Matrices & DeterminantshardFree

Matrices & Determinants: Vmatrix Vmatrix Equals

JEE Maths question with a full step-by-step solution.

Question

0 < \theta < \dfrac{\pi}{2}

and

\begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin4\theta \\ \sin^2\theta & 1+\cos^2\theta & 4\sin4\theta \\ \sin^2\theta & \cos^2\theta & 1+4\sin4\theta \end{vmatrix} = 0

, then

\theta

equals:

\dfrac{\pi}{24},\, \dfrac{5\pi}{24}

\dfrac{5\pi}{24},\, \dfrac{7\pi}{24}

\dfrac{7\pi}{24},\, \dfrac{11\pi}{24}

correct

DNone of these

Solution

Step 1: Apply

R_2 \to R_2 - R_1

and

R_3 \to R_3 - R_1

\begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin4\theta \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = 0

Step 2: Expand along row 1 using cofactors

A_{11} = (+1)\begin{vmatrix}1&0\\0&1\end{vmatrix} = +1, \quad A_{12} = (-1)^{1+2}\begin{vmatrix}-1&0\\-1&1\end{vmatrix} = (-1)(-1) = +1

\quad A_{13} = (+1)\begin{vmatrix}-1&1\\-1&0\end{vmatrix} = +1

Therefore:

(1+\sin^2\theta)(+1) + \cos^2\theta(+1) + 4\sin4\theta(+1) = 0

1 + \sin^2\theta + \cos^2\theta + 4\sin4\theta = 0 \implies 2 + 4\sin4\theta = 0

Step 3: Solve for

\theta

\sin4\theta = -\frac{1}{2} \implies 4\theta = n\pi+(-1)^n\!\left(-\frac{\pi}{6}\right)

For

\theta \in \left(0,\dfrac{\pi}{2}\right)

n=1

4\theta = \dfrac{7\pi}{6} \implies \theta = \dfrac{7\pi}{24}

\quad n=2

4\theta = \dfrac{11\pi}{6} \implies \theta = \dfrac{11\pi}{24}

. Answer: (3)

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