# How do I solve the system of matrix equations? [closed]

Knowing that $A=\begin{pmatrix} -1 &2 \\ 1&1 \end{pmatrix},B=\begin{pmatrix} 1 & 2\\ 0&1 \end{pmatrix},C=\begin{pmatrix} 2 &2 \\ 3 &1 \end{pmatrix}$ solve the system

$$\left\{\begin{matrix} AX+CY=BAC\\ CX-BY=CAB \end{matrix}\right.$$ So guys how do I solve this. I know how to find the inverse of matrix and I guess I should use it can you show me how to start and what to do after?

## closed as off-topic by Nick Peterson, martini, Shailesh, 2012ssohn, TastyRomeoJan 6 '17 at 0:53

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• Hello - any ideas? What have you tried? – Hans Engler Jan 5 '17 at 19:35
• Happy to help, but notice that accepting correct answers is highly appreciated, and probably the courtesy that is expected. Just think about the fact that "guys how do I solve this" is something that resonates with friends and family, but we don't have the pleasure to have built a personal relationship. – Antoni Parellada Jan 5 '17 at 19:36
• I tried this but I think it take too long and maybe there is other way $\left\{\begin{matrix} AX+CY=BAC/*C^{-1}\\ CX-BY=CAB/*B^{-1} \end{matrix}\right.=>\left\{\begin{matrix} C^{-1}AX+Y=C^{-1}BAC\\ B^{-1}CX-Y=B^{-1}CAB \end{matrix}\right.$ and I did first equation+second equation to get rid of Y. It take a little too long tho. – Ghost Jan 5 '17 at 19:38
• Well, I went as far as to use my own cell phone while in transit to make sure I hadn't misled you on my answer this morning, and I even posted a line of code for you to effortlessly verify the result as correct. Five points up, and the answer remains un-accepted. Totally uncool IMHO. – Antoni Parellada Jan 5 '17 at 19:43
• @OvyOvy Again, there is some etiquette to be followed: First of all, I am not "dude", and you should be more respectful. Second, you shouldn't post homework. – Antoni Parellada Jan 5 '17 at 20:23

$\begin{matrix} C^{-1}AX+Y=C^{-1}BAC\\ B^{-1}CX-Y=B^{-1}CAB \end{matrix}$

$(C^{-1}A+B^{-1}C)X=C^{-1}BAC + B^{-1}CAB\\ X=(C^{-1}A+B^{-1}C)^{-1}C^{-1}BAC + B^{-1}CAB$
$Y = C^{-1}BAC - C^{-1}AX$
• Nice solution! Just to add, since $C$ is invertible, you can simplify $(C^{-1}A + B^{-1}C)^{-1} C^{-1} BAC$ further into $(A + C B^{-1} C)^{-1} BAC$. Not much simpler, but one matrix less. ;-) – Florian Jan 5 '17 at 20:02
• Why not? $(C^{-1}A + B^{-1}C)^{-1} = (C^{-1}[A + CB^{-1}C])^{-1} = [A + CB^{-1}C]^{-1} C$, using $(XY)^{-1} = Y^{-1} X^{-1}$ in the second step. – Florian Jan 5 '17 at 20:13