CBSE NCERT Solution for Class 12 - Maths

Class 12 - Maths - Determinants

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Exercise 4.1

Evaluate the determinants in Exercises 1 and 2.

Question 1:

2 4

-5 -1

Answer:

Given, 2 4

-5 -1

= 2 * (−1) – 4 * (−5)

= -2 + 20

= 18

Question 2:

(i) cos θ -sin θ (ii) x² − x + 1 x - 1

sin θ cos θ x + 1 x + 1

Answer:

(i) Given, cos θ -sin θ

sin θ cos θ

= cos θ * cos θ – (-sin θ) * sin θ

= cos² θ + sin² θ

= 1

(ii) Given x² − x + 1 x - 1

x + 1 x + 1

= (x² − x + 1)(x + 1) − (x − 1)(x + 1)

= x³ − x² + x + x² − x + 1 − (x² − 1)

= x³ + 1 − x² + 1

= x³ − x² + 2

Question 3:

If A = 1 2

4 2 , then show that |2A| = 4|A|

Answer:

The given matrix is

A = 1 2

4 2

So, 2A = 2 1 2 = 2 4

4 2 8 4

LHS:

|2A| = 2 4

8 4

= 2 * 4 – 4 * 8

= 8 – 32

= -24

Now, |A| = 1 2

4 2

= 1 * 2 – 2 * 4

= 2 – 8

= -6

RHS:

4|A| = 4 * (-8) = -32

So, LHS = RHS

Question 4:

If 1 0 1

A = 0 1 2

0 0 4 , then show that |3A| = 27|A|

Answer:

The give matrix is:

1 0 1

A = 0 1 2

0 0 4

It can be observed that in the first column, two entries are zero.

Thus, we expand along the first column (C₁) for easier calculation.

Now, |A| = 1 1 2 - 0 0 1 + 0 0 1

0 4 0 4 1 2

=> |A| = 1(4 - 0) – 0 + 0

=> |A| = 4

So, 27|A| = 27 * 4 = 108 …………..1

1 0 1 3 0 3

3A =3 0 1 2 = 0 3 6

0 0 4 0 0 12

So, 3|A| = 3 3 6 - 0 0 3 + 0 0 3

0 12 0 12 3 6

=> 3|A| = 3(36 - 0) – 0 + 0

=> 3|A| = 108 ……….2

From equations 1 and 2, we get

|3A| = 27|A|

Hence, the given result is proved.

Question 5:

Evaluate the determinants

(i) 3 -1 -2 (ii) 3 -4 5

0 0 -1 1 1 -2

3 -5 0 2 3 1

(iii) 0 1 2 (iv) 2 -1 -2

-1 0 -3 0 2 -1

-2 3 0 3 -5 0

Answer:

(i) Let

3 -1 -2

A = 0 0 -1

3 -5 0

It can be observed that in the second row, two entries are zero.

Thus, we expand along the second row for easier calculation.

|A| = 0 -1 -2 + 0 3 -2 - (-1) 3 -1

-5 0 3 0 3 -5

=> |A| = 0(0 - 10) + 0(0 + 6) + 1(-15 + 3)

=> |A| = -12

(ii) Let

3 -4 5

A = 1 1 -2

2 3 1

By expanding along the first row, we have:

|A| = 3 1 -2 + 4 1 -2 + 5 1 1

3 1 2 1 2 3

=> |A| = 3(1 + 6) + 4(1 + 4) + 5(3 - 2)

=> |A| = 21 + 20 + 5

=> |A| = 46

(iii) Let

0 1 2

A = -1 0 -3

-2 3 0

By expanding along the first row, we have:

|A| = 0 0 -3 - 1 -1 -3 + 2 -1 0

3 0 -2 0 -2 3

=> |A| = 0(0 + 9) - 1(0 - 6) + 2(-3 + 0)

=> |A| = 6 - 6

=> |A| = 0

(iv) Let

2 -1 -2

A = 0 2 -1

3 -5 0

By expanding along the first column, we have:

|A| = 2 2 -1 - 0 -1 -2 + 3 -1 -2

-5 0 -5 0 2 -1

=> |A| = 2(0 - 5) - 0 + 3(1 + 4)

=> |A| = -10 + 15

=> |A| = 5

Question 6:

If 1 1 -2

A = 2 1 -3

5 4 -9 , then find |A|.

Answer:

The given matrix is:

1 1 -2

A = 2 1 -3

5 4 -9

By expanding along the first row, we have:

|A| = 1 1 -3 - 1 2 -3 -2 2 1

4 -9 5 -9 5 4

=> |A| = 1(-9 + 12) – 1(-18 + 15) - 2(8 - 5)

=> |A| = 3 + 3 - 6

=> |A| = 6 – 6

=> |A| = 0

Question 7:

Find the value of x, if

(i) 2 4 = 2x 4 (ii) 2 3 = x 3

2 1 6 x 4 5 2x 5

Answer:

(i) Given,

2 4 = 2x 4

2 1 6 x

=> 2 * 1 – 5 * 4 = 2x * x – 6 * 4

=> 2 – 20 = 2x² - 24

=> -18 = 2x² - 24

=> 2x² = 24 – 18

=> 2x² = 6

=> x² = 3

=> x = ±√3

So, the value of x is ±√3

(ii) Given,

2 3 = x 3

4 5 2x 5

=> 2 * 5 – 3 * 4 = x * 5 – 3 * 2x

=> 10 – 12 = 5x – 6x

=> -2 = -x

=> x = 2

So, the value of x is 2

Question 8:

If x 2 = 6 2

18 x 18 6 , then x is equal to

(A) 6 (B) ±6 (C) −6 (D) 0

Answer:

Given,

x 2 = 6 2

18 x 18 6

=> x * x – 2 * 18 = 6 * 6 – 18 * 2

=> x² = 36

=> x = ±6

Hence, the correct answer is B.

Exercise 4.2

Question 1:

Using the property of determinants and without expanding, prove that:

x a x + a

y b y + b

z c z + c

Answer:

Given,

x a x + a x a x x a a

y b y + b = y b y + y b b

z c z + c z c z z c c

= 0 + 0 [Since two columns of the determinant are identical]

= 0

Question 2:

Using the property of determinants and without expanding, prove that:

a - b b - c c - a

b - c c- a a - b = 0

c - a a - b b - c

Answer:

Let

a - b b - c c - a

Δ = b - c c- a a - b

c - a a - b b - c

Applying R₁ -> R₁ + R₂, we get

a - c b - a c - a

Δ = b - c c- a a - b

-(a – c) -(b – a) -(c – b)

a - c b - a c - a

Δ =- b - c c- a a - b

a – c b – a c – b

Here, the two rows R₁ and R₃ are identical.

So, ∆ = 0

Question 3:

Using the property of determinants and without expanding, prove that:

2 7 65

3 8 75 = 0

5 9 86

Answer:

Given,

2 7 65 2 7 63 + 2

3 8 75 = 3 8 72 + 3

5 9 86 5 9 81 + 5

= 2 7 63 2 7 2

3 8 72 + 3 8 3

5 9 81 5 9 5

= 2 7 63

3 8 72 + 0 [Since two columns are identical]

5 9 81

= 2 7 9 * 7

3 8 9 * 8

5 9 9 * 9

= 9 2 7 7

3 8 8 [Since two columns are identical]

5 9 9

= 9 * 0

= 0

Question 4:

Using the property of determinants and without expanding, prove that:

1 bc a(b + c)

1 ca b(c + a) = 0

1 ab c(a + b)

Answer:

Given,

1 bc a(b + c)

1 ca b(c + a)

1 ab c(a + b)

Applying C₃ -> C₃ + C₂, we get

= 1 bc ab + bc + ca

1 ca ab + bc + ca

1 ab ab + bc + ca

= (ab + bc + ca) 1 bc 1

1 ca 1

1 ab 1

= (ab + bc + ca) * 0 [Since two columns are identical]

= 0

Question 5:

Using the property of determinants and without expanding, prove that:

b + c q + r y + z a p x

c + a r + p z + x = 2 b q y

a + b p + q x + y c r z

Answer:

Let

b + c q + r y + z

Δ = c + a r + p z + x

a + b p + q x + y

b + c q + r y + z b + c q + r y + z

Δ = c + a r + p z + x + c + a r + p z + x

a p x b q x

Δ = Δ₁ + Δ₂ ………….1

Now,

b + c q + r y + z

Δ₁ = c + a r + p z + x

a p x

Applying R₂ -> R₂ + R₃, we get

b + c q + r y + z

Δ₁ = c r z

a p x

Applying R₁ -> R₁ – R₂, we get

b q y

Δ₁ = c r z

a p x

Applying R₁ R₃ and R₂ R₃, we get

a p x

Δ₁ = (-1)² b q y

c r z

a p x

Δ₁ = b q y

c r z ………….2

Now,

b + c q + r y + z

Δ₂ = c + a r + p z + x

b q y

Applying R₁ -> R₁ - R₃, we get

c r z

Δ₂ = c + a r + p z + x

a p x

Applying R₂ -> R₂ – R₁, we get

c r z

Δ₂ = a p x

b q y

Applying R₁ R₂ and R₂ R₃, we get

a p x

Δ₂ = (-1)² b q y

c r z

a p x

Δ₂ = b q y

c r z ………….3

From equations 1, 2 and 3, we get

a p x

Δ = 2 b q y

c r z

Hence, the given result is proved.

Question 6:

By using the property of determinants, show that:

0 a -b

-a 0 -c = 0

b c 0

Answer:

Given,

0 a -b

Δ = -a 0 -c

b c 0

Applying R₁ -> cR₁, we get

0 ac -bc

Δ = (1/c) -a 0 -c

b c 0

Applying R₁ -> R₁ – bR₂, we get

ab ac 0

Δ = (1/c) -a 0 -c

b c 0

Δ = (a/c) -a 0 -c

b c 0

Δ = (a/c) * 0 [Since two rows are identical]

Δ = 0

Question 7:

By using the property of determinants, show that:

-a² ab ac

ba -b² bc = 4a²b²c²

ca cb -c²

Answer:

Given,

-a² ab ac

Δ = ba -b² bc

ca cb -c²

-a b c

Δ = abc a -b c

a b -c [taking out factors a, b, c from R₁, R₂ and R₃]

-1 1 1

Δ = a²b²c² 1 -1 1

1 1 -1 [taking out factors a, b, c from C₁, C₂ and C₃]

Applying R₂ -> R₂ + R₁ and R₃ -> R₃ + R₁, we get

-1 1 1

Δ = a²b²c² 0 0 2

0 2 0

Δ = a²b²c² *(-1) 0 2

Δ = -a²b²c² (0 - 4)

Δ = 4a²b²c²

Hence, the given result is proved.

Question 8:

By using the property of determinants, show that:

(i) 1 a a²

1 b b² = (a - b)(b - c)(c - a)

1 c c²

(ii) 1 1 1

a b c = (a - b)(b - c)(c - a)(a + b + c)

a³ b³ c³

Answer:

(i) Given,

1 a a²

Δ = 1 b b²

1 c c²

Applying R₁ -> R₁ – R₃ and R₂ -> R₂ – R₃, we get

0 a - c a² - c²

Δ = 0 b - c b² – c²

1 c c²

0 -1 -(a + c)

Δ = (b - c)(c - a) 0 1 b + c

1 c c²

Applying R₁ -> R₁ + R₂, we get

0 0 -a + b

Δ = (b - c)(c - a) 0 1 b + c

1 c c²

0 0 -1

Δ = (b - c)(c - a)(c - a) 0 1 b + c

1 c c²

Expanding along C₁, we get

Δ = (b - c)(c - a)(c - a) 0 -1

1 b + c

Δ = (b - c)(c - a)(c - a)(0 + 1)

Δ = (b - c)(c - a)(c - a)

Hence, the given result is proved.

(ii) Given,

1 1 1

Δ = a b c

a² b² c²

Applying C₁ -> C₁ – C₃ and C₂ -> C₂ – C₃, we get

0 0 1

Δ = a - c b - c c

a³ –c³ b³ –c³ c³

0 0 1

Δ = a - c b - c c

(a - c)(a² + b² + ab) (b - c)(b² + c² + bc) c³

0 0 1

Δ = (c - a)(b - c) -1 1 c

-(a² + b² + ab) (b² + c² + bc) c³

Applying C₁ -> C₁ + C₂, we get

0 0 1

Δ = (c - a)(b - c) 0 1 c

(b² - a²) + (bc - ac) (b² + c² + bc) c³

0 0 1

Δ = (c - a)(b - c)(a - b) 0 1 c

-(a + b + c) (b² + c² + bc) c³

0 0 1

Δ = (c - a)(b - c)(a - b)(a + b + c) 0 1 c

-1 (b² + c² + bc) c³

Expanding along C₁, we get

Δ = (c - a)(b - c)(a - b)(a + b + c)(-1) 0 1

1 c

Δ = (c - a)(b - c)(a - b)(a + b + c)(-1)(0 - 1)

Δ = (c - a)(b - c)(a - b)(a + b + c)

Hence, the given result is proved.

Question 9:

By using the property of determinants, show that:

x x² yz

y y² zx = (x - y)(y - z)(z - x)(xy + yz + zx)

z z² xy

Answer:

Let

x x² yz

Δ = y y² zx

z z² xy

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

x x² yz

Δ = y - x y² – x² zx - yz

z - x z² – x² xy - yz

x x² yz

Δ = y - x -(x - y)(x + y) z(x – y)

z - x (z - x)(z + x) -y(z – x)

x x² yz

Δ = (x - y)(z - x) -1 -x - y z

1 z + x -y

Applying R₃ -> R₃ + R₂, we get

x x² yz

Δ = (x - y)(z - x) -1 -x - y z

0 z - y z - y

x x² yz

Δ = (x - y)(z - x)(z - y) -1 -x - y z

0 1 1

Expanding along R₃, we get

Δ = [(x - y)(z - x)(z - y)] (-1) x yz + 1 x x²

-1 z -1 -x - y

Δ = [(x - y)(z - x)(z - y)][-xz – yz + (-x² – xy + x²)]

Δ = -(x - y)(z - x)(z - y)(xy + yz + zx)

Δ = (x - y)(y - z)(z - x)(xy + yz + zx)

Hence, the given result is proved.

Question 10:

By using properties of determinants, show that:

(i) x + 4 2x 2x

2x x + 4 2x = (5x + 4)(4 - x)²

2x 2x x + 4

(ii) y + k y y

y y + k y = k²(3y + k)

y y y + k

Answer:

(i) Let

x + 4 2x 2x

Δ = 2x x + 4 2x

2x 2x x + 4

Applying R₁ → R₁ + R₂ + R₃, we get

5x + 4 5x + 4 5x + 4

Δ = 2x x + 4 2x

2x 2x x + 4

1 1 1

Δ = (5x + 4) 2x x + 4 2x

2x 2x x + 4

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we get

1 0 0

Δ = (5x + 4) 2x -x + 4 0

2x 0 -x + 4

1 0 0

Δ = (5x + 4)(4 - x)(4 - x) 2x 1 0

2x 0 1

Expanding along C₃, we get

Δ = (5x + 4)(4 - x)(4 - x) 1 0

2x 1

Δ = (5x + 4)(4 - x)(4 - x)(1 - 0)

Δ = (5x + 4)(4 - x)²

Hence, the given result is proved.

(ii) Let

y + k y y

Δ = y y + k y

y y y + k

Applying R₁ → R₁ + R₂ + R₃, we get

3y + k 3y + k 3y + k

Δ = y y + k y

y y y + k

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we get

1 0 0

Δ = (3y + k) y k 0

y 0 k

1 0 0

Δ = k²(3y + k) y 1 0

y 0 1

Expanding along C₃, we get

Δ = k²(3y + k) 1 0

y 1

Δ = k²(3y + k)(1 - 0)

Δ = k²(3y + k)

Hence, the given result is proved.

Question 11:

By using properties of determinants, show that:

(i) a – b - c 2a 2a

2b b – c - a 2b = (a + b + c)²

2c 2c c – a - b

(ii) x + y + 2z x y

z y + z + 2x y = 2(x + y + z)²

z x z + x + 2y

Answer:

(i) Let

a – b - c 2a 2a

Δ = 2b b – c - a 2b

2c 2c c – a - b

Applying R₁ → R₁ + R₂ + R₃, we get

a + b + c a + b + c a + b + c

Δ = 2b b – c - a 2b

2c 2c c – a - b

1 1 1

Δ = (a + b + c) 2b b – c - a 2b

2c 2c c – a - b

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we get

1 0 0

Δ = (a + b + c) 2b -(a + b + c) 0

2c 0 -(a + b + c)

1 0 0

Δ = (a + b + c)³ 2b -1 0

2c 0 -1

Expanding along C₃, we get

Δ = (a + b + c)³[-1 * (-1) - 0]

Δ = (a + b + c)³

Hence, the given result is proved.

(ii) Let

x + y + 2z x y

Δ = z y + z + 2x y

z x z + x + 2y

Applying C₁ → C₁ + C₂ + C₃, we get

2(x + y + z) x y

Δ = 2(x + y + z) y + z + 2x y

2(x + y + z) x z + x + 2y

1 x y

Δ = 2(x + y + z) 1 y + z + 2x y

1 x z + x + 2y

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we get

1 x y

Δ = 2(x + y + z) 0 x + y + z 0

0 0 x + y + z

1 x y

Δ = 2(x + y + z)³ 0 1 0

0 0 1

Expanding along R₃, we get

Δ = 2(x + y + z)³(1 - 0)

Δ = 2(x + y + z)³

Hence, the given result is proved.

Question 12:

By using the property of determinants, show that:

1 x x²

x² 1 x = (1 – x³)²

x x² 1

Answer:

Let

1 x x²

Δ = x² 1 x

x x² 1

Applying R₁ → R₁ + R₂ + R₃, we get

1 + x + x² 1 + x + x² 1 + x + x²

Δ = x² 1 x

x x² 1

1 1 1

Δ = (1 + x + x²) x² 1 x

x x² 1

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we get

1 0 0

Δ = (1 + x + x²) x² 1 - x² x – x²

x x² - x 1 - x

1 0 0

Δ = (1 + x + x²)(1 - x)(1 - x) x² 1 + x x

x - x 1

1 0 0

Δ = (1 – x³)(1 - x) x² 1 + x x

x - x 1

Expanding along R₁, we get

Δ = (1 – x³)(1 - x)[1(1 + x) – x * (-x)]

Δ = (1 – x³)(1 - x)[1 + x + x²]

Δ = (1 – x³)(1 – x³)

Δ = (1 – x³)²

Hence, the given result is proved.

Question 13:

By using properties of determinants, show that:

1 + a² – b² 2ab -2b

2ab 1 - a² + b² 2a = (1 + a² + b²)³

2b -2a 1 - a² – b²

Answer:

Let

1 + a² – b² 2ab -2b

Δ = 2ab 1 - a² + b² 2a

2b -2a 1 - a² – b²

Applying R₁ → R₁ + bR₃ and R₂ → R₂ - aR₃, we get

1 + a² + b² 0 -b(1 + a² + b²)

Δ = 0 1 + a² + b² a(1 + a² + b²)

2b -2a 1 - a² – b²

1 0 -b

Δ = (1 + a² + b²)² 0 1 a

2b -2a 1 - a² – b²

Expanding along R₁, we get

Δ = (1 + a² + b²)²1 1 a - b 0 1

-2a 1 - a² – b² 2b -2a

Δ = (1 + a² + b²)²[(1 - a² – b² + 2a²) – b(0 – 2b)]

Δ = (1 + a² + b²)²(1 + a² + b²)

Δ = (1 + a² + b²)³

Question 14:

By using the property of determinants, show that:

a² + 1 ab ac

ab b² + 1 bc = 1 + a² + b² + c²

ca cb c² + 1

Answer:

Let

a² + 1 ab ac

Δ = ab b² + 1 bc

ca cb c² + 1

Taking out common factors a, b, and c from R₁, R₂, and R₃ respectively, we have:

a + 1/a b c

Δ = abc a b + 1/b c

a c c + 1/c

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

a + 1/a b c

Δ = abc -1/a 1/b c

-1/a 0 1/c

Applying C₁ -> aC₁, C₂ -> bC₂and C₃ -> cC₃, we get

a² + 1 b² c²

Δ = abc * 1/abc -1 1 0

-1 0 1

a² + 1 b² c²

Δ = -1 1 0

-1 0 1

Expanding along R₃, we get

Δ = (-1)[ b² * 0 – 1 * c²] + 0 + 1[a² + 1 + b²]

Δ = c² + a² + 1 + b²

Δ = 1 + a² + b² + c²

Hence, the given result is proved.

Question 15:

Choose the correct answer.

Let A be a square matrix of order 3 * 3, then |kA| is equal to

k|A| B. k²|A| C. k³|A| D. 3k|A|

Answer:

A is a square matrix of order 3 * 3.

Let

a₁ a₂ a₃

A = b₁ b₂ b₃

c₁ c₂ c₃

Then

ka₁ ka₂ ka₃

kA = kb₁ kb₂ kb₃

kc₁ kc₂ kc₃

Now,

ka₁ ka₂ ka₃

|kA| = kb₁ kb₂ kb₃

kc₁ kc₂ kc₃

a₁ a₂ a₃

|kA| = (k * k * k) b₁ b₂ b₃

c₁ c₂ c₃ [Taking out common factor k from each row]

a₁ a₂ a₃

|kA| = k³ b₁ b₂ b₃

c₁ c₂ c₃

=> kA| = k³|A|

Hence, the correct answer is C.

Question 16:

Which of the following is correct?

Determinant is a square matrix.
Determinant is a number associated to a matrix.
Determinant is a number associated to a square matrix.
None of these Answer

Answer:

We know that to every square matrix A = [a_ij] of order n. We can associate a number called the

determinant of square matrix A, where a_ij = (i, j)^th element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is option C.

Exercise 4.3

Question 1:

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:

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

1 0 1

Δ = (1/2) 6 0 1

4 3 1

=> Δ = (1/2)[1(0 - 3) – 0(6 - 4) + 1(18 - 0)]

=> Δ = (1/2)[-3 + 18]

=> Δ = 15/2

Hence, area of the triangle is 15/2 square units.

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

2 7 1

Δ = (1/2) 1 1 1

10 8 1

=> Δ = (1/2)[2(1 - 8) – 7(1 - 10) + 1(8 - 10)]

=> Δ = (1/2)[-14 + 63 - 2]

=> Δ = (1/2)[-16 + 63]

=> Δ = 47/2

Hence, area of the triangle is 47/2 square units.

(iii)The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation,

-3 -3 1

Δ = (1/2) 3 2 1

-1 -8 1

=> Δ = (1/2)[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)]

=> Δ = (1/2)[-20 + 12 - 22]

=> Δ = (1/2)[-42 + 12]

=> Δ = -30/2

=> Δ = -15

Hence, area of the triangle is |-15| = 15 square units.

Question 2:

Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Answer:

Area of ∆ABC is given by the relation,

a b + c 1

Δ = (1/2) b c + a 1

c a + b 1

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁

a b + c 1

Δ = (1/2) b - a a - b 1

c - a a - c 1

a b + c 1

Δ = (1/2) * (a - b) * (c - a) -1 1 1

1 -1 1

Applying R₃ -> R₃ + R₂

a b + c 1

Δ = (1/2) * (a - b) * (c - a) -1 1 0

0 0 0

Δ = (1/2) * (a - b) * (c - a) * 0 [Since all elements of R₃ are 0]

Δ = 0

Since the area of the triangle formed by points A, B, and C is zero.

Hence, the points A, B, and C are collinear.

Question 3:

Find values of k if area of triangle is 4 square units and vertices are

(i) (k, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)

Answer:

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is the

absolute value of the determinant (∆), where

x₁ y₁ 1

Δ = (1/2) x₂ y₂ 1

X₃ y₃ 1

It is given that the area of triangle is 4 square units.

So, ∆ = ±4

(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,

k 0 1

Δ = (1/2) 4 0 1

0 2 1

=> ±4 = (1/2)[k(0 - 2) - 0(4 - 0) + 1(8 - 0)]

=> ±4 = (1/2)[-2k + 8]

=> ±4 = -k + 4

=> -k + 4 = ±4

When −k + 4 = − 4, then k = 8

When −k + 4 = 4, then k = 0

Hence, k = 0, 8

(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,

-2 0 1

Δ = (1/2) 0 4 1

0 k 1

=> ±4 = (1/2)[-2(4 - k) - 0(0 - 0) + 1(0 - 0)]

=> ±4 = (1/2)[-8 +2k]

=> ±4 = k – 4

=> k – 4 = ±4

When k − 4 = − 4, then k = 0

When k − 4 = 4, then k = 8

Hence, k = 0, 8

Question 4:

(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:

(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B,

and P are collinear. Therefore, the area of triangle ABP will be zero.

1 0 1

=> (1/2) 3 6 1 = 0

x y 1

=> (1/2)[1(6 - y) - 2(3 - x) + 1(3y – 6x)] = 0

=> 6 – y – 6 + 2x + 3y – 6x = 0

=> 2y – 4x = 0

=> y – 2x = 0

=> y = 2x

Hence, the equation of the line joining the given points is y = 2x.

(ii) Let P (x, y) be any point on the line joining points A (3, 1) and B (9, 3). Then, the points A, B,

and P are collinear. Therefore, the area of triangle ABP will be zero.

3 1 1

=> (1/2) 9 3 1 = 0

x y 1

=> (1/2)[3(3 - y) - 1(9 - x) + 1(9y – 3x)] = 0

=> 9 – 3y – 9 + x + 9y – 3x = 0

=> 6y – 2x = 0

=> x – 3y = 0

=> y = 2x

Hence, the equation of the line joining the given points is x − 3y = 0.

Question 5:

If the area of triangle is 35 square units with vertices (2, −6), (5, 4) and (k, 4). Then k is

12 B. −2 C. −12, −2 D. 12, −2

Answer:

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

2 -6 1

Δ = (1/2) 5 4 1

k 4 1

=> Δ = (1/2)[2(4 - 4) + 6(5 - k) + 1(20 – 4k)]

=> Δ = (1/2)[30 – 6k + 20 – 4k]

=> Δ = (1/2)[50 – 10k]

=> Δ = 25 – 5k

It is given that the area of the triangle is ±35.

Therefore, we have:

=> 25 – 5k = ±35

=> 5(5 – k) = ±35

=> 5 – k = ±7

When 5 − k = −7, then k = 5 + 7 = 12

When 5 − k = 7, then k = 5 − 7 = −2

So, k = 12, −2

Hence, the correct answer is option D.

Exercise 4.4

Write Minors and Cofactors of the elements of following determinants:

Question 1:

(i) 2 -4 (ii) a c

0 3 b d

Answer:

(i) The given determinant is

2 -4

0 3

Minor of element a_ij is M_ij

M₁₁ = minor of element a₁₁ = 3

M₁₂ = minor of element a₁₂ = 0

M₂₁ = minor of element a₂₁ = −4

M₂₂ = minor of element a₂₂ = 2

Cofactor of a_ij is A_ij = (−1)ⁱ ^{+ j} M_ij

A₁₁ = (−1)¹⁺¹ M₁₁ = (−1)² * 3 = 3

A₁₂ = (−1)¹⁺² M¹² = (−1)³ * 0 = 0

A₂₁ = (−1)²⁺¹ M₂₁ = (−1)³ * (−4) = 4

A₂₂ = (−1)₂₊₂ M₂₂ = (−1)⁴ * (2) = 2

(ii) The given determinant is

a c

b d

Minor of element a_ij is M_ij

M₁₁ = minor of element a₁₁ = d

M₁₂ = minor of element a₁₂ = b

M₂₁ = minor of element a₂₁ = c

M₂₂ = minor of element a₂₂ = a

Cofactor of a_ij is A_ij = (−1)ⁱ ^{+ j} M_ij

A₁₁ = (−1)¹⁺¹ M₁₁ = (−1)² * d = d

A₁₂ = (−1)¹⁺² M¹² = (−1)³ * b = -b

A₂₁ = (−1)²⁺¹ M₂₁ = (−1)³ * c = -c

A₂₂ = (−1)₂₊₂ M₂₂ = (−1)⁴ * a = a

Question 2:

(i) 1 0 0 (ii) 1 0 4

0 1 0 3 5 -1

0 0 1 0 1 2

Answer:

(i) The give matrix is:

1 0 0

0 1 0

0 0 1

By the definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁ = 1 0 = 1

0 1

M₁₂ = minor of a₁₂ = 0 0 = 0

0 1

M₁₃ = minor of a₁₃ = 0 0 = 0

0 1

M₂₁ = minor of a₂₁ = 0 0 = 0

0 1

M₂₂ = minor of a₂₂ = 1 0 = 1

0 1

M₂₃ = minor of a₂₃ = 1 0 = 0

0 0

M₃₁ = minor of a₃₁ = 0 0 = 0

1 0

M₃₂ = minor of a₃₂ = 1 0 = 0

0 0

M₃₃ = minor of a₃₃ = 1 0 = 1

0 1

A₁₁ = cofactor of a₁₁ = (-1)^{1 + 1} M₁₁ = 1

A₁₂ = cofactor of a₁₂ = (-1)^{1 + 2} M₁₂ = 0

A₁₃ = cofactor of a₁₃ = (-1)^{1 + 3} M₁₃ = 0

A₂₁ = cofactor of a₂₁ = (-1)^{2 + 1} M₂₁ = 0

A₂₂ = cofactor of a₂₂ = (-1)^{2 + 2} M₂₂ = 1

A₂₃ = cofactor of a₂₃ = (-1)^{2 + 3} M₂₃ = 0

A₃₁ = cofactor of a₃₁ = (-1)^{3 + 1} M₃₁ = 0

A₃₂ = cofactor of a₃₂ = (-1)^{3 + 2} M₃₂ = 0

A₃₃ = cofactor of a₃₃ = (-1)^{3 + 3} M₃₃ = 1

(ii) The give matrix is:

1 0 4

3 5 -1

0 1 2

By the definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁ = 5 -1 = 10 + 1 = 11

1 2

M₁₂ = minor of a₁₂ = 3 -1 = 6 – 0 = 6

0 2

M₁₃ = minor of a₁₃ = 3 5 = 3 – 0 = 3

0 1

M₂₁ = minor of a₂₁ = 0 4 = 0 – 4 = -4

1 2

M₂₂ = minor of a₂₂ = 1 4 = 2 – 0 = 2

0 2

M₂₃ = minor of a₂₃ = 1 0 = 1 – 0 = 1

0 1

M₃₁ = minor of a₃₁ = 0 4 = 0 – 20 = -20

5 -1

M₃₂ = minor of a₃₂ = 1 4 = -1 – 12 = -13

3 -1

M₃₃ = minor of a₃₃ = 1 0 = 5 – 0 = 5

3 5

A₁₁ = cofactor of a₁₁ = (-1)^{1 + 1} M₁₁ = 11

A₁₂ = cofactor of a₁₂ = (-1)^{1 + 2} M₁₂ = -6

A₁₃ = cofactor of a₁₃ = (-1)^{1 + 3} M₁₃ = 3

A₂₁ = cofactor of a₂₁ = (-1)^{2 + 1} M₂₁ = 4

A₂₂ = cofactor of a₂₂ = (-1)^{2 + 2} M₂₂ = 2

A₂₃ = cofactor of a₂₃ = (-1)^{2 + 3} M₂₃ = -1

A₃₁ = cofactor of a₃₁ = (-1)^{3 + 1} M₃₁ = -20

A₃₂ = cofactor of a₃₂ = (-1)^{3 + 2} M₃₂ = 13

A₃₃ = cofactor of a₃₃ = (-1)^{3 + 3} M₃₃ = 5

Question 3:

Using Cofactors of elements of second row, evaluate

5 2 8

Δ = 2 0 1

1 2 3

Answer:

The given determinant is:

5 2 8

2 0 1

1 2 3

We have:

M₂₁ = minor of a₂₁ = 3 8 = 9 – 16 = -7

2 3

So, A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 7

M₂₁ = minor of a₂₁ = 5 8 = 15 – 8 = 7

1 3

So, A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 7

M₂₃ = minor of a₂₃ = 5 1 = 10 – 3 = 7

1 2

So, A₂₂ = cofactor of a₂₃ = (−1)²⁺³ M₂₁ = -7

We know that ∆ is equal to the sum of the product of the elements of the second row with

their corresponding cofactors.

∆ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

= 2 * 7 + 0 * 7 + 1 * (−7)

= 14 − 7

= 7

Question 4:

Using Cofactors of elements of third column, evaluate

5 2 8

Δ = 2 0 1

1 2 3

Answer:

The given determinant is:

5 2 8

2 0 1

1 2 3

We have:

M₁₃ = minor of a₂₁ = 1 y = z – y

1 z

So, A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = z - y

M₂₃ = minor of a₂₃ = 1 x = z - x

1 z

So, A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = -(z - x) = x - z

M₃₃ = minor of a₃₃ = 1 x = y - x

1 y

So, A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = y - x

We know that ∆ is equal to the sum of the product of the elements of the second row with

their corresponding cofactors.

∆ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

= yz(z - y) + zx(x - z) + xy(y - x)

= yz² – y²z + x²z – xz² + xy² – x²y

= (x²z – y²z) + (yz² – xz²) + (xy² – x²y)

= z(x² – y²) - z² (x – y) - xy(x – y)

= z(x – y)(x + y) - z²(x – y) - xy(x – y)

= (x - y)[z(x + y) - z² - xy]

= (x - y)[zx + zy - z² - xy]

= (x - y)[zx - z² + zy - xy]

= (x - y)[z(x – z) + y(z – x)]

= (x - y)[-z(z – x) + y(z – x)]

= (x - y)(y - z)(z - x)

So, Δ = (x - y)(y - z)(z - x)

Question 5:

For the matrices A and B, verify that (AB)′ = B’A’where

(i) A = 1

-4 , B = [-1 2 1]

(ii) A = 0

1 , B = [1 5 7]

Answer:

(i) AB = 1 -1 2 1

-4 [-1 2 1] = 4 -8 -4

3 -3 6 3

So, (AB)’ = -1 4 -3

2 -8 6

1 -4 3

Now, A’ = [1 -4 3], B’ = -1

So, B’A’ = -1 -1 4 -3

2 [1 -4 3] = 3 -8 6

1 1 -4 3

Hence, it is verified that (AB)’ = B’A’

(ii) AB = 0 0 0 0

1 [1 5 7] = 1 5 7

2 2 10 14

So, (AB)’ = 0 1 2

0 5 10

0 7 14

Now, A’ = [0 1 2], B’ = 1

So, B’A’ = 1 0 1 2

5 [0 1 2] = 0 5 10

7 0 7 14

Hence, it is verified that (AB)’ = B’A’

Exercise 4.5

Find adjoint of each of the matrices in Exercises 1 and 2.

Question 1:

1 2

3 4

Answer:

Let

A = 1 2

3 4

Now,

A₁₁ = 4, A₁₂ = -3, A₂₁ = -2, A₂₂ = 1

So, adj(A) = A₁₁ A₂₁ = 4 -2

A₁₂ A₂₂ -3 1

Question 2:

1 -1 2

2 3 5

-2 0 1

Answer:

Let

A = 1 -1 2

2 3 5

-2 0 1

Now,

A₁₁ = 3 5 = 3 – 0 = 3

0 1

A₁₂ = - 2 5 = -(2 + 10) = -12

-2 1

A₁₃ = 2 3 = 0 + 6 = 6

-2 0

A₂₁ = - -1 2 = -(-1 - 0) = 1

0 1

A₂₂ = 1 2 = 1 + 4 = 5

-2 1

A₂₃ = - 1 -1 = -(0 - 2) = 2

-2 0

A₃₁ = -1 2 = -5 – 6 = -11

3 5

A₃₂ = - 1 2 = -(5 - 4) = -1

2 5

A₃₃ = 1 -1 = 3 + 2 = 5

2 3

Hence, adj(A) = A₁₁ A₂₁ A₃₁ 3 1 -11

A₁₂ A₂₂ A₃₂ = -12 5 -1

A₁₃ A₂₃ A₃₃ 6 2 5

Verify A * (adj A) = (adj A) * A = | A | * I in Exercises 3 and 4.

Question 3:

2 3

-4 -6

Answer:

Let

A = 2 3

-4 -6

Now,

|A| = 2 * (-6) – (-4) * 3

= -12 + 12

= 0

So, |A|I = 0 * 1 0 = 0 0

0 1 0 0

Now, A₁₁ = -6, A₁₂ = 4, A₂₁ = -3, A₂₂ = 2

So, adj(A) = A₁₁ A₂₁ = -6 -3

A₁₂ A₂₂ 4 2

Now, A(adj A) = 2 3 -6 -3

-4 -6 4 2

= -12 + 12 -6 + 6

24 – 24 12 – 12

= 0 0

0 0

also, A(adj A) = 6 -3 2 3

4 2 -4 -6

= -12 + 12 -18 + 18

8 – 8 12 – 12

= 0 0

0 0

Hence, A * (adj A) = (adj A) * A = | A | * I

Question 4:

1 -1 2

3 0 -2

1 0 3

Answer:

Let

A = 1 -1 2

3 0 -2

1 0 3

Now,

|A| = 1(0 - 0) + 1(9 + 2) + 2(0 - 0) = 11

1 0 0 11 0 0

So, |A|I = 11 * 0 1 0 = 0 11 0

0 0 1 0 0 11

Now,

A₁₁ = 0, A₁₂ = -11, A₁₃ = 0

A₂₁ = 3, A₂₂ = 1, A₂₃ = -1

A₃₁ = 2, A₃₂ = 8, A₃₃ = 3

So, adj (A) = 0 3 2

-11 1 8

0 -1 3

Now,

1 -1 2 0 3 2

A(adj A) = 3 0 -2 -11 1 8

1 0 3 0 -1 3

= 0 + 11 + 0 3 – 1 – 2 2 – 8 + 6

0 + 0 + 0 9 + 0 + 2 6 + 0 – 6

0 + 0 + 0 3 + 0 – 3 2 + 0 + 9

= 11 0 0

0 11 0

0 0 11

Also,

0 3 2 1 -1 2

(adj A)A = -11 1 8 3 0 -2

0 -1 3 1 0 3

= 0 + 9 + 2 0 + 0 + 0 0 – 6 + 6

-11 + 3 + 8 11 + 0 + 0 -22 - 2 + 24

0 - 3 + 3 0 + 0 + 0 0 + 2 + 9

= 11 0 0

0 11 0

0 0 11

Hence, A * (adj A) = (adj A) * A = | A | * I

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

Question 5:

2 -2

4 3

Answer:

Let

A = 2 -2

4 3

|A| = 2 * 3 – (-2) * 4

= 6 + 8

= 14

Now,

A₁₁ = 3, A₁₂ = -4, A₂₁ = 2, A₂₂ = 2

So, adj(A) = A₁₁ A₂₁ = 3 2

A₁₂ A₂₂ -4 2

Now, A^-1 = (adj A)/|A|

= (1/14) 3 2

-4 2

Question 6:

-1 5

-3 2

Answer:

Let

A = -1 5

-3 2

|A| = (-1) * 2 – 5 * (-3)

= -2 + 15

= 13

Now,

A₁₁ = 2, A₁₂ = 3, A₂₁ = -5, A₂₂ = -1

So, adj(A) = A₁₁ A₂₁ = 2 -5

A₁₂ A₂₂ 3 -1

Now, A^-1 = (adj A)/|A|

= (1/13) 2 -5

3 -1

Question 7:

1 2 2

0 2 4

0 0 5

Answer:

Let

A = 1 2 3

0 2 4

0 0 5

|A| = 1(10 - 0) – 2(0 - 0) + 3(0 - 0) = 10

Now,

A₁₁ = 10 – 0 = 10, A₁₂ = -(0 - 0) = 0, A₁₃ = 0 – 0 = 0

A₂₁ = -(10 - 0) = -10, A₂₂ = 5 – 0 = 5, A₂₃ = -(0 - 0) = 0

A₃₁ = 8 – 6 = 2, A₃₂ = -(4 - 0) = -4, A₃₃ = 2 – 0 = 2

Now, (adj A) = 10 -10 2

0 -5 -4

0 0 2

So, A^-1 = (adj A)/|A| = (1/10) 10 -10 2

0 -5 -4

0 0 2

Question 8:

1 0 0

3 3 0

5 2 -1

Answer:

Let

A = 1 0 0

3 3 0

5 3 -1

|A| = 1(-3 - 0) – 0 + 0 = -3

Now,

A₁₁ = -3 – 0 = -3, A₁₂ = -(-3 - 0) = 3, A₁₃ = 6 – 15 = -9

A₂₁ = -(0 - 0) = 0, A₂₂ = -1 – 0 = -1, A₂₃ = -(2 - 0) = -2

A₃₁ = 0 – 0 = 0, A₃₂ = -(0 - 0) = 0, A₃₃ = 3 – 0 = 3

Now, (adj A) = -3 0 0

3 -1 0

-9 -2 3

So, A^-1 = (adj A)/|A| = (-1/3) -3 0 0

3 -1 0

-9 -2 3

Question 9:

2 1 3

4 -1 0

-7 2 1

Answer:

Let

A = 2 1 3

4 -1 0

-7 2 1

|A| = 2(-1 - 0) – 1(4 - 0) + 3(8 - 7)

= -2 – 4 + 3

= -3

Now,

A₁₁ = -1 – 0 = -1, A₁₂ = -(4 - 0) = -4, A₁₃ = 8 – 7 = 1

A₂₁ = -(1 - 6) = 5, A₂₂ = 2 + 21 = 23, A₂₃ = -(4 + 7) = -11

A₃₁ = 0 + 3 = 3, A₃₂ = -(0 - 12) = 12, A₃₃ = -2 – 4 = -6

Now, (adj A) = -1 5 3

-4 23 12

1 -11 -6

So, A^-1 = (adj A)/|A| = (-1/3) -1 5 3

-4 23 12

1 -11 -6

Question 10:

1 -1 2

0 2 -3

3 -2 4

Answer:

Let

A = 1 -1 2

0 2 -3

3 -2 4

|A| = 1(8 - 6) - 0 + 3(3 - 4)

= 2 –3

= -1

Now,

A₁₁ = 8 – 6 = 2, A₁₂ = -(0 + 9) = -9, A₁₃ = 0 – 6 = -6

A₂₁ = -(-4 + 4) = 0, A₂₂ = 4 - 6 = -2, A₂₃ = -(-2 + 3) = -1

A₃₁ = 3 - 4 = -1, A₃₂ = -(-3 - 0) = 3, A₃₃ = 2 – 0 = 2

Now, (adj A) = 2 0 -1

-9 -2 3

-6 -1 2

So, A^-1 = (adj A)/|A| = (-1) 2 0 -1 = -2 0 1

-9 -2 3 9 2 -3

-6 -1 2 6 1 -2

Question 11:

1 0 0

0 cos α sin α

0 sin α - cos α

Answer:

Let

A = 1 0 0

0 cos α sin α

0 sin α -cos α

|A| = 1(-cos² α – sin² α) - 0 + 0

= -(cos² α + sin² α)

= -1

Now,

A₁₁ = -cos² α – sin² α = -1, A₁₂ = 0, A₁₃ = 0

A₂₁ = 0, A₂₂ = - cos α, A₂₃ = - sin α

A₃₁ = 0, A₃₂ = - sin α, A₃₃ = cos α

Now, (adj A) = -1 0 0

0 -cos α -sin α

0 -sin α cos α

So, A^-1 = (adj A)/|A| = (-1) -1 0 0 = 1 0 0

0 -cos α -sin α 0 cos α sin α

0 -sin α cos α 0 sin α -cos α

Question 12:

Let A = 3 7 B = 6 8

2 5 and 7 9 , verify that (AB)^-1 = B^-1A^-1

Answer:

Given,

A = 3 7

2 5

Now, |A| = 3 * 5 – 7 * 2 = 15 – 14 = 1

A₁₁ = 5, A₁₂ = -2, A₂₁ = -7, A₂₂ = 3

So, adj A = 5 -7

-2 3

Now, A^-1 = (adj A)/|A| = 5 -7

2 3

Again given,

B = 6 8

7 9

Now, |B| = 6 * 9 – 8 * 7 = 52 – 56 = -2

B₁₁ = 9, B₁₂ = -7, B₂₁ = -8, B₂₂ = 6

So, adj B = 9 -8

-7 6

Now, B^-1 = (adj B)/|B| = (-1/2) 9 -8 = -9/2 4

-7 6 7/2 -3

Now, B^-1A^-1 = -9/2 4 5 -7

7/2 -3 -2 3

= -4/2 – 8 63/2 + 12 = -61/2 87/2

35/2 + 6 -49/2 – 9 47/2 -67/2 …………..1

Again, AB = 3 7 6 8

2 5 7 9

= 18 + 49 24 + 63

12 + 35 16 + 45

= 67 87

47 61

Now, |AB| = 67 * 61 – 87 * 47 = 4087 – 4089 = -2

So, adj (AB) = 61 -87

-47 67

Now, (AB)^-1 = (adj AB)/|AB| = (-1/2) 61 -87 = -61/2 87/2

-47 67 47/2 -67/2 …………..1

From equations 1 and 2, we get

(AB)^-1 = B^-1A^-1

Hence, the give result is proved.

Question 13:

If A = 3 7

-1 2 , show that A² – 5A + 7I = 0. Hence find A^-1

Answer:

Given,

A = 3 7

-1 2

A² = A * A = 3 1 3 1

-1 2 -1 2

= 9 – 1 3 + 2

-3 – 2 -1 + 4

= 8 5

-5 3

Now,

A² – 5A + 7I = 8 5 - 5 3 1 + 7 1 0

-5 3 -1 2 0 1

= 8 5 - 15 5 + 7 0

-5 3 -5 10 0 7

= 8 – 15 + 7 5 – 5 + 0

-5 + 5 + 0 3 – 10 + 7

= 0 0

0 0

Hence, A² – 5A + 7I = 0

=> A * A – 5A + 7I = 0

=> A * A – 5A = -7I

=> A * A(A^-1) – 5A(A^-1) = -7I(A^-1)

=> A * (AA^-1) – 5(AA^-1) = -7I(A^-1)

=> AI – 5I = -7A^-1

=> A – 5I = -7A^-1

=> A^-1 = (-1/7)(A – 5I)

=> A^-1 = (1/7)(5I - A)

=> A^-1 = (1/7) 5 0 - 3 1

0 5 -1 2

=> A^-1 = (1/7) 5 - 3 0 - 1

0 + 1 5 - 2

=> A^-1 = (1/7) 2 -1

1 3

Question 14:

For the matrix A = 3 2

1 1 , find the numbers a and b such that A² + aA + bI = O.

Answer:

Given,

A = 3 2

1 1

A² = A * A = 3 2 3 2

1 1 1 1

= 9 + 2 6 + 2

3 + 1 2 + 1

= 11 8

4 3

Now, A² + aA + bI = 0

=> A * A + aA + bI = 0

=> A * A + aA = -bI

=> A * A(A^-1) + aA(A^-1) = -bI(A^-1)

=> A * (AA^-1) + a(AA^-1) = -bI(A^-1)

=> AI + aI = -bA^-1

=> A + aI = -bA^-1

=> A^-1 = (-1/b)(A + aI) ……………1

Now, A^-1 = (adj A)/|A|

=> A^-1 = (1/1) 1 -2 = 1 -2

-1 3 -1 3

From equation 1, we have

=> 1 -2 = (-1/b) 3 2 - a 0

-1 3 1 1 0 a

=> 1 -2 = (-1/b) 3 + a 2

-1 3 1 1 + a

=> 1 -2 = (-3 – a)/b -2/b

-1 3 1/b (-1 – a)/2

Comparing the corresponding elements of the two matrices, we get

=> -1/b = -1

=> b = 1

And (-3 – a)/b = 1

=> -3 – a = b

=> -3 – a = 1

=> a = -3 – 1

=> a = -4

Hence, −4 and 1 are the required values of a and b respectively.

Question 15:

For the matrix A = 1 1 1

1 2 -3

2 -1 3 , show that A³ – 6A² + 5A + 11I = O. Hence find A^-1

Answer:

Given,

1 1 1

A = 1 2 -3

2 -1 3

A² = A * A

= 1 1 1 1 1 1

1 2 -3 1 2 -3

2 -1 3 2 -1 3

= 1 + 1 + 2 1 + 2 – 1 1 – 3 + 3

1 + 2 - 6 1 + 4 + 3 1 – 6 – 9

2 - 1 + 6 2 - 2 – 3 2 + 3 + 9

= 4 2 1

-3 8 -14

7 -3 14

A³ = A² * A

= 4 2 1 4 2 1

-3 8 -14 -3 8 -14

7 -3 14 7 -3 14

= 4 + 2 + 2 4 + 4 – 1 4 – 6 + 3

-3 + 8 - 28 -3 + 16 + 14 -3 – 24 – 42

7 - 3 + 28 7 - 6 – 14 7 + 9 + 42

= 8 7 1

-23 27 -69

32 -13 58

So, A³ – 6A² + 5A + 11I

= 8 7 1 4 2 1 1 1 1 1 0 0

-23 27 -69 - 6 -3 8 -14 + 5 1 2 -3 + 11 0 1 0

32 -13 58 7 -3 14 2 -1 3 0 0 1

= 8 7 1 24 12 6 5 5 5 11 0 0

-23 27 -69 - -18 48 -84 + 5 10 -15 + 0 11 0

32 -13 58 42 -18 84 10 -5 15 0 0 11

= 8 – 24 + 5 + 11 7 – 12 + 5 + 0 1 – 6 + 5 + 0

-23 + 18 + 5 + 0 27 – 48 + 5 + 11 -69 + 84 - 15 + 0

32 – 42 + 10 + 0 -13 + 18 - 5 + 0 58 – 84 + 15 + 11

= 0 0 0

0 0 0

Hence, A³ – 6A² + 5A + 11I = 0

Now, A³ – 6A² + 5A + 11I = 0

=> A * A * A – 6(A * A) + 5A + 11I = 0

=> (A * A * A) * A^-1 – 6(A * A) * A^-1 + 5A * A^-1 + 11I * A^-1 = 0

=> (A * A) * (A * A^-1) – 6A * (A * A^-1) + 5(A * A^-1) = -11(I * A^-1)

=> A² – 6A + 5I = -11A^-1

=> A^-1 = (-1/11)(A² – 6A + 5I) ………………1

Now, A² – 6A + 5I

= 4 2 1 1 1 1 1 0 0

-3 8 -14 - 6 1 2 -3 + 5 0 1 0

7 -3 14 2 -1 3 0 0 1

= 4 2 1 6 6 6 5 0 0

-3 8 -14 - 6 12 -18 + 0 5 0

7 -3 14 12 -6 18 0 0 5

= 4 - 6 + 5 2 - 6 + 0 1 - 6 + 0

-3 - 6 + 0 8 - 12 + 5 -14 + 18 + 0

7 - 12 + 0 -3 + 6 + 0 14 - 18 + 5

= 3 -4 -5

-9 1 4

-5 3 1

From equation 1, we get

A^-1 = (-1/11) 3 -4 -5 = (1/11) -3 4 5

-9 1 4 9 -1 -4

-5 3 1 5 -3 -1

Question 16:

For the matrix A = 2 -1 1

-1 2 -1

1 -1 2 , show that A³ – 6A² + 9A - 4I = O. Hence find A^-1

Answer:

Given,

2 -1 1

A = -1 2 -1

1 -1 2

A² = A * A

= 2 -1 1 2 -1 1

-1 2 -1 -1 2 -1

1 -1 2 1 -1 2

= 4 + 1 + 1 -2 - 2 – 1 2 + 1 + 2

-2 - 2 - 1 1 + 4 + 1 -1 – 2 – 2

2 + 1 + 2 -1 - 2 – 2 1 + 1 + 4

= 6 -5 5

-5 6 -5

5 -5 6

A³ = A² * A

= 6 -5 5 2 -1 1

-5 6 -5 -1 2 -1

5 -5 6 1 -1 2

= 12 + 5 + 5 -6 - 10 – 5 6 + 5 + 10

-10 - 6 - 5 5 + 11 + 5 -5 – 6 – 10

10 + 5 + 6 -5 - 10 – 6 5 + 5 + 12

= 22 -12 21

-21 22 -21

21 -21 22

So, A³ – 6A² + 9A - 4I

= 22 -21 21 6 -5 5 2 -1 1 1 0 0

-21 22 -21 - 6 -5 6 -5 + 5 -1 2 -1 + 4 0 1 0

21 -21 22 5 -5 6 1 -1 2 0 0 1

= 22 -21 21 36 -30 30 18 -9 9 4 0 0

-21 22 -21 - -30 36 -30 + -9 18 -9 + 0 4 0

21 -21 22 30 -30 36 9 -9 18 0 0 4

= 22 – 36 + 18 + 4 -21 + 30 - 9 + 0 21 – 30 + 9 + 0

-21 + 30 - 9 + 0 22 – 36 + 18 + 4 -21 + 30 - 9 + 0

21 – 30 + 9 + 0 -21 + 30 - 9 + 0 22 – 36 + 18 + 4

= 0 0 0

0 0 0

Hence, A³ – 6A² + 9A - 4I = 0

Now, A³ – 6A² + 9A - 4I = 0

=> A * A * A – 6(A * A) + 9A - 4I = 0

=> (A * A * A) * A^-1 – 6(A * A) * A^-1 + 9A * A^-1 - 4I * A^-1 = 0

=> (A * A) * (A * A^-1) – 6A * (A * A^-1) + 9(A * A^-1) = 4(I * A^-1)

=> A² – 6A + 9I = 4A^-1

=> A^-1 = (1/4)(A² – 6A + 9I) ………………1

Now, A² – 6A + 9I

= 6 -5 5 2 -1 1 1 0 0

-5 6 -5 - 6 -1 2 -1 + 9 0 1 0

5 -5 6 1 -1 2 0 0 1

= 6 -5 5 12 -6 6 9 0 0

-5 6 -5 - -6 12 -6 + 0 9 0

5 -5 6 6 -6 12 0 0 9

= 6 - 12 + 9 -5 + 6 + 0 5 - 6 + 0

-5 + 6 + 0 6 - 12 + 9 -5 + 6 + 0

5 - 6 + 0 -5 + 6 + 0 6 - 12 + 9

= 3 1 -1

1 3 1

-1 -1 3

From equation 1, we get

A^-1 = (1/4) 3 1 -1

1 3 1

-1 -1 3

Question 17:

Let A be a nonsingular square matrix of order 3 * 3. Then |adj A| is equal to

|A| B. |A|² C. |A|³ D. 3|A|

Answer:

We know that,

|A| 0 |A|

(adj A)A = |A|I = 0 |A| 0

0 0 |A|

|A| 0 |A|

=> |(adj A)A| = 0 |A| 0

0 0 |A|

1 0 0

=> |(adj A)||A| = |A|³ = 0 1 0 = |A|³ I

0 0 1

So, => |(adj A)| = |A|²

Hence, the correct answer is option B.

Question 18:

If A is an invertible matrix of order 2, then det (A^-1) is equal to

det (A) B. 1/ det (A) C. 1 D. 0

Answer:

Since A is an invertible matrix, then A^-1 exists and A^-1 = (adj A)/|A|

Given, the matrix A is of order 2,

Let A = a b

c d

Then |A| = ad – bc

And adj A = d -b

-c a

Now, A^-1 = (adj A)/|A|

= d/|A| -b/|A|

-c/|A| a/|A|

So, A^-1 = d/|A| -b/|A|

-c/|A| a/|A|

= (1/|A|²) d -b

-c a

= (1/|A|²) * (ad - bc)

= (1/|A|²) * |A|

= |A|

So, det(A^-1) = 1/det(A)

Hence, the correct answer is option B.

Exercise 4.6

Examine the consistency of the system of equations in Exercises 1 to 6.

Question 1:

x + 2y = 2

2x + 3y = 3

Answer:

The given system of equations is:

x + 2y = 2

2x + 3y = 3

The given system of equations can be written in the form of AX = B, where

A = 1 2 X = x and B = 2

2 3 , y 3

Now,

|A| = 1 * 3 – 2 * 2 = 3 – 4 = -1 ≠ 0

So, A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 2:

2x - y = 5

x + y = 4

Answer:

The given system of equations is:

2x - y = 5

x + y = 4

The given system of equations can be written in the form of AX = B, where

A = 2 -1 X = x and B = 5

1 1 , y 4

Now,

|A| = 2 * 1 – 1 * (-1) = 2 + 1 = 3 ≠ 0

So, A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 3:

x + 3y = 5

2x + 6y = 8

Answer:

The given system of equations is:

x + 3y = 5

2x + 6y = 8

The given system of equations can be written in the form of AX = B, where

A = 1 3 X = x and B = 5

2 6 , y 5

Now,

|A| = 1 * 6 – 2 * 3 = 6 – 6 = 0

So, A is a singular matrix.

Now,

adj(A) = 6 -3

-2 1

adj(A)B = 6 -3 5 = 30 – 24 = 6 ≠ O

-2 1 8 -10 + 8 -2

Thus, the solution of the given system of equations does not exist.

Hence, the system of equations is inconsistent.

Question 4:

Examine the consistency of the system of equations.

x + y + z = 1 2x + 3y + 2z = 2 ax + ay + 2az = 4

Answer:

The given system of equations is:

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

This system of equations can be written in the form AX = B, where

1 1 1 x 1

A = 2 3 2 , X = y and B = 2

a a 2a z 4

Now,

|A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)

= 4a – 2a – a

= a ≠ 0

So, A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 5:

Examine the consistency of the system of equations.

3x - y - 2z = 2 2y - z = -1 3x - 5y = 3

Answer:

The given system of equations is:

3x - y - 2z = 2

2y - z = -1

3x - 5y = 3

This system of equations can be written in the form AX = B, where

3 -1 -2 x 2

A = 0 2 -1 , X = y and B = -1

3 -5 0 z 3

Now,

|A| = 3(0 – 5) -0 + 3(1 + 4) = -15 + 15 = 0

So, A is a singular matrix.

Now,

adj(A) = -5 10 5 2 -10 – 10 + 15 -5

-3 6 3 -1 = -6 – 6 + 9 = -3 ≠ O

-6 12 6 3 -1 – 12 + 18 -6

Thus, the solution of the given system of equations does not exist.

Hence, the system of equations is inconsistent.

Question 6:

Examine the consistency of the system of equations.

5x − y + 4z = 5 2x + 3y + 5z = 2 5x − 2y + 6z = −1

Answer:

The given system of equations is:

5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1

This system of equations can be written in the form of AX = B, where

5 -1 4 x 5

A = 2 3 5 , X = y and B = 2

5 -2 6 z -1

Now,

|A| = 5(18 + 10) + 1(12 – 25) + 4(-4 – 15)

= 5 * 28 + 1 * (-13) + 4 * (-19)

= 140 – 13 – 76

= 51 ≠ 0

So, A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7:

5x + 2y = 4

7x + 3y = 5

Answer:

The given system of equations is:

5x + 2y = 4

7x + 3y = 5

The given system of equations can be written in the form of AX = B, where

A = 5 2 X = x and B = 4

7 3 , y 5

Now,

|A| = 5 * 3 – 7 * 2 = 15 – 14 = 1 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

Now, A^-1 = adj(A) /|A|

So, A^-1 = 3 -2

-7 5

Now,

X = A^-1B = 3 -2 4

-7 5 5

x = 12 - 10 = 2

y -28 + 25 -3

So, x = 2 and y = -3

Question 8:

2x - y = -2

3x + 4y = 3

Answer:

The given system of equations is:

2x - y = -2

3x + 4y = 3

The given system of equations can be written in the form of AX = B, where

A = 2 -1 X = x and B = -2

3 4 , y 3

Now,

|A| = 2 * 4 – 3 * (-1) = 8 + 3 = 11 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

Now, A^-1 = adj(A) /|A|

So, A^-1 = (1/11) 4 1

-3 2

Now,

X = A^-1B = (1/11) 4 1 -2

-3 2 3

x = (1/11) -8 + 3 = (1/11) -5 = -5/11

y 6 + 6 12 12/11

So, x = -5/11 and y = 12/11

Question 9:

4x - 3y = 3

3x - 5y = 7

Answer:

The given system of equations is:

4x - 3y = 3

3x - 5y = 7

The given system of equations can be written in the form of AX = B, where

A = 4 -3 X = x and B = 3

3 -5 , y 7

Now,

|A| = (-5) * 4 – 3 * (-3) = -20 + 9 = -11 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

Now, A^-1 = adj(A) /|A|

So, A^-1 = (-1/11) -5 3 = (1/11) 5 -3

-3 4 3 -4

Now,

X = A^-1B = (1/11) 5 -3 3

3 -4 7

x = (1/11) 15 - 21 = (1/11) -6 = -6/11

y 9 - 28 -19 -19/11

So, x = -6/11 and y = -19/11

Question 10:

5x + 2y = 3

3x + 2y = 5

Answer:

The given system of equations is:

5x + 2y = 3

3x + 2y = 5

The given system of equations can be written in the form of AX = B, where

A = 5 2 X = x and B = 3

3 2 , y 5

Now,

|A| = 5 * 2 – 3 * 2 = 10 - 6 = 4 ≠ 0

So, A is non-singular matrix.

Therefore, its inverse exists.

Question 11:

2x + y + z = 1 x – 2y – z = 3/2 3y – 5z = 9

Answer:

The given system of equations can be written in the form of AX = B, where

2 1 1 x 1

A = 1 -2 -1 , X = y and B = 3/2

0 3 -5 z 9

Now,

|A| = 2(10 + 3) – 1(-5 - 3) + 0

= 2 * 13 – 1 * (-8)

= 26 + 8

= 34 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A₁₁ = 13, A₁₂ = 5, A₁₃ = 3

A₂₁ = 8, A₂₂ = -10, A₂₃ = -6

A₃₁ = 1, A₃₂ = 3, A₃₃ = -5

So, A^-1 = adj(A) /|A| = (1/34) 13 8 1

5 -10 3

3 -6 -5

So, X = A^-1B = (1/34) 13 8 1 1

5 -10 3 3/2

3 -6 -5 9

x = (1/34) 13 + 12 + 9

y 5 – 15 + 27

z 3 – 9 - 45

x = (1/34) 34 = 1

y 17 1/2

z -51 -3/2

So, x = 1, y = 1/2 and z = -3/2

Question 12:

x − y + z = 4 2x + y − 3z = 0 x + y + z = 2

Answer:

The given system of equations can be written in the form of AX = B, where

1 -1 1 x 4

A = 2 1 -3 , X = y and B = 0

1 1 1 z 2

Now,

|A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1)

= 4 + 5 + 1

= 10 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A₁₁ = 4, A₁₂ = -5, A₁₃ = 1

A₂₁ = 2, A₂₂ = 0, A₂₃ = -2

A₃₁ = 2, A₃₂ = 5, A₃₃ = 3

So, A^-1 = adj(A) /|A| = (1/10) 4 2 2

-5 0 5

1 -2 3

So, X = A^-1B = (1/10) 4 2 2 4

-5 0 5 0

1 -2 3 2

x = (1/10) 16 + 0 + 4

y -20 + 0 + 10

z 4 + 0 + 6

x = (1/10) 20 = 2

y -10 -1

z 10 1

So, x = 2, y = -1 and z = 1

Question 13:

2x + 3y + 3z = 5 x − 2y + z = −4 3x − y − 2z = 3

Answer:

The given system of equations can be written in the form of AX = B, where

2 3 3 x 5

A = 1 -2 1 , X = y and B = -4

3 -1 -2 z 3

Now,

|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6)

= 2 * 5 – 3 * (-5) + 3 * 5

= 10 + 15 + 15

= 40 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A₁₁ = 5, A₁₂ = 5, A₁₃ = 5

A₂₁ = 3, A₂₂ = -13, A₂₃ = 11

A₃₁ = 9, A₃₂ = 1, A₃₃ = -7

So, A^-1 = adj(A) /|A| = (1/40) 5 3 9

5 -13 1

5 11 -7

So, X = A^-1B = (1/40) 5 3 9 5

5 -13 1 -4

5 11 -7 3

x = (1/40) 25 - 12 + 27

y 25 + 52 + 3

z 25 - 44 - 21

x = (1/40) 40 = 1

y 80 2

z -40 -1

So, x = 1, y = 2 and z = -1

Question 14:

x − y + 2z = 7 3x + 4y − 5z = −5 2x − y + 3z = 12

Answer:

The given system of equations can be written in the form of AX = B, where

1 -1 2 x 7

A = 3 4 -5 , X = y and B = -5

2 -1 3 z 12

So, |A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8)

= 7 + 19 + 2 * (-11)

= 7 + 19 - 22

= 4 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A₁₁ = 7, A₁₂ = -19, A₁₃ = -11

A₂₁ = 1, A₂₂ = -1, A₂₃ = -1

A₃₁ = -3, A₃₂ = 11, A₃₃ = 7

So, A^-1 = adj(A) /|A| = (1/4) 7 1 -3

-19 -1 11

-11 -1 7

So, X = A^-1B = (1/4) 7 1 3 7

-19 -1 11 -5

-11 -1 7 12

x = (1/4) 49 - 5 - 36

y -133 + 5 + 132

z -77 + 5 + 84

x = (1/4) 8 = 2

y 4 1

z 12 3

So, x = 2, y = 1 and z = 3

Question 15:

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

Answer:

Given,

2 -3 5

A = 3 2 -4

1 1 -2

So, |A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2)

= 0 - 6 + 5

= -1 ≠ 0

Now,

A₁₁ = 0, A₁₂ = 2, A₁₃ = 1

A₂₁ = -1, A₂₂ = -9, A₂₃ = -5

A₃₁ = 2, A₃₂ = 23, A₃₃ = 13

So, A^-1 = adj(A) /|A| = (-1) 0 -1 2 0 1 -2

2 -9 23 = -2 9 -23

1 -5 13 -1 5 -13 ……………1

The given system of equations can be written in the form of AX = B, where

2 -3 5 x 11

A = 3 2 -4 , X = y and B = -5

1 1 -2 z -3

The solution of the system of equations is given by

X = A^-1B

So, X = A^-1B = 0 1 -2 11

-2 9 -23 -5 [From equation 1]

-1 5 -13 -3

x = 0 – 5 + 6

y -22 - 45 + 69

z -11 - 25 + 39

x = 1

y 2

z 3

So, x = 1, y = 2 and z = 3

Question 16:

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Answer:

Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.

Then, the given situation can be represented by a system of equations as:

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

This system of equations can be written in the form of AX = B, where

4 3 2 x 60

A = 2 4 6 , X = y and B = 90

6 2 3 z 70

|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24)

= 0 – 3 * (-30) + 2 * (-20)

= 90 - 40

= 50 ≠ 0

Now,

A₁₁ = 0, A₁₂ = 30, A₁₃ = -20

A₂₁ = -5, A₂₂ = 0, A₂₃ = 10

A₃₁ = 10, A₃₂ = -20, A₃₃ = 10

So, A^-1 = adj(A) /|A| = (1/50) 0 -5 10

30 0 -20

-20 10 10

So, X = A^-1B = (1/50) 0 -5 10 60

30 0 -20 90

-20 10 10 70

x = (1/50) 0 - 450 + 700

y 1800 + 0 - 1400

z -1200 + 900 + 700

x = (1/50) 250 = 5

y 400 8

z 400 8

So, x =5, y = 8 and z = 8

Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is

Rs 8 per kg.

Miscellaneous Exercise on Chapter 4

Question 1:

Prove that the determinant x sin θ cos θ

-sin θ -x 1

cos θ 1 x is independent of θ.

Answer:

Let

x sin θ cos θ

Δ = -sin θ -x 1

cos θ 1 x

Δ = x(x² - 1) – sin θ(-x * sin θ – cos θ) + cos θ(-sin θ + x * cos θ)

Δ = x³ - x + x * sin² θ – sin θ * cos θ - sin θ *cos θ + x * cos² θ

Δ = x³ - x + x * (sin² θ + cos² θ)

Δ = x³ - x + x

Δ = x³ [Independent of θ]

Hence the given determinant is independent of θ.

Question 2:

Without expanding the determinant, prove that

a a² bc 1 a² a³

b b² ca = 1 a² b³

c c² ab 1 a² c³

Answer:

LHS:

a a² bc

b b² ca

c c² ab

= (1/abc) a² a³ abc

b² b³ abc

c² c³ abc [Apply R₁ -> aR₁, R₂ -> bR₂, R₃ -> cR₃]

= (abc/abc) a² a³ 1

b² b³ 1

c² c³ 1 [Taking out factor abc from C₃]

= a² a³ 1

b² b³ 1

c² c³ 1

= 1 a² a³

1 b² b³

1 c² c³ [Apply C₁ C₃ and C₂ C₃]

= RHS

Hence, the give result is proved.

Question 3:

Evaluate cos α * cos β cos α * cos β -sin α

-sin β cos β 0

sin α * cos β sin α * sin β cos α

Answer:

Given,

cos α * cos β cos α * cos β -sin α

Δ = -sin β cos β 0

sin α * cos β sin α * sin β cos α

Expanding along C₃, we get

Δ = -sin α(-sin α * sin² β – cos² β * sin α) + 0 + cos α(cos α * cos² β + cos α * sin² β)

Δ = sin² α(sin² β + cos² β) + cos² α(cos² β + sin² β)

Δ = sin² α * 1 + cos² α * 1

Δ = sin² α + cos² α

Δ = 1

Question 4:

If a, b and c are real numbers and

b + c c + a a + b

Δ = c + a a + b b + c = 0

a + b b + c c + a

Show that either a + b + c = 0 or a = b = c.

Answer:

Given,

b + c c + a a + b

Δ = c + a a + b b + c

a + b b + c c + a

Applying R₁ -> R₁ + R₂ + R₃, we get

2(a + b + c) 2(a + b + c) 2(a + b + c)

Δ = c + a a + b b + c

a + b b + c c + a

1 1 1

Δ = 2(a + b + c) c + a a + b b + c

a + b b + c c + a

Applying C₂ -> C₂ – C₁ and C₃ -> C₃ – C₁, we get

1 0 0

Δ = 2(a + b + c) c + a b - c b - a

a + b c - a c - b

Expanding along R₁, we get:

Δ = 2(a + b + c)[1 * {(b - c)(c - b) – (b - a)(c - a)}]

Δ = 2(a + b + c)[-b² – c² + 2bc – bc + ba + ca – a²]

Δ = 2(a + b + c)[ab + bc + ca + ba – a² - b² – c²]

It is given that Δ = 0

=> 2(a + b + c)[ab + bc + ca + ba – a² - b² – c²] = 0

Either a + b + c = 0 or ab + bc + ca + ba – a² - b² – c² = 0

Now, 2ab + 2bc + 2ca + 2ba – 2a² - 2b² – 2c² = 0

=> (a - b)² + (b - c)² + (c - a)² = 0

=> (a - b)² = (b - c)² = (c - a)² = 0 [Since (a - b)², (b - c)², (c - a)² are non negative]

=> (a - b) = (b - c) = (c - a) = 0

=> a = b = c

Hence, if ∆ = 0 then either a + b + c = 0 or a = b = c.

Question 5:

Solve the equations x + a x x

x x + a x = 0, a ≠ 0

x x x + a

Answer:

Given,

x + a x x

x x + a x = 0

x x x + a

Applying R₁ -> R₁ + R₂ + R₃, we get

3x + a 3x + a 3x + a

x x + a x = 0

x x x + a

1 1 1

(3x + a) x x + a x = 0

x x x + a

Applying C₂ -> C₂ – C₁ and C₃ -> C₃ – C₁, we get

1 0 0

(3x + a) x a 0 = 0

x 0 a

Expanding along R₁, we get:

(3x + a)(1 * a² – 0) = 0

=> a²(3x + a) = 0

=> 3x + a = 0 [Since a ≠ 0]

=> x = -a/3

Question 6:

Prove that a² bc ac + c²

a² + ab b² ac = 4a²b²c²

ab b² + bc c²

Answer:

Given,

a² bc ac + c²

Δ = a² + ab b² ac

ab b² + bc c²

Taking out common factors a, b and c from C₁, C₂ and C₃, we get

a c a + c

Δ = abc a + b b a

b b + c c

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

a c a + c

Δ = abc b b - c -c

b - a b -a

Applying R₂ -> R₂ + R₁, we get

a c a + c

Δ = abc a + b b a

b - a b -a

Applying R₃ -> R₃ + R₂, we get

a c a + c

Δ = abc a + b b a

2b 2b 0

a c a + c

Δ = 2ab²c a + b b a

1 1 0

Applying C₂ -> C₂ – C₁, we get

a c - a a + c

Δ = 2ab²c a + b -a a

1 1 0

Expanding along R₃, we get:

Δ = 2ab²c[a(c - a) + a(a + c)]

Δ = 2ab²c[ac – a² + a² + ac]

Δ = 2ab²c * 2ac

Δ = 4a²b²c²

Hence, the given result is proved.

Question 7:

If A^-1 = 3 -1 1 1 2 -2

-15 6 -5 and B = -1 3 0

5 -2 2 0 -2 1 , find (AB)^-1

Answer:

We know that (AB)^-1 = B^-1A^-1

1 2 -2

B = -1 3 0

0 -2 1

|B| = 1 * 3 – 2 * (-1) – 2 * 2

= 3 + 2 - 4

= 1

Now,

B₁₁ = 3, B₁₂ = 1, B₁₃ = 2

B₂₁ = 2, B₂₂ = 1, B₂₃ = 2

B₃₁ = 6, B₃₂ = 2, B₃₃ = 5

So, B^-1 = adj(B) /|B| = 3 2 6

1 1 2

2 2 5

Now, (AB)^-1 = B^-1A^-1

= 3 2 6 3 -1 -1

1 1 2 -15 6 -5

2 2 5 5 -2 2

= 9 – 30 + 30 -3 + 12 – 12 3 – 10 + 12

3 – 15 + 10 -1 + 6 – 4 1 – 5 + 4

6 – 30 + 25 -2 + 12 – 10 2 – 10 + 10

= 9 -3 5

-2 1 0

1 0 2

Question 8:

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

Answer:

Given,

1 -2 1

A = -2 3 1

1 1 5

|A| = 1(15 - 1) + 2(-10 - 1) + 1 (-2 - 3)

= 14 – 22 - 5

= -13

Now,

A₁₁ = 14, A₁₂ = 11, A₁₃ = -5

A₂₁ = 11, A₂₂ = 4, A₂₃ = -3

A₃₁ = -5, A₃₂ = -3, A₃₃ = -1

So,

14 11 -5

adj(A) = 11 4 -3

-5 -3 -1

So, A^-1 = adj(A) /|A| = (-1/13) 14 11 -5 = (1/13) -14 -11 5

11 4 -3 -11 -4 3

-5 -3 -1 5 3 1

(i) |adj(A)| = 14(-4 - 9) – 11(-11 - 15) – 5(-33 + 20)

= 14 * (-13) – 11 * (-26) – 5 * (-13)

= -182 + 286 + 65

= 169

We have

-13 26 -13

adj[adj(A)] = 26 -39 -13

-13 -13 -65

So, [adj(A)]^-1 = adj[adj(A)]/|adj(A)| = (1/169) -13 26 -13

26 -39 -13

-13 -13 -65

= (1/13) -1 2 -1

2 -3 -1

-1 -1 -5

Now, A^-1 = (1/13) -14 -11 5

-11 -4 3

5 3 1

Now, A^-1 = -14/13 -11/13 5/13

-11/13 -4/13 3/13

5/13 3/13 1/13

(-4/169 – 9/169) -(-11/169 – 15/169) (-33/169 + 20/169)

adj(A^-1 ) = -(-11/169 – 15/169) (-14/169 – 25/169) -(-42/169 + 55/169)

(-33/169 + 20/169) -(-42/169 + 55/169) (56/169 – 121/169)

= (1/169) -13 26 -13

26 -39 -13

-13 -13 -65

= (1/13) -1 2 -1

2 -3 -1

-1 -1 -5

Hence, [adj(A)]^-1 = adj(A^-1)

(ii) We have

A^-1 = (1/13) -14 -11 5

-11 -4 3

5 3 1

And

adj(A^-1)= (1/13) -1 2 -1

2 -3 -1

-1 -1 -5

Now, | A^-1| = (1/13)³[-14 * (-13) + 11 * (-26) + 5 * (-13)]

= (1/13)³(-169)

= -1/13

So, (A^-1)^-1 = adj(A^-1) /| A^-1| = [1/(-13)] * (1/13) -1 2 -1

2 -3 -1

-1 -1 -5

= 1 -2 1

-2 3 1

1 1 5

= A

So, (A^-1)^-1 = A

Question 9:

Evaluate x y x + y

y x + y x

x + y x y

Answer:

Given,

x y x + y

Δ = y x + y x

x + y x y

Applying R₁ -> R₁ + R₂ + R₃, we get

2(x + y) 2(x + y) 2(x + y)

Δ = y x + y x

x + y x y

1 1 1

Δ = 2(x + y) y x + y x

x + y x y

Applying C₂ -> C₂ – C₁ and C₃ -> C₃ – C₁, we get

1 0 0

Δ = 2(x + y) y x x - y

x + y -y -x

Expanding along R₁, we get:

Δ = 2(x + y)[-x² + y(x - y)]

=> Δ = 2(x + y)[-x² + xy – y²]

=> Δ = -2(x + y)[x² + y² – xy]

=> Δ = -2(x³ + y³)

Question 10:

Evaluate 1 x y

1 x + y y

1 x x + y

Answer:

Given,

1 x y

Δ = 1 x + y y

1 x x + y

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

1 x y

Δ = 0 y 0

0 0 x

Expanding along C₁, we get:

Δ = 1(xy - 0) = xy

Using properties of determinants in Exercises 11 to 15, prove that:

Question 11:

α α² β + γ

β β² γ + α = (α - β)( β - γ)(γ - α)(α + β + γ)

γ γ² α + β

Answer:

Given,

α α² β + γ

Δ = β β² γ + α

γ γ² α + β

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

α α² β + γ

Δ = β - α β² – α² α - β

γ - α γ² – α² α - γ

α α² β + γ

Δ = (β - α)( γ - α) 1 β + α -1

1 γ + α -1

Applying R₃ -> R₃ – R₂, we get

α α² β + γ

Δ = (β - α)( γ - α) 1 β + α -1

0 γ - β 0

Expanding along R₃, we get:

Δ = (β - α)( γ - α)[-( γ - β)(-α – β - γ)]

Δ = (β - α)(γ - α)(γ - β)(α + β + γ)

Δ = (α - β)( β - γ)(γ - α)(α + β + γ)

Hence, the given result is proved.

Question 12:

x x² 1 + px³

y y² 1 + py³ = (1 + pxyz)(x - y)(y - z)(z - x)

z z² 1 + pz³

Answer:

Given,

x x² 1 + px³

Δ = y y² 1 + py³

z z² 1 + pz³

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

x x² 1 + px³

Δ = y - x y² - x² p(y³ - x³)

z - x z² - x² p(z³ - x³)

x x² 1 + px³

Δ = (y – x)(z - x) 1 y + x p(y² + x² + xy)

1 z + x p(z² + x² + zx)

Applying R₃ -> R₃ – R₂, we get

x x² 1 + px³

Δ = (y – x)(z - x) 1 y + x p(y² + x² + xy)

0 z - y p(z - y)(x + y + z)

x x² 1 + px³

Δ = (y – x)(z - x)(z - y) 1 y + x p(y² + x² + xy)

0 1 p(x + y + z)

Expanding along R₃, we get:

Δ = (y – x)(z - x)(z - y)[(-1) * p * (xy² + x³ + x²y) + 1 + px³ + p(x + y + z)(xy)]

Δ = (y – x)(z - x)(z - y)[-pxy² - px³ - px²y + 1 + px³ + px²y + pxy² + pxyz]

Δ = (x – y)(y - z)(z - x)(1 + pxyz)

Hence, the given result is proved.

Question 13:

3a -a + b -a + c

-b + a 3b -b + c = 3(a + b + c)(ab + bc + ca)

-c + a -c + b 3c

Answer:

Given,

3a -a + b -a + c

Δ = -b + a 3b -b + c

-c + a -c + b 3c

Applying C₁ -> C₁ + C₂ + C₃, we get

a + b + c -a + b -a + c

Δ = a + b + c 3b -b + c

a + b + c -c + b 3c

1 -a + b -a + c

Δ = (a + b + c) 1 3b -b + c

1 -c + b 3c

Applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

1 -a + b -a + c

Δ = (a + b + c) 0 2b + a a - b

0 a - c 2c + a

Expanding along C₁, we get:

Δ = (a + b + c)[(2b + a)(2c + a) – (a - b)(a - c)]

Δ = (a + b + c)[4bc + 2ab + 2ac + a² – a² + ac + ab - bc]

Δ = (a + b + c)(3ab + 3bc + 3ca)

Δ = 3(a + b + c)(ab + bc + ca)

Hence, the given result is proved.

Question 14:

1 1 + p 1 + p + q

2 3 + 2p 4 + 3p + 2q = 1

3 6 + 3p 10 + 6p + 3q

Answer:

Given,

1 1 + p 1 + p + q

Δ = 2 3 + 2p 4 + 3p + 2q

3 6 + 3p 10 + 6p + 3q

Applying R₂ -> R₂ – 2R₁ and R₃ -> R₃ – 3R₁, we get

1 1 + p 1 + p + q

Δ = 0 1 2 + p

0 3 7 + 3p

Applying R₃ -> R₃ – 3R₂, we get

1 1 + p 1 + p + q

Δ = 0 1 2 + p

0 0 1

Expanding along C₁, we get:

Δ = 0 + 0 + 1[1 * 1 – 0 *(2 + p)]

Δ = 1

Hence, the given result is proved.

Question 15:

sin α cos α cos(α + δ)

sin β cos β cos(β + δ) = 0

sin γ cos γ cos(γ + δ)

Answer:

Given,

sin α cos α cos(α + δ)

Δ = sin β cos β cos(β + δ)

sin γ cos γ cos(γ + δ)

sin α * sin δ cos α * cos δ cos α * cos δ - sin α * sin δ

Δ = 1/(sin δ * cos δ) sin β * sin δ cos β * cos δ cos β * cos δ - sin β * sin δ

sin γ * sin δ cos γ * cos δ cos γ * cos δ - sin γ * sin δ

Applying C₁ -> C₁ + C₃, we get

cos α * cos δ cos α * cos δ cos α * cos δ - sin α * sin δ

Δ = 1/(sin δ * cos δ) cos β * cos δ cos β * cos δ cos β * cos δ - sin β * sin δ

cos γ * cos δ cos γ * cos δ cos γ * cos δ - sin γ * sin δ

Δ = 0 [Since two columns C₁ and C₂ are identical]

Hence, the given result is proved.

Question 16:

Solve the system of the following equations

2/x + 3/y + 10/z = 4

4/x – 6/y + 5/z = 1

6/x + 9/y – 20/z = 2

Answer:

Let 1/x = p, 1/y = q and 1/z = r

Then the given system of equations is as follows:

2p + 3q + 10r = 4

4p – 6q + 5r = 1

6p + 9q – 20r = 2

This system of equations can be written in the form of AX = B, where

2 3 10 p 4

A = 4 -6 5 , X = q and B = 1

6 9 -20 r 2

|A| = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36)

= 150 + 330 + 720

= 1200

Now,

A₁₁ = 75, A₁₂ = 110, A₁₃ = 72

A₂₁ = 150, A₂₂ = -100, A₂₃ = 0

A₃₁ = 75, A₃₂ = 30, A₃₃ = -24

So, A^-1 = adj(A) /|A| = (1/1200) 75 150 75

110 -100 30

72 0 -24

So, X = A^-1B = (1/1200) 75 150 75 4

110 -100 30 1

72 0 -24 2

p = (1/1200) 300 + 150 + 150

q 400 - 100 + 60

r 288 + 0 - 48

x = (1/1200) 600 = 1/2

y 400 1/3

z 200 1/5

So, p =1/2, q = 1/3 and r = 1/5

Hence, x = 2, y = 3 and z = 5

Question 17:

Choose the correct answer.

If a, b, c, are in A.P., then the determinant

x + 2 x + 3 x + 2a

x + 3 x + 4 x + 2b

x + 4 x + 5 x + 2c

0 B. 1 C. x D. 2x

Answer:

Given,

x + 2 x + 3 x + 2a

Δ = x + 3 x + 4 x + 2b

x + 4 x + 5 x + 2c

x + 2 x + 3 x + 2a

Δ = x + 3 x + 4 x + (a + c)

x + 4 x + 5 x + 2c [If a, b, c are in AP then 2b = a + c]

Applying R₁ -> R₁ – R₂ and R₃ -> R₃ – R₂, we get

-1 -1 a - c

Δ = x + 3 x + 4 x + (a + c)

1 1 c - a

Applying R₁ -> R₁ + R₃, we get

0 0 0

Δ = x + 3 x + 4 x + (a + c)

1 1 c - a

Δ = 0 [Since all the elements of the first row (R₁) are zero]

Hence, the correct answer is option A.

Question 18:

If x, y, z are nonzero real numbers, then the inverse of matrix

A = x 0 0

0 y 0

0 0 z is

(A) x^-1 0 0 (B) x^-1 0 0

0 y^-1 0 xyz 0 y^-1 0

0 0 z^-1 0 0 z^-1

(1/xyz) 0 y 0 (1/xyz) 0 1 0

0 0 z 0 0 1

Answer:

Given,

A = x 0 0

0 y 0

0 0 z

So, |A| = x(yz - 0) + 0 + 0

= xyz

Now,

A₁₁ = yz, A₁₂ = 0, A₁₃ = 0

A₂₁ = 0, A₂₂ = xz, A₂₃ = 0

A₃₁ = 0, A₃₂ = 0, A₃₃ = xy

So, A^-1 = adj(A) /|A| = (1/xyz) yz 0 0

0 xz 0

0 0 xy

= 1/x 0 0

0 1/y 0

0 0 1/z

= x^-1 0 0

0 y^-1 0

0 0 z^-1

Hence, the correct answer is option A.

Question 19:

Choose the correct answer.

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) є (2, ∞) C). Det (A) є (2, 4) D). Det (A) є [2, 4]

Answer:

Given,

1 sin θ 1

A = -sin θ 1 sin θ

-1 -sin θ 1

Now, |A| = 1(1 + sin² θ) – sin θ(-sin θ + sin θ) + 1(sin² θ + 1)

=> |A| = 1 + sin² θ + sin² θ + 1

=> |A| = 2 + 2 sin² θ

=> |A| = 2(1 + sin² θ)

Now, 0 ≤ θ ≤ 2π

=> sin 0 ≤ sin θ ≤ sin 2π

=> 0 ≤ sin θ ≤ 1

=> 0 ≤ sin² θ ≤ 1

=> 1 + 0 ≤ 1 + sin² θ ≤ 1 + 1

=> 1 ≤ 1 + sin² θ ≤ 2

=> 2 * 1 ≤ 2(1 + sin² θ) ≤ 2 * 2

=> 2 ≤ 2(1 + sin² θ) ≤ 4

So, Det (A) є [2, 4]

Hence, the correct answer is option D.

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