<s>[INST] Existence of $A$ and $B$ such that $A^2-4A+4I=0$, $A+B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}$ and $AB=$... 
Question: Prove or disprove that $A$ and $B$ exist such that
  \begin{align}
&A^2-4A+4I=0\\
&A+B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}\\
&AB=\begin{pmatrix} 1 & 1 \\ -9 & 3\end{pmatrix}\\
\end{align}


(*) I've found what I've done wrongly below. (see the red and blue).
If I follow abnry's method instead then I get the right answer because $C-4I$ has its inverse matrix in this case.

What I did is as follows:
\begin{align}
&\text{Assume that such }A\text{ and }B\text{ exist.}\\
&\text{Then }A\text{ satisfies both of below equations:}\\
&X^2-4IX+4I=0\tag1\\
&\color{red}{(X-A)(X-B)=X^2-(A+B)X+AB=0\leftarrow Wrong}\\
&\color{blue}{(X-A)(X-B)=X^2-(AX+XB)+AB=0}\\
&\\
&(1)-(2):\\
&(A+B-4I)X=AB-4I\\
&\begin{pmatrix} 0 & 1 \\ -3 & 0\end{pmatrix}X=\begin{pmatrix} -3 & 1 \\ -9 & -1\end{pmatrix}\\
&X=\begin{pmatrix} 0 & 1 \\ -3 & 0\end{pmatrix}^{-1}\begin{pmatrix} -3 & 1 \\ -9 & -1\end{pmatrix}=\begin{pmatrix} 0 & -\frac13 \\ 1 & 0\end{pmatrix}\begin{pmatrix} -3 & 1 \\ -9 & -1\end{pmatrix}=\begin{pmatrix} 3 & \frac13 \\ -3 & 1\end{pmatrix}\\
&\\
&\text{As }X\text{ is uniquely found, so }A=X\\
&B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}-A=\begin{pmatrix} 1 & \frac23 \\ 0 & 3\end{pmatrix}\\
&\text{Now calculating }AB,\\
&\\
&AB= \begin{pmatrix} 3 & \frac13 \\ -3 & 1\end{pmatrix}\begin{pmatrix} 1 & \frac23 \\ 0 & 3\end{pmatrix}=\begin{pmatrix} 3 & 3 \\ -3 & 1\end{pmatrix}\ne\begin{pmatrix} 1 & 1 \\ -9 & 3\end{pmatrix}\\
&\\
&\text{Therefore, such }A\text{ and }B\text{ do not exist.}\\
\end{align}
But as I don't have much knowledge on linear algebra (or at least I forgot all of them), I don't know a good explanation on why this happened, and on what condition $A$ and $B$ exist or not. I think there must be much better way to know it only by looking at those two equations without actually trying to find $A$ and $B$. Could someone help here?
 [/INST] The second two equations are of the form $A+B=C$ and $AB=D$.
Then $A^2+AB=AC$ and so $A^2=AC-AB=AC-D$.
Substitute for $A^2$ in the original equation to get
$$AC-D-4A+4I=0$$
or
$$A(C-4I)=D-4I.$$
So if the inverse of $C-4I$ exists, this gives you a unique $A$. From this you can compute $B=C-A$ (and verify that indeed $AB=D$).
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