Question

If the system of linear equations $(\sin\theta)x - y + z = 0$, $x - (\cos\theta)y + z = 0$, $x + y + (\sin\theta)z = 0$ has a non-trivial solution, then the least positive value of $\theta$ is

• A. $\pi/6$
• B. $\pi/4$
• C. $\pi/3$
• D. $\pi/2$

Hint:

For a homogeneous system of equations $AX = 0$, a non-trivial solution (a solution other than all variables being zero) exists only if $\det(A) = 0$. This is a fundamental concept in linear algebra.

Answer 1

The Correct Option is D

A system of homogeneous linear equations has a non-trivial solution if and only if the determinant of the coefficient matrix is zero.
The given system of equations is:
$(\sin\theta)x - y + z = 0$
$x - (\cos\theta)y + z = 0$
$x + y + (\sin\theta)z = 0$
The determinant of the coefficient matrix is:
$\Delta = \begin{vmatrix} \sin\theta & -1 & 1 \\ 1 & -\cos\theta & 1 \\ 1 & 1 & \sin\theta \end{vmatrix}$
Setting the determinant to zero for a non-trivial solution:
$\sin\theta(-\cos\theta\sin\theta - 1) - (-1)(\sin\theta - 1) + 1(1 - (-\cos\theta)) = 0$
$-\sin^2\theta\cos\theta - \sin\theta + \sin\theta - 1 + 1 + \cos\theta = 0$
$-\sin^2\theta\cos\theta + \cos\theta = 0$
Factor out $\cos\theta$:
$\cos\theta(1 - \sin^2\theta) = 0$
Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get $1 - \sin^2\theta = \cos^2\theta$.
$\cos\theta(\cos^2\theta) = 0$
$\cos^3\theta = 0$
$\cos\theta = 0$.
The general solution for $\cos\theta = 0$ is $\theta = (2n + 1)\frac{\pi}{2}$, where $n$ is an integer.
We are looking for the least positive value of $\theta$.
Let $n = 0$, then $\theta = (2(0) + 1)\frac{\pi}{2} = \frac{\pi}{2}$.
This is the smallest positive value for $\theta$.