First slide

Introduction to Determinants

Question

$The solutions of the equation |\begin{matrix} 1 + \sin^{2} x & \sin^{2} x & \sin^{2} x \\ \cos^{2} x & 1 + \cos^{2} x & \cos^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0, (0 < x < π) a r e$

Moderate

A

$\frac{π}{12}, \frac{π}{6}$
B

$\frac{7 π}{12}, \frac{11 π}{12}$
C

$\frac{5 π}{12}, \frac{7 π}{12}$
D

$\frac{π}{6}, \frac{5 π}{6}$

Solution

$\begin{array}{l} The given determinant is |\begin{matrix} 1 + \sin^{2} x & \sin^{2} x & \sin^{2} x \\ \cos^{2} x & 1 + \cos^{2} x & \cos^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0, (0 < x < π) \\ use the column operations C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1} \\ The determinat becomes \\ |\begin{matrix} 1 + \sin^{2} x & - 1 & - 1 \\ \cos^{2} x & 1 & 0 \\ 4 \sin 2 x & 0 & 1 \end{matrix}| = 0 \\ by expanding the above determinant \\ (1 + \sin^{2} x) (1) + 1 (\cos^{2} x) - 1 (- 4 \sin 2 x) = 0 \\ 1 + \sin^{2} x + \cos^{2} x + 4 \sin 2 x = 0 \\ 2 + 4 \sin 2 x = 0 \\ \sin 2 x = \frac{- 1}{2} \\ 2 x = π + \frac{π}{6}, 2 π - \frac{π}{6} \\ x = \frac{7 π}{12}, \frac{11 π}{12} \end{array}$

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