# Solving a matrix equation.

Questions in this general form have been asked a lot here, but I've searched for hours and I haven't found any that I can generalize to my problem, so I've asked it again:

I've been given three matrices, $A = \begin{bmatrix} 5 & 3 \\ 3 & 2 \\ \end{bmatrix}$, $B = \begin{bmatrix} 6 & 2 \\ 2 & 4 \\ \end{bmatrix}$, and have to solve the equation AX+B = X. I've no clue where to start at all, so any help would be appreciated.

• Copy-paste error, fixed. – user3564783 Apr 10 '15 at 23:47
• What about re-arranging, $AX-X=-B$ then $(A-I)X=-B$ so $X=-(A-I)^{-1}B$. This of course assumes that you can find the inverse of $A-I$. – TravisJ Apr 10 '15 at 23:49
• what about solving $(I-A)X = B?$ – abel Apr 10 '15 at 23:49
• Can you explain how you moved from AX−X=−B to (A−I)X=−B? – user3564783 Apr 10 '15 at 23:55
• @user $M = IM=MI$ for every matrix $M$ and appropriately shaped $I$., So you have $AX - X = AX - IX = (A-I)X$ since distribution holds in matrix arithmetic as well. – JMoravitz Apr 10 '15 at 23:56

$AX+B-X=0$
$(A-I)X=-B$
$X=-(A-I)^{-1}B$