Question

If $z=\left|\begin{array}{ccc}1& 1+2i& -5i\\ 1-2i& -3& 5+3i\\ 5i& 5-3i& 7\end{array}\right|$, then $i=\left(\sqrt{-1}\right)$

• A. $z$ is purely real
• B. $z$ is purely imaginary
• C. $z+\overline{z}=0$
• D. $(z-\overline{z})i$ is purely imaginery

Answer 1

The correct option is A $z$ is purely real

Given $z=\left|\begin{array}{ccc}1& 1+2i& -5i\\ 1-2i& -3& 5+3i\\ 5i& 5-3i& 7\end{array}\right|$
$=1(-21-(25+9\left)\right)-(1+2i)(7-14i-(25i-15\left)\right)-5i\left[\right(-1-13i)+15i]$
$-55-(1+2i)(22-39i)-5i(-1+2i)$
$=-55-100-5i+5i+10$
$=-145$
Hence, $z$ is purely real.