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Question 1:

Consider the following system of equations:

x + 2y – 3z = a

2x + 6y – 11 z = b

x – 2y + 7z = c

Where a, b and c are real constants. Then, the system of equations

(a) has a unique solution when 5a = 2b + c

(b) has an infinite number of solutions when 5a = 2b + c

(c) has no solution for all a, b and c

(d) has a unique solution for all a, b and c

Solution:

Given equations are x + 2y – 3z = a

2x + 6y – 11 z = b

x – 2y + 7z = c

= 20 – 50 + 30

= 0

= 20a – 2(7b + 11c) – 3(- 2b – 6c)

= 20a – 14b – 22c + 6b + 18c

= 20a – 8b – 4c

= 4(5a – 2b – c)

= 7b+ 11c – a(25) – 3(2c – b)

= 7b+ 11c – 25a – 6c + 3b

= – 25a + 10b+ 5c

= – 5(5a – 2b- c)

= 6c + 2b- 2(2c – b) – 10a

= – 10a + 4b+ 2c

= – 2(5a – 2b- c)

For infinite solution,

Δ = Δ₁ = Δ₂ = Δ₃ = 0

⇒ 5a = 2b + c

Then, the system of equations has an infinite number of solutions when 5a = 2b + c.

Hence, option (b) is the answer.

Question 2:

Let

Solution:

|P^-1AP – I|²

= |(P^-1AP – I) (P^-1AP – I)|

= |P^-1APP^-1 AP – 2P^-1AP + I|

= |P^-1A²P – 2P^-1AP + P^-1IP|

= |P^-1(A² – 2A + I) P|

= |P^-1(A – I)² P|

= |P^-1| |A – I|² |P|

= |A – I|²

= (1 (ω (ω + 1) + ω) – 7ω + ω² . ω)²

= (ω² + 2ω – 7ω + 1)²

= (ω² – 5ω + 1)²

= (- 6ω)²

= 36ω²

⇒ ɑ = 36

Hence, the value of ɑ is equal to 36.

Question 3:

The maximum value of

(a) √7

(b) √5

(c) 5

(d) 3 / 4

Solution:

Given that

C₁ → C₁ + C₂

R₁ → R₁ – R₂

= (–1)[2 sin2x – cos2x] = cos2x – 2sin2x

Maximum value = √5

Hence, option (b) is the answer.

Question 3:

The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k² has no solution if k is equal to

(a) – 2

(b) – 1

(c) 1

(d) 0

Solution:

Given equations are kx + y + z = 1

x + ky + z = k

x + y + zk = k²

These equations can be written as:

The condition for no solution to the given system of equations is |A| = 0.

⇒ k(k² – 1) – (k – 1) + (1 – k) = 0

⇒ (k – 1) (k² + k – 1 – 1) = 0

⇒ (k – 1) (k² + k – 2) = 0

⇒ (k –1) (k –1) (k + 2) = 0

⇒ k = 1, k = -2

For k = 1, the given system of equations is identical. So, the given equations have no solution when k = -2.

Hence, option (a) is the answer.

Question 4:

If

Solution:

|A¹⁰ – adj (2A)¹⁰| = 0

|A|⁴ = 16

Hence, the value of det (A⁴) + det (A¹⁰ – Adj (2A))¹⁰) = 16

Question 5:

If x, y, z are in arithmetic progression with common difference d, x ≠ 3d, and the determinant of the matrix

(a) 6

(b) 36

(c) 72

(d) 12

Solution:

x, y, z are in AP, so x + z = 2y

And

Hence, option (c) is the answer.

Question 6:

If 1, log₁₀ (4^x – 2) and log₁₀ [4^x + (18 / 5)] are in arithmetic progression for a real number x, then the value of the determinant

Solution:

1, log₁₀ (4^x – 2) and log₁₀ [4^x + (18 / 5)] are in arithmetic progression.

2. log₁₀ (4^x – 2) = 1 + log₁₀ [4^x + (18 / 5)]

log₁₀ [4^x – 2]² = log₁₀ [10 . [4^x + (18 / 5)]

[4^x – 2]² = 10 . [4^x + (18 / 5)]

(4^x)² + 4 – 4 . 4^x = 10 . 4^x + 36

(4^x)² – 14 . 4^x – 32 = 0

(4^x)² + 2 . 4^x – 16 . 4^x – 32 = 0

4^x [4^x + 2] – 16 [4^x + 2] = 0

4^x = – 2 or 4^x = 16

Rejected the value of 4^x = – 2

Therefore,

= 3 (-2) – 1 (0 – 4) + 4 (1 – 0)

= – 6 + 4 + 4

= 2

Question 7:

Let α, β, γ be the roots of the equations, x³ + ax² + bx + c = 0, (a, b, c ∈ R and a, b and a, b ≠ 0). The system of the equations (in u, v, w) given by αu + βv + γw = 0; βu + γv + αw = 0; γu + αv + βw = 0 has non-trivial solutions, then the value of a² / b is

(a) 5

(b) 1

(c) 0

(d) 3

Solution:

Let α, β, γ be the roots of the equations, x³ + ax² + bx + c = 0

Given that αu + βv + γw = 0; βu + γv + αw = 0; γu + αv + βw = 0 has non-trivial solutions,

For non-trivial solutions,

α³ + β³ + γ³ – 3αβγ = 0

(α + β + γ) ((α + β + α)² – 3 ∑αβ) = 0

(-a) [a² – 3b] = 0

a² = 3b [because a ≠ 0]

a²/b = 3

Hence, option (d) is the answer.

Question 8:

The solutions of the equation

(a) π/6, 5π/6

(b) 7π/12, 11π/12

(c) 5π/12, 7π/12

(d) π/12, π/6

Solution:

Therefore, 2 + 8 sin 2x – 4 sin 2x = 0

⇒ sin 2x = – 1 / 2

⇒ x = 7π/12, 11π/12

Hence, option (b) is the answer.

Question 9:

Let the system of linear equations

4x + λy + 2z = 0

2x – y + z = 0

μx + 2y + 3z = 0, λ, μ ∈ R

has a non-trivial solution. Then which of the following is true?

(a) μ = 6, λ ∈ R

(b) λ = 2, μ ∈ R

(c) λ = 3, μ ∈ R

(d) μ = – 6, λ ∈ R

Solution:

Given, 4x + λy + 2z = 0

2x – y + z = 0

μx + 2y + 3z = 0

For a non-trivial solution,

Δ = 0

4 (- 3 – 2) – λ (6 – μ) + 2 (4 + μ) = 0

– 20 – 6λ + λμ + 8 + 2μ = 0

– 12 – 6λ + λμ + 2μ = 0

– 6 (λ + 2) + μ (λ + 2) = 0

(λ + 2) (μ – 6) = 0

μ = 6, λ ∈ R [since for λ = -2 the first two equations will become identical]

Hence, option (a) is the answer.

Question 10:

Let A = [a_ij] and B = [b_ij] be two 3 × 3 real matrices such that b_ij = (3)^(i+j−2)a_ji, where i, j = 1, 2, 3. If the determinant of B is 81, then the determinant of A is

(a) 1/9

(b) 1/81

(c) 1/3

(d) 3

Solution:

b_i = (3)^(i+j−2)a_ji

B =

Taking 3² common each from C₃ and R₃

Taking 3 common each from C₂ and R₂

Given ǀBǀ = 81

81 = 81×9 ǀAǀ [since |A| = |A|^T]

ǀAǀ = 1/9

Hence, option (a) is the answer.

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