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 ____________.

Answer 1

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


$$\begin{gathered} 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...\left(1\right) \\ \mathrm{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|\mathrm{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| \\ \mathrm{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| \\ \mathrm{Taking}\mathrm{out}\sqrt{3},\sqrt{5}\mathrm{and}\sqrt{5}\mathrm{common}\mathrm{from}{C}_1,C_2\mathrm{and}{C}_3,\mathrm{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}\left(0\right)\left(\because C_1\mathrm{and}{C}_2\mathrm{are}\mathrm{identical},\therefore \mathrm{determinant}\mathrm{is}\mathrm{zero}\right) \\ =0...\left(2\right) \\ 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| \\ \mathrm{Taking}\mathrm{out}\sqrt{23},\sqrt{5}\mathrm{and}\sqrt{5}\mathrm{common}\mathrm{from}{C}_1,C_2\mathrm{and}{C}_3,\mathrm{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}\left(0\right)\left(\because C_1\mathrm{and}{C}_3\mathrm{are}\mathrm{identical},\therefore \mathrm{determinant}\mathrm{is}\mathrm{zero}\right) \\ =0...\left(3\right) \\ \mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ A=0+0 \\ =0 \end{gathered}$$

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 0.