Let $\mathrm{A}=\left(\begin{array}{cc}4& -2\\ \mathrm{\alpha }& \mathrm{\beta }\end{array}\right)$ and the trace of $A$ is zero, If $A^2=2I$ ,then the value of $\alpha$  is 

Since the trace of the matrix is zero, so that $\beta =-4$

Now the matrix is $\mathrm{A}=\left(\begin{array}{cc}4& -2\\ \mathrm{\alpha }& -4\end{array}\right)$

Hence, $\begin{array}{rcl}{\mathrm{A}}^2& =& \left(\begin{array}{cc}4& -2\\ \mathrm{\alpha }& -4\end{array}\right)\left(\begin{array}{cc}4& -2\\ \mathrm{\alpha }& -4\end{array}\right)=\left(\begin{array}{cc}16-2\alpha & 0\\ 0& -2\alpha +16\end{array}\right)\\ & & \end{array}$

Since $A^2=2I$, 

$16-2\alpha =2\Rightarrow 2\alpha =14\Rightarrow \alpha =7$