If z 1 = 2 + 3 i ,   z 2 = 3 + 2 i where i = - 1 then z 1 z 2 - z 2 ¯ z 1 ¯ z 1 ¯ - z 2…

$z_1=2+3i \therefore \overline{z_1}=2-3i$

$z_2=3+2i \therefore \overline{z_2}=3-2i$

Now, $\left[\begin{array}{cc}{z}_1& {\mathrm{z}}_2\\ -\overline{{\mathrm{z}}_2}& \overline{{\mathrm{z}}_1}\end{array}\right]\left[\begin{array}{cc}\overline{{\mathrm{z}}_1}& -{\mathrm{z}}_2\\ \overline{{\mathrm{z}}_2}& {\mathrm{z}}_1\end{array}\right]=\left[\begin{array}{cc}{z}_1\overline{{\mathrm{z}}_1}+{\mathrm{z}}_2\overline{{\mathrm{z}}_2}& -z_1{\mathrm{z}}_2+{\mathrm{z}}_1{\mathrm{z}}_2\\ -{\overline{{\mathrm{z}}_1\mathrm{z}}}_2+\overline{{\mathrm{z}}_1{\mathrm{z}}_2}& z_1\overline{{\mathrm{z}}_1}+{\mathrm{z}}_2\overline{{\mathrm{z}}_2}\end{array}\right]$

$=\left[\begin{array}{cc}(2+3i)(2-3i) + (3-2i)(3+2i)& 0\\ 0& (2+3i)(2-3i) + (3-2i)(3+2i)\end{array}\right]$

$=\left[\begin{array}{cc}4-9i^2+9-4i^2& 0\\ 0& 4-9i^2+9-4i^2\end{array}\right]$

$=\left[\begin{array}{cc}26& 0\\ 0& 26\end{array}\right]$

$=26\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] = 26I \because I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$