If the matrix $\mathrm{A}=\left(\begin{array}{cc}0& 2\\ \mathrm{k}& -1\end{array}\right)$ satisfies $\mathrm{A}({\mathrm{A}}^3+3\mathrm{I})=2\mathrm{I},$ then the value of k is :

$\begin{array}{rcl}\mathrm{A}& =& \left[\begin{array}{cc}0& 2\\ \mathrm{k}& -1\end{array}\right], {\mathrm{A}}^2=\left[\begin{array}{cc}0& 2\\ \mathrm{k}& -1\end{array}\right]\left[\begin{array}{cc}0& 2\\ \mathrm{k}& -1\end{array}\right]\\ & =& \left[\begin{array}{cc}2\mathrm{k}& -2\\ -\mathrm{k}& 2\mathrm{k}+1\end{array}\right]\\ {\mathrm{A}}^4& =& {\mathrm{A}}^2{\mathrm{A}}^2\\ & =& \left[\begin{array}{cc}2\mathrm{k}& -2\\ -\mathrm{k}& 2\mathrm{k}+1\end{array}\right]\left[\begin{array}{cc}2\mathrm{k}& -2\\ -\mathrm{k}& 2\mathrm{k}+1\end{array}\right]\\ & =& \left[\begin{array}{cc}4{\mathrm{k}}^2+2\mathrm{k}& -4\mathrm{k}-4\mathrm{k}-2\\ -2{\mathrm{k}}^2+2{\mathrm{k}}^2+\mathrm{k}& 2\mathrm{k}+{(2\mathrm{k}+1)}^2\end{array}\right]\\ \therefore {\mathrm{A}}^4+3\mathrm{A}& =& 2\mathrm{I}\\ \therefore 4{\mathrm{k}}^2+2\mathrm{k}& =& 2\\ 2{\mathrm{k}}^2+\mathrm{k}-1& =& 0\\ \mathrm{k}& =& -1, \frac{1}{2}\end{array}$