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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.

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  • $\begingroup$ Copy-paste error, fixed. $\endgroup$ – user3564783 Apr 10 '15 at 23:47
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    $\begingroup$ 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 $A-I$. $\endgroup$ – TravisJ Apr 10 '15 at 23:49
  • $\begingroup$ what about solving $(I-A)X = B?$ $\endgroup$ – abel Apr 10 '15 at 23:49
  • $\begingroup$ Can you explain how you moved from AX−X=−B to (A−I)X=−B? $\endgroup$ – user3564783 Apr 10 '15 at 23:55
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    $\begingroup$ @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. $\endgroup$ – JMoravitz Apr 10 '15 at 23:56
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$AX+B-X=0$

$(A-I)X=-B$

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

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And, here's an example to model this process:

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  • $\begingroup$ why is the font so huge? $\endgroup$ – abel Apr 11 '15 at 1:39

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