Question

The value of q for which the system of equations

(sin 3q) x – 2y + 3z = 0 cos 2q) x + 8y – 7z = 0

2x + 14y – 11z = 0 has a non-trivial solution is –

• A. np
• B. np + (–1)ⁿp/3
• C. np + (–1)ⁿp/8
• D. None

Answer 1

Answer: (A)

np

Answer 2

The system of equations has a non-trivial solution if and only if $\left| \begin{matrix} \sin 3\theta & - 2 & 3 \\ \cos 2\theta & 8 & - 7 \\ 2 & 14 & - 11 \end{matrix} \right|$= 0

Applying R₂ ® R₂ + 4R₁, R₃ ® R₃ + 7R₁, we get

$\left| \begin{matrix} \sin 3\theta & - 2 & 3 \\ \cos 2\theta + 4\sin 3\theta & 0 & 5 \\ 2 + 7\sin 3\theta & 0 & 10 \end{matrix} \right|$= 0

Expanding along C₂, we get

2(cos 2q + 4 sin 3q) – (2 + 7 sin 3q) = 0

Ž 2 – 2 cos 2q – sin 3q = 0

Ž 4 sin² q – (3 sin q – 4 sin³ q) = 0

Ž sin q (4 sin² q + 4 sin q – 3) = 0

Ž sin q (2 sin q – 1) (2 sin q + 3) = 0

Ž sin q = 0 or sin q = 1/2.

[Q sin q = –3/2 is non possible]

\ For, q = np the system of equations has a non-trivial

solution.