The number of solutions of the matrix equation ${\mathrm{X}}^2=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right]$ is

Let $\mathrm{X}=\left(\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right)\Rightarrow {\mathrm{X}}^2=\left(\begin{array}{l}{\mathrm{a}}^2+\mathrm{bc}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{ab}+\mathrm{bd}\\ \mathrm{ac}+\mathrm{cd}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{bc}+{\mathrm{d}}^2\end{array}\right)$
$\begin{array}{l}\Rightarrow {\mathrm{a}}^2+\mathrm{bc}=1\text{ and }\mathrm{ab}+\mathrm{bd}=1\Rightarrow \mathrm{b}(\mathrm{a}+\mathrm{d})=1\\ \Rightarrow \left({\mathrm{b}}^2+\frac{1}{4{\mathrm{b}}^2}+1\right)+2{\mathrm{b}}^2=3\Rightarrow 3{\mathrm{b}}^2+\frac{1}{4{\mathrm{b}}^2}=2\\ \Rightarrow 3\mathrm{x}+\frac{1}{4\mathrm{x}}=2\text{ or }\mathrm{b}=\pm \frac{1}{\sqrt{6}}\text{ or }\mathrm{b}=\pm \frac{1}{\sqrt{2}}\end{array}$
Therefore, matrices are
$\left(\begin{array}{cc}0& 1/\sqrt{2}\\ \sqrt{2}& \sqrt{2}\end{array}\right),\left(\begin{array}{cc}0& -1/\sqrt{2}\\ -\sqrt{2}& -\sqrt{2}\end{array}\right),\left(\begin{array}{cc}2/\sqrt{6}& -1/\sqrt{6}\\ 2/\sqrt{6}& 4\sqrt{6}\end{array}\right)$