# 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?

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 $AB=D$).

• I appreciate, thanks. – Kay K. Sep 12 '16 at 1:54
• Wait a second. I think it is same to what I did, but I got $AB\ne D$... – Kay K. Sep 12 '16 at 1:59
• I found one error in the above. It should have been $A(C-4I)$, and then I think $AB=D$ holds. Thanks. – Kay K. Sep 12 '16 at 3:49
• Yes, thank you. – abnry Sep 12 '16 at 3:55