The determinant

Question:

The determinant $A=\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5\end{array}\right|$ is equal to is equal to

Solution:

Let $A=\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5\end{array}\right|$

$A=\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5\end{array}\right|$

$=\left|\begin{array}{ccc}\sqrt{3}+\sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{5} \sqrt{3}+\sqrt{2} \sqrt{23} & 5 & \sqrt{10} \\ \sqrt{3} \sqrt{3}+\sqrt{5} \sqrt{23} & \sqrt{15} & 5\end{array}\right|$

$=\left|\begin{array}{ccc}\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{5} \sqrt{3} & 5 & \sqrt{10} \\ \sqrt{3} \sqrt{3} & \sqrt{15} & 5\end{array}\right|+\left|\begin{array}{ccc}\sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{2} \sqrt{23} & 5 & \sqrt{10} \\ \sqrt{5} \sqrt{23} & \sqrt{15} & 5\end{array}\right|$

$=P+Q$.....(1)

where

$P=\left|\begin{array}{ccc}\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{5} \sqrt{3} & 5 & \sqrt{10} \\ \sqrt{3} \sqrt{3} & \sqrt{15} & 5\end{array}\right|$ and $Q=\left|\begin{array}{ccc}\sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{2} \sqrt{23} & 5 & \sqrt{10} \\ \sqrt{5} \sqrt{23} & \sqrt{15} & 5\end{array}\right|$

Now,

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

Taking out $\sqrt{3}, \sqrt{5}$ and $\sqrt{5}$ common from $C_{1}, C_{2}$ and $C_{3}$, respectively

$=\sqrt{3} \times \sqrt{5} \times \sqrt{5}\left|\begin{array}{ccc}1 & 1 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5}\end{array}\right|$

$=5 \sqrt{3}(0) \quad\left(\because C_{1}\right.$ and $C_{2}$ are identical, $\therefore$ determinant is zero)

$=0 \quad \ldots(2)$

$Q=\left|\begin{array}{ccc}\sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{2} \sqrt{23} & 5 & \sqrt{10} \\ \sqrt{5} \sqrt{23} & \sqrt{15} & 5\end{array}\right|$

Taking out $\sqrt{23}, \sqrt{5}$ and $\sqrt{5}$ common from $C_{1}, C_{2}$ and $C_{3}$, respectively

$=\sqrt{23} \times \sqrt{5} \times \sqrt{5}\left|\begin{array}{ccc}1 & 1 & 1 \\ \sqrt{2} & \sqrt{5} & \sqrt{2} \\ \sqrt{5} & \sqrt{3} & \sqrt{5}\end{array}\right|$

$=5 \sqrt{23}(0)$   $\left(\because C_{1}\right.$ and $C_{3}$ are identical, $\therefore$ determinant is zero)

$=0 \quad \ldots(3)$

From $(1),(2)$ and $(3)$, we get

\begin{aligned} A &=0+0 \\ &=0 \end{aligned}

Hence, the determinant $A=\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5\end{array}\right|$ is equal to $\underline{0}$.