<s>[INST] Finding real matrices of order $2\times 2$ in matrix equation 
Finding all real matrices $X$ of order $2\times 2$ which satisfy the equation $X^2=\begin{pmatrix}
1 & 2 \\ 
 3 & 7
\end{pmatrix}$

My Try: Let $\displaystyle X=\begin{pmatrix} a & b\\
c & d \end{pmatrix}$. Then $\displaystyle X^2=\begin{pmatrix}a^2+bc & b(a+d)\\
c(a+d) & bc+d^2\end{pmatrix}=\begin{pmatrix}1 & 2\\ 3 & 7\end{pmatrix}$
So $a^2+bc=1\cdots (1)$ and $b(a+d)=2\cdots (2)$
And $c(a+d)=3\cdots (3)$ and $bc+d^2=7\cdots (4)$
But this is very complex method 
Could some help me to solve it some easy way .Thanks
 [/INST] The characteristic equation of $X$ is, by definition, $\det (X - t I_2) = 0$. The unknown is $t$, and $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Since $X$ is square of dimension $2$, this equation will be of degree $2$; more specifically, it will be
$$t^2 - (\operatorname{tr} X) t + \det X = 0 \ .$$
By the Cayley-Hamilton theorem, $X$ too will verify the above equation, therefore
$$X^2 - (\operatorname{tr} X) X + (\det X) I_2 = 0 \ .$$
In this equation (in $X$, this time), we already know $X^2$. On the other hand, we have
$$(\det X)^2 = \det (X^2) = \det \begin{pmatrix} 1 & 2 \\  3 & 7 \end{pmatrix} = 1 \ ,$$
whence we deduce that $\det X = \pm 1$.
This means that our equation in $X$ becomes
$$\begin{pmatrix} 1 & 2 \\  3 & 7 \end{pmatrix} \pm \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = (\operatorname{tr} X) X \ .$$
This has two possible cases:


*

*$\det X = -1$, therefore $(\operatorname{tr} X) X = \begin{pmatrix} 0 & 2 \\  3 & 6 \end{pmatrix}$. Taking the trace, we get $(\operatorname{tr} X) ^2 = 6$, whence $\operatorname{tr} X = \pm \sqrt 6$, so $X = \pm \frac 1 {\sqrt 6} \begin{pmatrix} 0 & 2 \\  3 & 6 \end{pmatrix}$;

*$\det X = 1$, therefore $(\operatorname{tr} X) X = \begin{pmatrix} 2 & 2 \\  3 & 8 \end{pmatrix}$. Taking the trace, we get $(\operatorname{tr} X) ^2 = 10$, whence $\operatorname{tr} X = \pm \sqrt {10}$, so $X = \pm \frac 1 {\sqrt {10}} \begin{pmatrix} 2 & 2 \\  3 & 8 \end{pmatrix}$.
It is now easy to check that all the above 4 matrices are indeed solutions of the given equation. They are also the only ones, by the above analysis.
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