 |
1 | 3 |
Q14. For the matrix A =
3 | 2 |
1 | 1 |
find the numbers a and b such that A² + aA + bI = O.
Ans. A² =
11 | 8 |
4 | 3 |
A² + aA + bI =
11 + 3a + b | 8 + 2a |
4 + a | 3 + a + b |
= O → 4 + a = 0 → a = –4; 3 + a + b = 0 → b = 1 (check: 11 – 12 + 1 = 0, 8 – 8 = 0). So a = –4, b = 1.
Q15. For the matrix A =
1 | 1 | 1 |
1 | 2 | –3 |
2 | –1 | 3 |
Show that A³ – 6A² + 5A + 11I = O. Hence, find A⁻¹.
Ans. A² =
4 | 2 | 1 |
–3 | 8 | –14 |
7 | –3 | 14 |
A³ = A² × A =
8 | 7 | 1 |
–23 | 27 | –69 |
32 | –13 | 58 |
A³ – 6A² + 5A + 11I =
0 | 0 | 0 |
0 | 0 | 0 |
0 | 0 | 0 |
= O. Proved.
Multiply by A⁻¹: A² – 6A + 5I + 11A⁻¹ = O → A⁻¹ = (–1/11)(A² – 6A + 5I). A² – 6A + 5I =
3 | –4 | –5 |
–9 | 1 | 4 |
–5 | 3 | 1 |
So A⁻¹ = (–1/11) ×
3 | –4 | –5 |
–9 | 1 | 4 |
–5 | 3 | 1 |
= (1/11) ×
–3 | 4 | 5 |
9 | –1 | –4 |
5 | –3 | –1 |
Q16. If A =
2 | –1 | 1 |
–1 | 2 | –1 |
1 | –1 | 2 |
Verify that A³ – 6A² + 9A – 4I = O and hence find A⁻¹
Ans. A² =
6 | –5 | 5 |
–5 | 6 | –5 |
5 | –5 | 6 |
A³ =
22 | –21 | 21 |
–21 | 22 | –21 |
21 | –21 | 22 |
A³ – 6A² + 9A – 4I =
0 | 0 | 0 |
0 | 0 | 0 |
0 | 0 | 0 |
= O. Verified.
Multiply by A⁻¹: A² – 6A + 9I – 4A⁻¹ = O → A⁻¹ = (1/4)(A² – 6A + 9I) = (1/4) ×
3 | 1 | –1 |
1 | 3 | 1 |
–1 | 1 | 3 |
Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to (A) |A| (B) |A|² (C) |A|³ (D) 3|A|
Answer
A (adj A) = |A| I
⇒ |A| |adj A| = |A|³
⇒ |adj A| = |A|ⁿ⁻¹ = |A|². Correct answer: (B).
If A is an invertible matrix of order 2, then det (A⁻¹) is equal to (A) det (A) (B) 1/det (A) (C) 1 (D) 0
Answer
AA⁻¹ = I
⇒ |A| |A⁻¹| = 1
⇒ det(A⁻¹) = 1/det(A). Correct answer: (B).
Exercise 4.5
Examine the consistency of the system of equations in Exercises 1 to 6.
x + 2y = 2; 2x + 3y = 3
Answer
A =
1 | 2 |
2 | 3 |
|A| = 3 – 4 = –1 ≠ 0, so A⁻¹ exists and the system has a unique solution. The system is consistent.
2x – y = 5; x + y = 4
Answer
A =
2 | –1 |
1 | 1 |
|A| = 2 + 1 = 3 ≠ 0
⇒ unique solution
⇒ consistent.
x + 3y = 5; 2x + 6y = 8
Answer
A =
1 | 3 |
2 | 6 |
|A| = 6 – 6 = 0, so check (adj A)B. adj A =
6 | –3 |
–2 | 1 |
(adj A)B =
6 |
–2 |
≠ O, so the system has no solution. The system is inconsistent.
x + y + z = 1; 2x + 3y + 2z = 2; ax + ay + 2az = 4
Answer
A =
1 | 1 | 1 |
2 | 3 | 2 |
a | a | 2a |
|A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) = 4a – 2a – a = a.
For a ≠ 0, |A| ≠ 0, so the system has a unique solution and is consistent (this is the NCERT answer). (If a = 0, the third equation reads 0 = 4, which is impossible.)
3x – y – 2z = 2; 2y – z = –1; 3x – 5y = 3
Answer
A =
3 | –1 | –2 |
0 | 2 | –1 |
3 | –5 | 0 |
|A| = expanding along R₁: 3 × (–5) – (–1) × 3 + (–2) × (–6) = 0. adj A =
–5 | 10 | 5 |
–3 | 6 | 3 |
–6 | 12 | 6 |
(adj A)B =
–5 |
–3 |
–6 |
≠ O, so no solution exists. The system is inconsistent.
5x – y + 4z = 5; 2x + 3y + 5z = 2; 5x – 2y + 6z = –1
Answer
A =
5 | –1 | 4 |
2 | 3 | 5 |
5 | –2 | 6 |
|A| = expanding along R₁: 5 × 28 – (–1) × (–13) + 4 × (–19) = 51 ≠ 0. A unique solution exists, so the system is consistent.
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
5x + 2y = 4; 7x + 3y = 5
Answer
Write AX = B with A =
5 | 2 |
7 | 3 |
and B =
4 |
5 |
|A| = 5 × 3 – 2 × 7 = 1 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
adj A =
3 | –2 |
–7 | 5 |
X = (1/|A|)(adj A)B = (1) ×
=
Hence x = 2, y = –3.
2x – y = –2; 3x + 4y = 3
Answer
Write AX = B with A =
2 | –1 |
3 | 4 |
and B =
–2 |
3 |
|A| = 2 × 4 – (–1) × 3 = 11 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
adj A =
4 | 1 |
–3 | 2 |
X = (1/|A|)(adj A)B = (1/11) ×
–5 |
12 |
=
–5/11 |
12/11 |
Hence x = –5/11, y = 12/11.
4x – 3y = 3; 3x – 5y = 7
Answer
Write AX = B with A =
4 | –3 |
3 | –5 |
and B =
3 |
7 |
|A| = 4 × (–5) – (–3) × 3 = –11 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
adj A =
–5 | 3 |
–3 | 4 |
X = (1/|A|)(adj A)B = (–1/11) ×
6 |
19 |
=
–6/11 |
–19/11 |
Hence x = –6/11, y = –19/11.
5x + 2y = 3; 3x + 2y = 5
Answer
Write AX = B with A =
5 | 2 |
3 | 2 |
and B =
3 |
5 |
|A| = 5 × 2 – 2 × 3 = 4 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
adj A =
2 | –2 |
–3 | 5 |
X = (1/|A|)(adj A)B = (1/4) ×
–4 |
16 |
=
–1 |
4 |
Hence x = –1, y = 4.
2x + y + z = 1; x – 2y – z = 3/2; 3y – 5z = 9
Answer
Write AX = B with A =
2 | 1 | 1 |
1 | –2 | –1 |
0 | 3 | –5 |
and B =
1 |
3/2 |
9 |
|A| = expanding along R₁: 2 × 13 – 1 × (–5) + 1 × 3 = 34 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
Cofactor matrix:
13 | 5 | 3 |
8 | –10 | –6 |
1 | 3 | –5 |
adj A =
13 | 8 | 1 |
5 | –10 | 3 |
3 | –6 | –5 |
X = (1/|A|)(adj A)B = (1/34) ×
34 |
17 |
–51 |
=
1 |
1/2 |
–3/2 |
Hence x = 1, y = 1/2, z = –3/2.
x – y + z = 4; 2x + y – 3z = 0; x + y + z = 2
Answer
Write AX = B with A =
1 | –1 | 1 |
2 | 1 | –3 |
1 | 1 | 1 |
and B =
4 |
0 |
2 |
|A| = expanding along R₁: 1 × 4 – (–1) × 5 + 1 × 1 = 10 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
Cofactor matrix:
4 | –5 | 1 |
2 | 0 | –2 |
2 | 5 | 3 |
adj A =
4 | 2 | 2 |
–5 | 0 | 5 |
1 | –2 | 3 |
X = (1/|A|)(adj A)B = (1/10) ×
20 |
–10 |
10 |
=
2 |
–1 |
1 |
Hence x = 2, y = –1, z = 1.
2x + 3y + 3z = 5; x – 2y + z = –4; 3x – y – 2z = 3
Answer
Write AX = B with A =
2 | 3 | 3 |
1 | –2 | 1 |
3 | –1 | –2 |
and B =
5 |
–4 |
3 |
|A| = expanding along R₁: 2 × 5 – 3 × (–5) + 3 × 5 = 40 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
Cofactor matrix:
5 | 5 | 5 |
3 | –13 | 11 |
9 | 1 | –7 |
adj A =
5 | 3 | 9 |
5 | –13 | 1 |
5 | 11 | –7 |
X = (1/|A|)(adj A)B = (1/40) ×
40 |
80 |
–40 |
=
1 |
2 |
–1 |
Hence x = 1, y = 2, z = –1.
x – y + 2z = 7; 3x + 4y – 5z = –5; 2x – y + 3z = 12
Answer
Write AX = B with A =
1 | –1 | 2 |
3 | 4 | –5 |
2 | –1 | 3 |
and B =
7 |
–5 |
12 |
|A| = expanding along R₁: 1 × 7 – (–1) × 19 + 2 × (–11) = 4 ≠ 0, so A⁻¹ exists and X = A⁻¹B.
Cofactor matrix:
7 | –19 | –11 |
1 | –1 | –1 |
–3 | 11 | 7 |
adj A =
7 | 1 | –3 |
–19 | –1 | 11 |
–11 | –1 | 7 |
X = (1/|A|)(adj A)B = (1/4) ×
8 |
4 |
12 |
=
2 |
1 |
3 |
Hence x = 2, y = 1, z = 3.
Q15. If A =
2 | –3 | 5 |
3 | 2 | –4 |
1 | 1 | –2 |
find A⁻¹. Using A⁻¹ solve the system of equations 2x – 3y + 5z = 11; 3x + 2y – 4z = –5; x + y – 2z = –3
Ans. |A| = expanding along R₁: 2 × 0 – (–3) × (–2) + 5 × 1 = –1.
Cofactors Aᵢⱼ (placed in position (i, j)):
0 | 2 | 1 |
–1 | –9 | –5 |
2 | 23 | 13 |
adj A = transpose of the cofactor matrix =
0 | –1 | 2 |
2 | –9 | 23 |
1 | –5 | 13 |
A⁻¹ = (1/|A|) adj A = (–1) ×
0 | –1 | 2 |
2 | –9 | 23 |
1 | –5 | 13 |
The system is AX = B with B =
11 |
–5 |
–3 |
X = A⁻¹B = (–1) ×
–1 |
–2 |
–3 |
=
1 |
2 |
3 |
Hence x = 1, y = 2, z = 3.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.
Answer
Let the cost per kg of onion, wheat and rice be ₹x, ₹y, ₹z. Then 4x + 3y + 2z = 60, 2x + 4y + 6z = 90, 6x + 2y + 3z = 70, i.e. AX = B with A =
4 | 3 | 2 |
2 | 4 | 6 |
6 | 2 | 3 |
and B =
60 |
90 |
70 |
|A| = expanding along R₁: 4 × 0 – 3 × (–30) + 2 × (–20) = 50 ≠ 0. Cofactor matrix:
0 | 30 | –20 |
–5 | 0 | 10 |
10 | –20 | 10 |
adj A =
0 | –5 | 10 |
30 | 0 | –20 |
–20 | 10 | 10 |
X = (1/50) × (adj A)B = (1/50) ×
250 |
400 |
400 |
=
5 |
8 |
8 |
Onion ₹5 per kg, wheat ₹8 per kg, rice ₹8 per kg.
Miscellaneous Exercises on Chapter 4
Q1. Prove that the determinant
x | sin θ | cos θ |
–sin θ | –x | 1 |
cos θ | 1 | x |
is independent of θ.
Ans. Expanding along R₁: Δ = x(–x² – 1) – sin θ(–x sin θ – cos θ) + cos θ(–sin θ + x cos θ)
= –x³ – x + x sin²θ + sin θ cos θ – sin θ cos θ + x cos²θ = –x³ – x + x(sin²θ + cos²θ) = –x³.
Δ = –x³ does not contain θ, so it is independent of θ.
Q2. Evaluate
cos α cos β | cos α sin β | –sin α |
–sin β | cos β | 0 |
sin α cos β | sin α sin β | cos α |
Ans. Expanding along C₃: Δ = –sin α × [–sin β · sin α sin β – cos β · sin α cos β] – 0 + cos α × [cos α cos²β + cos α sin²β]
= –sin α × (–sin α)(sin²β + cos²β) + cos α × cos α = sin²α + cos²α = 1.
Q3. If A⁻¹ =
3 | –1 | 1 |
–15 | 6 | –5 |
5 | –2 | 2 |
and B =
1 | 2 | –2 |
–1 | 3 | 0 |
0 | –2 | 1 |
find (AB)⁻¹
Ans. (AB)⁻¹ = B⁻¹A⁻¹. For B: |B| = expanding along R₁: 1 × 3 – 2 × (–1) + (–2) × 2 = 1. Cofactor matrix of B:
3 | 1 | 2 |
2 | 1 | 2 |
6 | 2 | 5 |
adj B =
3 | 2 | 6 |
1 | 1 | 2 |
2 | 2 | 5 |
B⁻¹ = (1/1) adj B =
3 | 2 | 6 |
1 | 1 | 2 |
2 | 2 | 5 |
(AB)⁻¹ = B⁻¹A⁻¹ =
9 | –3 | 5 |
–2 | 1 | 0 |
1 | 0 | 2 |
Q4. Let A =
1 | 2 | 1 |
2 | 3 | 1 |
1 | 1 | 5 |
Verify that (i) [adj A]⁻¹ = adj (A⁻¹) (ii) (A⁻¹)⁻¹ = A
Ans. |A| = expanding along R₁: 1 × 14 – 2 × 9 + 1 × (–1) = –5. adj A =
14 | –9 | –1 |
–9 | 4 | 1 |
–1 | 1 | –1 |
A⁻¹ = (–1/5) adj A =
–14/5 | 9/5 | 1/5 |
9/5 | –4/5 | –1/5 |
1/5 | –1/5 | 1/5 |
(i) |adj A| = |A|² = 25. [adj A]⁻¹ = (1/|adj A|) adj(adj A) =
–1/5 | –2/5 | –1/5 |
–2/5 | –3/5 | –1/5 |
–1/5 | –1/5 | –1 |
adj(A⁻¹) =
–1/5 | –2/5 | –1/5 |
–2/5 | –3/5 | –1/5 |
–1/5 | –1/5 | –1 |
Both are equal. Verified.
(ii) |A⁻¹| = 1/|A| = –1/5. (A⁻¹)⁻¹ = (1/|A⁻¹|) adj(A⁻¹) = (–5) ×
–1/5 | –2/5 | –1/5 |
–2/5 | –3/5 | –1/5 |
–1/5 | –1/5 | –1 |
=
1 | 2 | 1 |
2 | 3 | 1 |
1 | 1 | 5 |
= A. Verified.
Q5. Evaluate
x | y | x + y |
y | x + y | x |
x + y | x | y |
Ans. Apply R₁ → R₁ + R₂ + R₃: each entry of R₁ becomes 2(x + y). Taking 2(x + y) common: Δ = 2(x + y) ×
1 | 1 | 1 |
y | x + y | x |
x + y | x | y |
Apply C₂ → C₂ – C₁, C₃ → C₃ – C₁: Δ = 2(x + y) ×
1 | 0 | 0 |
y | x | x – y |
x + y | –y | –x |
= 2(x + y)[x(–x) – (x – y)(–y)] = 2(x + y)(–x² + xy – y²) = –2(x + y)(x² – xy + y²) = –2(x³ + y³).
Q6. Evaluate
1 | x | y |
1 | x + y | y |
1 | x | x + y |
Ans. Apply R₂ → R₂ – R₁ and R₃ → R₃ – R₁: Δ =
1 | x | y |
0 | y | 0 |
0 | 0 | x |
Expanding along C₁: Δ = 1 × (y · x – 0) = xy.
Solve the system of equations 2/x + 3/y + 10/z = 4; 4/x – 6/y + 5/z = 1; 6/x + 9/y – 20/z = 2
Answer
Put 1/x = u, 1/y = v, 1/z = w: 2u + 3v + 10w = 4, 4u – 6v + 5w = 1, 6u + 9v – 20w = 2, i.e. AX = B with A =
2 | 3 | 10 |
4 | –6 | 5 |
6 | 9 | –20 |
|A| = expanding along R₁: 2 × 75 – 3 × (–110) + 10 × 72 = 1200 ≠ 0. Cofactor matrix:
75 | 110 | 72 |
150 | –100 | 0 |
75 | 30 | –24 |
adj A =
75 | 150 | 75 |
110 | –100 | 30 |
72 | 0 | –24 |
X = (1/1200) (adj A)B = (1/1200) ×
600 |
400 |
240 |
=
1/2 |
1/3 |
1/5 |
So u = 1/2, v = 1/3, w = 1/5
⇒ x = 2, y = 3, z = 5.
Choose the correct answer in Exercises 8 and 9.
If x, y, z are nonzero real numbers, then the inverse of matrix A =
Answer
x | 0 | 0 |
0 | y | 0 |
0 | 0 | z |
is (A) the diagonal matrix with entries x⁻¹, y⁻¹, z⁻¹ (B) xyz × diag(x⁻¹, y⁻¹, z⁻¹) (C) (1/xyz) × diag(x, y, z) (D) (1/xyz) × I
Ans. |A| = xyz ≠ 0 and adj A = diag(yz, xz, xy). A⁻¹ = (1/xyz) adj A =
x⁻¹ | 0 | 0 |
0 | y⁻¹ | 0 |
0 | 0 | z⁻¹ |
Correct answer: (A).
Q9. Let A =
1 | sin θ | 1 |
–sin θ | 1 | sin θ |
–1 | –sin θ | 1 |
where 0 ≤ θ ≤ 2π. Then (A) Det(A) = 0 (B) Det(A) ∈ (2, ∞) (C) Det(A) ∈ (2, 4) (D) Det(A) ∈ [2, 4]
Ans. Det(A) = 1(1 + sin²θ) – sin θ(–sin θ + sin θ) + 1(sin²θ + 1) = 2 + 2sin²θ.
Since 0 ≤ sin²θ ≤ 1, 2 ≤ Det(A) ≤ 4, i.e. Det(A) ∈ [2, 4]. Correct answer: (D).
Important Extra Questions
Q1. Find the value of x if
x | 2 |
3 | x |
= 10.
Ans. x² – 6 = 10 → x² = 16 → x = ±4.
If A is a square matrix of order 3 and |A| = 5, find |2A| and |adj A|.
Answer
|kA| = kⁿ|A|
⇒ |2A| = 2³ × 5 = 40. |adj A| = |A|ⁿ⁻¹ = 5² = 25.
Find the area of the triangle with vertices (1, 1), (2, 3) and (4, 1).
Answer
Δ =
1 | 1 | 1 |
2 | 3 | 1 |
4 | 1 | 1 |
Δ = expanding along R₁: 1 × 2 – 1 × (–2) + 1 × (–10) = –6
⇒ Area = (1/2)|–6| = 3 sq units.
Q4. Find the inverse of A =
2 | 5 |
1 | 3 |
Ans. |A| = 6 – 5 = 1 ≠ 0. adj A =
A⁻¹ = (1/1) adj A =
Solve by matrix method: x + y = 5, x – y = 1.
Answer
A =
1 | 1 |
1 | –1 |
|A| = –1 – 1 = –2 ≠ 0. adj A =
–1 | –1 |
–1 | 1 |
X = (–1/2) ×
–6 |
–4 |
=
3 |
2 |
So x = 3, y = 2.