Question

The absolute value of the determinant $\left|\begin{array}{ccc}-1& 2& 1\\ 3+2\sqrt{2}& 2+2\sqrt{2}& 1\\ 3-2\sqrt{2}& 2-2\sqrt{2}& 1\end{array}\right|$ is $\frac{k}{\sqrt{2}}$ then $k=$

Answer 1

Let $\Delta =\left|\begin{array}{ccc}-1& 2& 1\\ 3+2\sqrt{2}& 2+2\sqrt{2}& 1\\ 3-2\sqrt{2}& 2-2\sqrt{2}& 1\end{array}\right|$

Applying $C_1\to C_1-(C_2+C_3),$ then

$\Delta =\left|\begin{array}{ccc}-4& 2& 1\\ 0& 2+\sqrt{2}& 1\\ 0& 2-\sqrt{2}& 1\end{array}\right|=(-4)\left(4\sqrt{2}\right)=2[-4(2+2\sqrt{2}-2)]=-16\sqrt{2}$

Hence absolute value $=16\sqrt{2}$

Therefore $k=16$