Question

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

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

Answer 1

The correct option is A

$z$ is purely real

Explanation for the correct option.

Find the value of $z$.

Evaluate the determinant and find $z$.

$\begin{array}{rcl}z& =& \left|\begin{array}{ccc}1& 1+2i& -5i\\ 1-2i& -3& 5+3i\\ 5i& 5-3i& 7\end{array}\right|\\ & =& 1\left[-21-\left(5-3i\right)\left(5+3i\right)\right]-\left(1+2i\right)\left[7\left(1-2i\right)-5i\left(5+3i\right)\right]+\left(-5i\right)\left[\left(1-2i\right)\left(5-3i\right)-\left(-15i\right)\right]\\ & =& -21-\left(25-9i^2\right)-\left(1+2i\right)\left[7-14i-25i-15i^2\right]-5i\left[5-3i-10i+6i^2+15i\right]\\ & =& -21-\left(25+9\right)-\left(1+2i\right)\left(7-39i-15\left(-1\right)\right)-5i\left[5+2i+6\left(-1\right)\right]\left[i^2=-1\right]\\ & =& -21-34-\left(1+2i\right)\left(22-39i\right)-5i\left(-1+2i\right)\\ & =& -55-22+39i-44i+78i^2+5i-10i^2\\ & =& -77+78\left(-1\right)-10\left(-1\right)\left[i^2=-1\right]\\ & =& -77-78+10\\ & =& -145\end{array}$

So, $z$ has no imaginary part so it is purely real.

Hence, the correct option is A.