Question

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

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

Answer 1

The correct option is A $z$ is purely real

$z=\left|\begin{array}{ccc}1& 1+2i& -5i\\ 1-2i& -3& 5+3i\\ 5i& 5-3i& 7\end{array}\right|=1(-21-36)-(1+2i)(7-14i-25i+15)-5i(-13i-1)=-57-(1+2i)(22-39i)-65+5i=-57-(22-39i+44i+78)-65+5i=-57-100-5i-65+5i=-222$

$\therefore z$ is purely real.