Q. Determinant $\left| \begin{matrix} 1 & x & y \\ 2 & \sin{x + 2x} & \sin{y + 3y} \\ 3 & \cos{x + 3x} & \cos{y + 3y} \end{matrix} \right|$ is equal to

A) $\sin{(x - y)}$
B) $\cos{(x - y)}$
C) $\cos{(x + y)}$
D) $xy(\sin{(x - y)}$
Explanation

(a)
Let $\mathrm{\Delta} = \left| \begin{matrix}
1 & x & y \\
2 & \sin{x + 2x} & \sin{y + 3y} \\
3 & \cos{x + 3x} & \cos{y + 3y}
\end{matrix} \right|$
$= \left| \begin{matrix}
1 & x & y \\
0 & \sin x & \sin y \\
0 & \cos x & \cos y
\end{matrix} \right|$ $\ \binom{R_{2} \rightarrow R_{2} - 2R_{1},}{R_{3} \rightarrow R_{3} - 3R_{1}}$
$= \sin x\ \cos y - \cos x\ \sin y = \sin{(x - y)}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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