If $A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$  and ${\left(a{1}_2+bA\right)}^2=A$ , then

$a{I}_2+bA=a\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+b\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

$=\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right]+\left[\begin{array}{cc}0& b\\ -b& 0\end{array}\right]$

$=\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]$

${\left(a{I}_2+bA\right)}^2=A$

$\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

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

$a^2-b^2=0,2ab=1$

$\left(a+b\right)\left(a-b\right)=0$

$a=b$   or  $a=-b$

If $a=b\Rightarrow 2a^2=1\Rightarrow a^2=\frac{1}{2}\Rightarrow a=\frac{1}{\sqrt{2}}$

If $a=-b\Rightarrow -2b^2=1\Rightarrow b^2=-\frac{1}{2}$  not possible

$\therefore a=\frac{1}{\sqrt{2}}$  and $b=\frac{1}{\sqrt{2}}$