Question 1:

Prove that the determinant  is independent of θ.

 

Answer:

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

= x³ – x + x sin θ + sin θ cos θ – sin θ cos θ + x cos² θ

= x³ – x + x (sin² θ + cos² θ)

= x³ – x + x

= x³ (Independent of θ)

Hence, Δ is independent of θ.

 

Question 2: Without expanding the determinant, prove that

Answer

Hence, the given result is proved.

 

Question 3:

Answer

Expanding along C₃, we have:

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

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

= sin² α (1) + cos² α (1)

= 1

 

Question 4:

If a, b and c are real numbers, and ,

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

Answer

Applying R₁→R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ - C₁ and C₃ → C₃ - C₁ we have:

Expanding along R₁, we have:

Δ = 2(a + b + c) (1) [(b - c) (c -b) (b - a) (c - a)]

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

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

If is given that Δ = 0.

(a + b + c) [Ab + bc + ca – a² – b² – c²] = 0

⇒ Either a + b + c = 0, or ab + bc + ca – a² – b² – c² = 0.

Now,

ab + bc + ca – a² – b² – c² = 0

⇒ - 2ab – 2bc – 2ca + 2a² + 2b² + 2c² = 0

⇒ (a - b)² + (b - c)² + (c - a)² = 0

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

⇒ (a - b) = (b - c) = (c - a) = 0

⇒ a  = b = c

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

 

Question 5:

Solve the equations

Answer

Applying R₁ → R₁ + R₂ + R₃, we get:

Applying C₂→ C₂ – C₂ and C₃→C₃ – C₁, we have:

Expanding along R₁, we have:

(3x + a) [1 x a²] = 0

⇒ a² (3x + a) = 0

But a ≠ 0.

Therefore, we have:

3x + a = 0

⇒ x = -(a/3)

Question 6:

Answer:

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

Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we have:

Applying R₂ → R₂ + R₁, we have:

Applying R₃ → R₃ + R₂, we have:

Applying C₂ → C₂ + C₁, we have:

Expanding along R₃, we have:

∆ = 2ab²c [a(c - a) + a(a + c)]

= 2ab²c [ac – a² + a² + ac]

= 2ab²c (2ac)

= 4 a²b²c²

Hence, the given result is proved.

 

Question 8:

Let verify that

(i) [adjA]^-1 = adj(A^-1)

(ii) (A^-1)^-1 = A

Answer:

∴ |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

(i)

|adjA| = 14 (-4 - 9) – 11 (-11 - 5) -5 (-33 + 20)

= 14 (-13) -11 (-26) -5 (-13)

= -182 + 286 + 65 = 169

We have,

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

(ii)

We have shown that:

Now,

|A^-1| = (1/13)³ [-14 x (-13) + 11 x (-26) + 5 x (-13)] = (1/13)3 x (-169) = -(1/13)

 

Question 9:

Answer

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ – C₁ and C₃ → C₃ – C₂, we have:

Expanding along R₁, we have:

Δ = 2(x + y) [-x² + y (x - y)]

= -2 (x + y) (x² + y² - yx)

= -2 (x³ + y³)

 

Question 10:

Answer:

Expanding along C₁, we have:

 Δ = 1 (xy - 0) = xy

 

Question 11:

Using properties of determinants, prove that:

Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₂, we have:

Applying R₃ → R₃ – R₂, we have:

Expanding along R₃, we have:

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

= (β - α) (γ - α) (γ - α) (α + β + γ)

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

Hence, the given result is proved.

 

Question 12:

Using properties of determinants, prove that:

Answer:

Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₂, we have:

Applying R₃ → R₃ – R₂, we have:

 Expanding along R₃, we have:

Δ = (x - y) (y – z) (z - x) [(-1) (p) (xy² + x² + x²y) + 1 + px³ + p(x + y + z) (xy)] 

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

= (x - y) (y – z) (z - x) (1 + pxyz)

Hence, the given result is proved.

 

Question 13:

Using properties of determinants, prove that:

Answer

Applying C₁ → C₁ + C₂ + C₃, we have:

Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₂, we have:

Expanding along C₁, we have:

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

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

= (a + b + c) (3ab + 3bc + 3ac)

= 3(a + b + c) (ab + bc + ac)

Hence, the given result is proved.

 

Question 14:

Using properties of determinants, prove that:

Answer

Applying R₂ → R₂ – 2R₁ and R₃ → R₃ – 3R₁, we have:

Applying R₃ → R₃ – 3R₂, we have:

Expanding along C₁, we have:

Hence, the given result is proved.

 

Question 15:

Using properties of determinants, prove that:

Answer

Applying C₁ → C₁ + C₃, we have:

Here, two columns C₁ and C₂ are identical.

∴ Δ = 0.

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, (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 can be written in the form of AX = B, where 

Now,

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

= 150 + 330 + 720

= 1200

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

Now,

A₁₁ = 75, A₁₂ = 110, A₁₃ = 72

A₂₁ = 150, A₂₂ = −100, A₂₃ = 0

A₃₁ = 75, A₃₂ = 30, A₃₃ = − 24

∴ A^-1 = (1 / |A|) adjA

Now,

X = A^-1B

∴ p = ½,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

A. 0                      B. 1                      C. x                      D. 2x

Answer

Answer: A

Applying R₁ → R₁ – R₂ and R₃ → R₃ – R₂, we have:

Applying R₁ → R₁ – R₃, we have:

 Here, all the elements of the first row (R₁) are zero.

Hence, we have Δ = 0.

The correct answer is A.

 

Question 18:

Choose the correct answer.

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

Answer: A

∴ |A| = x(yz - 0) = xyz ≠ 0

Now,  A₁₁ = yz, A₁₂ = 0, A₁₃ = 0

A₂₁ = 0, A₂₂ = −1xz, A₂₃ = 0

A₃₁ = 0, A₃₂ = 0, A₃₃ = xy

∴ A^-1 = (1 / |A|)adjA

The correct answer is A.

 

Question 19:

Choose the correct answer.

Let , where 0 ≤ θ ≤ 2n, then 

A). Det (A) = 0

B.) Det (A) ∈ (2, ∞)

C.) Det (A) ∈ (2, 4)

D.) Det (A) ∈ [2, 4]

Answer: D

∴ |A| = 1(1 + sin² θ) – sin θ(sin θ + sin θ) + 1 (sin² θ + 1)

= 1 + sin² θ + sin² θ + 1

= 2 + 2sin² θ

= 2(1 + sin² θ)

Now, 0 ≤ θ ≤ 2π

⇒ 0 ≤ sin θ ≤ 1

⇒ 0 ≤ sin² θ ≤ 1

⇒ 1 ≤ 1 + sin² θ ≤ 2

⇒ 2 ≤ 2 (1+ sin² θ) ≤ 4

∴ Det (A) ∈ [2, 4]

The correct answer is D.