Question

Prove that $\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3}& \sqrt{5}& \sqrt{5}\\ \sqrt{15}+\sqrt{46}& 5& \sqrt{10}\\ 3+\sqrt{115}& \sqrt{15}& 5\end{array}\right|=0$

Answer 1

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

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

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

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

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

In the first determinant using $C_1\to \frac{C_1}{\sqrt{23}},C_2\to \frac{C_2}{\sqrt{5}}$ and $C_3\to \frac{C_2}{\sqrt{5}}$

In the second determinant using $C_1\to \frac{C_1}{\sqrt{3}},C_2\to \frac{C_2}{\sqrt{5}}$ and $C_3\to \frac{C_2}{\sqrt{5}}$

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

$C_1\to C_1-C_3$ in the first determinant and $C_1\to C_1-C_2$

$5\sqrt{23}\left|\begin{array}{ccc}0& 1& 1\\ 0& \sqrt{5}& \sqrt{2}\\ 0& \sqrt{3}& \sqrt{5}\end{array}\right|+5\sqrt{3}\left|\begin{array}{ccc}0& 1& 1\\ 0& \sqrt{5}& \sqrt{2}\\ 0& \sqrt{3}& \sqrt{5}\end{array}\right|$

$=0$ since one column in both the determinants are zero

Hence Proved