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 is real and imaginary both
• D. $(z+\overline{z})=0$

Answer 1

The correct option is A z is purely real

From given, we have,

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

$=1\times \left|\begin{array}{cc}-3& 5+3i\\ 5-3i& 7\end{array}\right|-(1+2i)\times \left|\begin{array}{cc}1-2i& 5+3i\\ 5i& 7\end{array}\right|-5i\times \left|\begin{array}{cc}1-2i& -3\\ 5i& 5-3i\end{array}\right|$

$=1\cdot (-55)-(1+2i)(22-39i)-5i(-1+2i)$

$=-145$

Hence $z$ is purely real