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Suppose $B$ is a $2 \times 2$ matrix satisfying $AB = B+I$, where $$A = \begin{bmatrix}4&2\\2&3\end{bmatrix}$$ Find B.

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$$AB = B+I \implies AB - B = I \implies AB - IB = I \implies (A-I)B=I \\ \implies (A-I)^{-1} = B$$

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    Just want to make sure, how did you go from AB-BI=I to the next one? – 2017-02-21
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    @JanoyCresva At no point did I write $AB-BI = I$. I wrote $AB-IB = I$. It doesn't make a difference in this case because $I$ commutes with all square matrices, but remember that matrix multiplication isn't commutative in general so usually you can't just switch two factors. But as for how I got the part after that, I factored out the $B$ on the right. – 2017-02-22
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$AB=B+I$

-> $(A-I)B=I$

$A-I=\begin{bmatrix}3&2\\2&2\end{bmatrix}$

$B=\begin{bmatrix}1&-1\\-1&\frac{3}{2}\end{bmatrix}$

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    I'm confused how you got (A-I)B – 2017-02-21
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    $AB=B+I$ implies $AB-B=I$; but, $AB-B=(A-I)B$. – 2017-02-21
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A summary of the other answers. Hopefully this clears things up.

\begin{equation} \begin{split} AB = B+I &\iff AB - B = I \\ &\iff AB - IB = I \\ &\iff (A-I)B=I. \end{split} \end{equation} $\det(A - I) =6-4=2\neq 0$, and therefore $A-I$ has an inverse. Hence $$B = (A-I)^{-1}.$$