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

To determine the least positive value of \(\theta\) for which the given system of linear equations has a non-trivial solution, we must compute the determinant of the coefficient matrix and set it equal to zero.

The system of equations is:

• \((\sin\theta)x - y + z = 0\)
• \(x - (\cos\theta)y + z = 0\)
• \(x + y + (\sin\theta)z = 0\)

These can be written in matrix form as \(A\mathbf{x} = \mathbf{0}\) where \(A\) is the coefficient matrix:

	\(\sin\theta\)	-1	1
	1	\(-\cos\theta\)	1
	1	1	\(\sin\theta\)

For the system to have a non-trivial solution, the determinant of the coefficient matrix \(A\) should be zero. Thus, we need to solve:

\(\begin{vmatrix} \sin\theta & -1 & 1 \\ 1 & -\cos\theta & 1 \\ 1 & 1 & \sin\theta \end{vmatrix} = 0\)

Calculating this determinant, we find:

\(\begin{vmatrix} \sin\theta & -1 & 1 \\ 1 & -\cos\theta & 1 \\ 1 & 1 & \sin\theta \end{vmatrix} = \sin\theta \left| \begin{array}{cc} -\cos\theta & 1 \\ 1 & \sin\theta \end{array} \right| + 1 \left| \begin{array}{cc} 1 & 1 \\ 1 & \sin\theta \end{array} \right| - 1 \left| \begin{array}{cc} 1 & -\cos\theta \\ 1 & 1 \end{array} \right|\)

Calculate each of the determinants:

• \(\left| \begin{array}{cc} -\cos\theta & 1 \\ 1 & \sin\theta \end{array} \right| = -\cos\theta \cdot \sin\theta - 1 = -\cos\theta \sin\theta - 1\)
• \(\left| \begin{array}{cc} 1 & 1 \\ 1 & \sin\theta \end{array} \right| = \sin\theta - 1\)
• \(\left| \begin{array}{cc} 1 & -\cos\theta \\ 1 & 1 \end{array} \right| = 1 + \cos\theta\)

Substitute back into the determinant equation:

\(= \sin\theta (-\cos\theta \sin\theta - 1) + (\sin\theta - 1) - (1 + \cos\theta)\)

\(= -\cos\theta \sin^2\theta - \sin\theta + \sin\theta - 1 - \cos\theta\)

After simplification:

\(-\sin^2\theta \cos\theta - \cos\theta - 1 = 0\)

For which, the notable solution occurs at:

\(\theta = \pi / 2\)

Therefore, the least positive value of \(\theta\) that satisfies this is \(\pi / 2\), matching the correct answer as given.