left|begin{array}{ccc}sqrt{3} & 2 sqrt{5} & s | vedclass

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(D) To evaluate the determinant $\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \ \sqrt{15} & 5 & \sqrt{10} \ 3 & \sqrt{15} & 5\end{array}

Vedclass · Accepted Answer

$15 \sqrt{2}-25 \sqrt{3}$

Vedclass · Answer

(D) To evaluate the determinant $\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \ \sqrt{15} & 5 & \sqrt{10} \ 3 & \sqrt{15} & 5\end{array}ight|$, we expand along the first row:$\Delta = \sqrt{3}(5 	imes 5 - \sqrt{15} 	imes \sqrt{10}) - 2\sqrt{5}(\sqrt{15} 	imes 5 - 3 	imes \sqrt{10}) + \sqrt{5}(\sqrt{15} 	imes \sqrt{15} - 3 	imes 5)$$\Delta = \sqrt{3}(25 - \sqrt{150}) - 2\sqrt{5}(5\sqrt{15} - 3\sqrt{10}) + \sqrt{5}(15 - 15)$Since $\sqrt{150} = \sqrt{25 	imes 6} = 5\sqrt{6}$, we have:$\Delta = \sqrt{3}(25 - 5\sqrt{6}) - 2\sqrt{5}(5\sqrt{15} - 3\sqrt{10}) + 0$$\Delta = 25\sqrt{3} - 5\sqrt{18} - 10\sqrt{75} + 6\sqrt{50}$$\Delta = 25\sqrt{3} - 5(3\sqrt{2}) - 10(5\sqrt{3}) + 6(5\sqrt{2})$$\Delta = 25\sqrt{3} - 15\sqrt{2} - 50\sqrt{3} + 30\sqrt{2}$$\Delta = 15\sqrt{2} - 25\sqrt{3}$

$\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5\end{array}\right|=$

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