Question

If $\Delta =\left|\begin{array}{ccc}\sqrt{6}& 2i& 3+\sqrt{6}\\ \sqrt{12}& \sqrt{3}+\sqrt{8i}& 3\sqrt{2}+\sqrt{6i}\\ \sqrt{18}& \sqrt{2}+\sqrt{12}i& \sqrt{27}+2i\end{array}\right|$
then $\Delta$ is

• A. an integer
• B. a rational number
• C. an irrational number
• D. an imaginary number

Answer 1

The correct option is A an integer

Taking $\sqrt{6}$ common from $C_1$ , we get
$\Delta =\sqrt{6}\left|\begin{array}{ccc}1& 2i& 3+\sqrt{6}\\ \sqrt{2}& \sqrt{3}+2\sqrt{2i}& 3\sqrt{2}+\sqrt{6i}\\ \sqrt{3}\sqrt{2}& \sqrt{2}+2\sqrt{3}i& 3\sqrt{3}+2i\end{array}\right|$
Applying $R_2\to R_2-\sqrt{2}{R}_1$ and $R_3\to R_3-\sqrt{3}{R}_1$
$=\sqrt{6}\left|\begin{array}{ccc}1& 2i& 3+\sqrt{6}\\ 0& \sqrt{3}& \sqrt{6i}-2\sqrt{3}\\ 0& \sqrt{2}& 2i-3\sqrt{2}\end{array}\right|$
$=\sqrt{6}\left|\begin{array}{cc}\sqrt{3}& \sqrt{6}i-2\sqrt{3}\\ \sqrt{2}& 2i-3\sqrt{2}\end{array}\right|=\sqrt{6}\left|\begin{array}{cc}\sqrt{3}& -2\sqrt{3}\\ \sqrt{2}& -3\sqrt{2}\end{array}\right|$
Using $C_2\to C_2-\sqrt{2}i{C}_1$
$=\sqrt{6}(-3\sqrt{6}+2\sqrt{6})=-6$ which is an integer