Question

Find the values of the following determinant.
$\left|\begin{array}{cc}1+3i& i-2\\ -i-2& 1-3i\end{array}\right|$, where $i=\sqrt{-1}$.

Answer 1

Given, $\left|\begin{array}{cc}1+3i& i-2\\ -i-2& 1-3i\end{array}\right|$, where $i=\sqrt{-1}$.

Expanding the determinant we get,

$=(1+3i)(1-3i)-(i-2)(-i-2)$

$=1-9i^2+(i-2)(i+2)$

$=1-9i^2+i^2-4$

$=-3-8i^2$

$=-3+8=5$