<s>[INST] solving $X^2 - 3X - A = 0$ where $A,X \in \mathbb{M_2(\mathbb{R})}$ Given $A = \begin{pmatrix} 7 & 3 \\ 3 & 7 \end{pmatrix}$ find a $2\times 2$ matrix $X$ s.t. $X^2 - 3X - A = 0$, in the previous parts I have diagonalised $A$ and got $P^{-1}AP = \begin
{pmatrix} 4 & 0 \\ 0 & 10 \end{pmatrix}$, but I'm not sure how this relates to finding $X$.
An idea was that because $A$ has $7,3$ repeating then $X = \begin{pmatrix} a & b \\ b & a \end{pmatrix}$ but this was only an idea and I'm not sure why $X$ would be in that form.
 [/INST] $$X^2-3X-A=0$$
$$P(X^2-3X-A)P^{-1}=0$$
Which using the distruibitive property and the fact that $PX^2P^{-1}=(PXP^{-1})^2$, yields:
$$(PXP^{-1})^2-3(PXP^{-1})-PAP^{-1}=0$$
Now complete the square:
$$(PXP^{-1})^2-3(PXP^{-1})+\frac{9}{4}I_2=\frac{9}{4}I_2+PAP^{-1}$$
$$(PXP^{-1}-\frac{3}{2}I_2)^2=\frac{9}{4}I_2+PAP^{-1}= \begin{pmatrix} \frac{9}{4} & 0 \\ 0 & \frac{9}{4} \end{pmatrix}+\begin{pmatrix} 4 & 0 \\ 0 & 10 \end{pmatrix}$$
$$(PXP^{-1}-\frac{3}{2}I_2)^2= \begin{pmatrix} 6.25 & 0 \\ 0 & 12.25 \end{pmatrix}=\begin{pmatrix} 2.5 & 0 \\ 0 & 3.5 \end{pmatrix}^2$$
Thus, one of the solutions is obtained by solving 
$$PXP^{-1}-\frac{3}{2}I_2= \begin{pmatrix} 2.5 & 0 \\ 0 & 3.5 \end{pmatrix}$$
Which is easy to solve.
To determine all solutions you just need to find all square roots of the diagonal matrix:$$\begin{pmatrix} 6.25 & 0 \\ 0 & 12.25 \end{pmatrix}$$
I put one of four square roots I know, I am not sure if there are others. I think you can easily determine this
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