If $A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$, the values of $\alpha ,\beta$such that $\left(\alpha I + \beta A^2\right)=A^2$ are

$A^2=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=-1$

Also, ${\left(\alpha I+\beta A\right)}^2=\left(\alpha I+\beta A\right)\left(\alpha I+\beta A\right)$

$={\alpha }^2{I}^2+\alpha \beta IA+\beta \alpha AI+{\beta }^2{A}^2$

$={\alpha }^{2}I+2\alpha \beta A+{\beta }^2{\alpha }^2 \left(\because I^2=I,IA=AI\right)$

$={\alpha }^{2}I+2\alpha \beta A+{\beta }^{2}I \left(\because A^2=-I\right)$

$=\left({\alpha }^2-{\beta }^2\right)I+2\alpha \beta A$

$=\left[\begin{array}{cc}{\alpha }^2-{\beta }^2& 0\\ 0& {\alpha }^2-{\beta }^2\end{array}\right]+2\alpha \beta \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

$=\left[\begin{array}{cc}{\alpha }^2-{\beta }^2& 2\alpha \beta \\ -2\alpha \beta & {\alpha }^2-{\beta }^2\end{array}\right]$

On equating corresponding elements in ${\left(\alpha I+\beta A\right)}^2 and A^2$2
we get
${\alpha }^2-{\beta }^2=0 ....................... \left(1\right)$
$2\alpha \beta =1...................... \left(2\right)$

From (1),

$\alpha =\pm \beta . If \alpha =\beta , then from \left(2\right),$

$2{\beta }^2=1$

$\Rightarrow \beta =\pm \frac{1}{\sqrt{2}}where \alpha =\pm \frac{1}{\sqrt{2}}$

Again, if $\alpha =-\beta$, then from (2)
.$\Rightarrow \beta =+\frac{i}{\sqrt{2}}where \alpha =\pm \frac{i}{\sqrt{2}}$
The correct option is (A) and (D)