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Find matrix such that: $$\begin{pmatrix} 3 & 2 & 3 \\ 3 & 6 & 3 \\ 1 & 2 & 4 \end{pmatrix} X+ \begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & 2 \\ 1 & 0 & 1 \end{pmatrix}=2X $$ inverse of which matrix i have to find? How to approach that?

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    $\begingroup$ $2X=2EX$, where $E$ is identity matrix. $\endgroup$
    – sas
    Jan 8, 2015 at 12:23

2 Answers 2

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Hint:

Notice that $$ AX +B = 2X \implies (A-2I)X = -B \implies X = - (A-2I)^{-1}B.$$

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Hint: you should write it like that:

$\begin{pmatrix} 3 & 2 & 3 \\ 3 & 6 & 3 \\ 1 & 2 & 4 \end{pmatrix} X+ \begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & 2 \\ 1 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}X$

or

$\begin{pmatrix} 1 & 2 & 3 \\ 3 & 4 & 3 \\ 1 & 2 & 2 \end{pmatrix} X=\begin{pmatrix} -1 & -1 & 0 \\ -1 & 1 & -2 \\ -1 & 0 & -1 \end{pmatrix}$

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