${\left(2^3\right)}^2=2^3\cdot 2^3=2^{3+3}=2^6$

$\left|\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right|=\mathrm{a}\mathrm{d}-\mathrm{b}\mathrm{c}$

${\displaystyle \frac{1}{2}}\cdot \left|\begin{array}{cc}1& 3\\ -6& 0\end{array}\right|={\displaystyle \frac{1}{2}}[1\cdot 0-3\cdot (-6)]={\displaystyle \frac{1}{2}}\cdot [0+18]=9$