2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $AB=D$).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.