Question

Find the value of the determinant $\left|\begin{array}{ccc}\sqrt{13}+\sqrt{3}& 2\sqrt{5}& \sqrt{5}\\ \sqrt{15}+\sqrt{26}& 5& \sqrt{10}\\ 3+\sqrt{65}& \sqrt{15}& 5\end{array}\right|$ = $a\sqrt{2}-b\sqrt{3}$, then find $b-a?$

Answer 1

Let $\Delta =\left|\begin{array}{ccc}\sqrt{13}+\sqrt{3}& 2\sqrt{5}& \sqrt{5}\\ \sqrt{15}+\sqrt{26}& 5& \sqrt{10}\\ 3+\sqrt{65}& \sqrt{15}& 5\end{array}\right|$
$=\left|\begin{array}{ccc}\sqrt{3}& 2\sqrt{5}& \sqrt{5}\\ \sqrt{15}& 5& \sqrt{10}\\ 3& \sqrt{15}& 5\end{array}\right|+\left|\begin{array}{ccc}\sqrt{13}& 2\sqrt{5}& \sqrt{5}\\ \sqrt{26}& 5& \sqrt{10}\\ \sqrt{65}& \sqrt{15}& 5\end{array}\right|$
Taking common from 1st determinant $\sqrt{3},\sqrt{5},\sqrt{5}$ from $C_1,C_2,C_3$ respectively and
Taking common from 2nd determinant $\sqrt{13},\sqrt{5},\sqrt{5}$ from $C_1,C_2,C_3$ respectively, we get
$=\sqrt{3}\times \sqrt{5}\times \sqrt{5}\left|\begin{array}{ccc}1& 2& 1\\ \sqrt{5}& \sqrt{5}& \sqrt{2}\\ \sqrt{3}& \sqrt{3}& \sqrt{5}\end{array}\right|+\sqrt{13}\times \sqrt{5}\times \sqrt{5}\left|\begin{array}{ccc}1& 2& 1\\ \sqrt{2}& \sqrt{5}& \sqrt{2}\\ \sqrt{5}& \sqrt{3}& \sqrt{5}\end{array}\right|$
$=\sqrt{3}\times 5\left|\begin{array}{ccc}1& 2& 1\\ \sqrt{5}& \sqrt{5}& \sqrt{2}\\ \sqrt{3}& \sqrt{3}& \sqrt{5}\end{array}\right|+0$ ($\because C_1$ and $C_2$
are indentical)
$=5\sqrt{3}\left|\begin{array}{ccc}1& 2& 1\\ \sqrt{5}& \sqrt{5}& \sqrt{2}\\ \sqrt{3}& \sqrt{3}& \sqrt{5}\end{array}\right|$
Applying $C_2\to C_2-C_1$
$\therefore \Delta =5\sqrt{3}\left|\begin{array}{ccc}1& 1& 1\\ \sqrt{5}& 0& \sqrt{2}\\ \sqrt{3}& 0& \sqrt{5}\end{array}\right|$
Expanding along $C_2$
$\therefore \Delta =5\sqrt{3}\cdot \left(01\right)\left|\begin{array}{cc}\sqrt{5}& \sqrt{2}\\ \sqrt{3}& \sqrt{5}\end{array}\right|$
$=-5\sqrt{3}(5-\sqrt{6})$
$=-25\sqrt{3}+15\sqrt{2}$
$=15\sqrt{2}-25\sqrt{3}$
$\Rightarrow b-a=10$