Q. The value of the determinant, $\left| \begin{matrix} \sqrt{13} + \sqrt{3} & 2\sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{26} & 5 & \sqrt{10} \\ 3 + \sqrt{65} & \sqrt{15} & 5 \end{matrix} \right|$ is

A) $5(\sqrt{6} - 5)$
B) $5\sqrt{3}\ (\sqrt{6} - 5)$
C) $\sqrt{5}\ (\sqrt{6} - \sqrt{3}\ )$
D) $\sqrt{2}\ (\sqrt{7} - \sqrt{5}\ )$
Explanation

(b)
$\left| \begin{matrix}
\sqrt{13} + \sqrt{3} & 2\sqrt{5} & \sqrt{5} \\
\sqrt{15} + \sqrt{26} & 5 & \sqrt{10} \\
3 + \sqrt{65} & \sqrt{15} & 5
\end{matrix} \right|$
$= \left| \begin{matrix}
\sqrt{13} & 2\sqrt{5} & \sqrt{5} \\
\sqrt{16} & 5 & \sqrt{10} \\
\sqrt{65} & \sqrt{15} & 5
\end{matrix} \right| + \left| \begin{matrix}
\sqrt{3} & 2\sqrt{5} & \sqrt{5} \\
\sqrt{15} & 5 & \sqrt{10} \\
3 & \sqrt{15} & 5
\end{matrix} \right|$
$= \sqrt{13}.\ \sqrt{5}.\sqrt{5}\ \left| \begin{matrix}
1 & 2 & 3 \\
\sqrt{2} & \sqrt{5} & \sqrt{2} \\
\sqrt{5} & \sqrt{3} & \sqrt{5}
\end{matrix} \right|$
$+ \sqrt{3.}\sqrt{5}.\sqrt{5}\left| \begin{matrix}
1 & 2 & 3 \\
\sqrt{5} & \sqrt{5} & \sqrt{2} \\
\sqrt{3} & \sqrt{3} & \sqrt{5}
\end{matrix} \right|$
$= 0 + 5\sqrt{3}\left| \begin{matrix}
- 1 & 2 & 1 \\
0 & \sqrt{5} & \sqrt{2} \\
0 & \sqrt{3} & \sqrt{5}
\end{matrix} \right| = 5\sqrt{3\ }(\sqrt{6} - 5)$

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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