Question

The determinant $ \Delta=\begin{vmatrix} \sqrt{13}+\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{26} & 5 & \sqrt{10} \\ 3+\sqrt{65} & \sqrt{15} & 5 \end{vmatrix}\text { equals }$

• A. $15 \sqrt{2}-25 \sqrt{3}$
• B. $25 \sqrt{3}-15 \sqrt{2}$
• C. $3 \sqrt{5}$
• D. $-15 \sqrt{2}+7 \sqrt{3}$

Answer 1

Answer: (A)

Taking $\sqrt{5}$ common from $C_2$ and $C_3$, we get
$\Delta=(\sqrt{5})^2\begin{vmatrix}\sqrt{13}+\sqrt{3} & 2 & 1 \\\sqrt{15}+\sqrt{26} & \sqrt{5} & \sqrt{2} \\3+\sqrt{65} & \sqrt{3} & \sqrt{5}\end{vmatrix}$
Applying $C_1 \rightarrow C_1-\sqrt{13} C_3-\sqrt{3} C_2$, we get
$\Delta =(5)\begin{vmatrix}-\sqrt{3} & 2 & 1 \\0 & \sqrt{5} & \sqrt{2} \\0 & \sqrt{3} & \sqrt{5}\end{vmatrix}=5(-\sqrt{3})(5-\sqrt{6}) $
$ =5(\sqrt{18})-25 \sqrt{3}=15 \sqrt{2}-25 \sqrt{3}$