Question

The value of the determinant $\Delta =\begin{vmatrix} \sqrt{13}+\sqrt{3} & 2\sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{26} & 5 & \sqrt{10} \\ 3+\sqrt{65} & \sqrt{15} & 5 \end{vmatrix}$ is equal to

• 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
$\Delta =\left(\sqrt{5}\right)^{2}\begin{vmatrix} \sqrt{13}+\sqrt{3} & 2 & 1 \\ \sqrt{15}+\sqrt{26} & \sqrt{5} & \sqrt{2} \\ 3+\sqrt{65} & \sqrt{3} & \sqrt{5} \end{vmatrix}$
Applying $C_{1} \rightarrow C_{1}-\sqrt{13}C_{3}-\sqrt{3}C_{2},$ we get
$\Delta =\left(5\right)\begin{vmatrix} -\sqrt{3} & 2 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{vmatrix}=5\left(- \sqrt{3}\right)\left(5 - \sqrt{6}\right)$
$=5\left(\sqrt{18}\right)-25\sqrt{3}=15\sqrt{2}-25\sqrt{3}$