The determinant Δ=13+325515+265103+65155 equals

Taking 5 common from C2 and C3, we get Δ=(5)213+32115+26523+6535Applying C1→C1−13C3−3C2, we get Δ=(5)−321052035=5(−3)(5−6) =5(18)−253=152−253

The determinant Δ=13+325515+265103+65155 equals

Taking 5 common from C2 and C3, we get Δ=(5)213+32115+26523+6535Applying C1→C1−13C3−3C2, we get Δ=(5)−321052035=5(−3)(5−6) =5(18)−253=152−253

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The determinant $Δ = |\begin{array}{ccc} \sqrt{13} + \sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{26} & 5 & \sqrt{10} \\ 3 + \sqrt{65} & \sqrt{15} & 5 \end{array}|$ equals

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$- 15 \sqrt{2} + 7 \sqrt{3}$

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

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

$3 \sqrt{5}$

answer is A.

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Detailed Solution

Taking $\sqrt{5}$ common from $C_{2}$ and $C_{3}$ , we get
$Δ = (\sqrt{5})^{2} |\begin{array}{ccc} \sqrt{13} + \sqrt{3} & 2 & 1 \\ \sqrt{15} + \sqrt{26} & \sqrt{5} & \sqrt{2} \\ 3 + \sqrt{65} & \sqrt{3} & \sqrt{5} \end{array}|$
Applying $C_{1} \to C_{1} - \sqrt{13} C_{3} - \sqrt{3} C_{2}$ , we get

$\begin{array}{l} Δ = (5) |\begin{array}{ccc} - \sqrt{3} & 2 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}| = 5 (- \sqrt{3}) (5 - \sqrt{6}) \\ = 5 (\sqrt{18}) - 25 \sqrt{3} = 15 \sqrt{2} - 25 \sqrt{3} \end{array}$