Question

If \( \begin{bmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{bmatrix} \) is a singular matrix, then find all values of \( x \) where \( x \in [0, \pi/2] \).

Hint:

For singular matrices, \( |A| = 0 \).
When solving trigonometric equations, always filter solutions based on the given interval \( [0, \pi/2] \).

Answer 1

Concept:
• Singular Matrix: A matrix whose determinant is zero.
• Trigonometric Identities: \( \sin 3x = 3\sin x - 4\sin^3 x, \cos 2x = 1 - 2\sin^2 x \).

Step 1: Expand the determinant and set it to zero
\[ 3(8 - 7\cos 2x) - (-1)(-14 + 11\cos 2x) + \sin 3x(-49 + 44) = 0 \] \[ 24 - 21\cos 2x - 14 + 11\cos 2x - 5\sin 3x = 0 \] \[ 10 - 10\cos 2x - 5\sin 3x = 0 \implies 2 - 2\cos 2x - \sin 3x = 0 \]

Step 2: Substitute identities and simplify
\[ 2 - 2(1 - 2\sin^2 x) - (3\sin x - 4\sin^3 x) = 0 \] \[ 2 - 2 + 4\sin^2 x - 3\sin x + 4\sin^3 x = 0 \] \[ 4\sin^3 x + 4\sin^2 x - 3\sin x = 0 \]

Step 3: Solve the equation
\[ \sin x (4\sin^2 x + 4\sin x - 3) = 0 \]
• \( \sin x = 0 \implies x = 0 \in [0, \pi/2] \).
• \( 4\sin^2 x + 4\sin x - 3 = 0 \implies (2\sin x - 1)(2\sin x + 3) = 0 \).
• \( \sin x = 1/2 \implies x = \pi/6 \in [0, \pi/2] \).
• \( \sin x = -3/2 \) (Impossible). The values of \( x \) are \( 0 \) and \( \pi/6 \).