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\begin{equation} A= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \\ \end{bmatrix} \\ \end{equation}

Solve the matrix equation: \begin{equation} XA=A+2X \end{equation} Where $X$ is a matrix. If I have understood the problem correctly $X$ is $3x3$.

I was thinking maybe I could use the Identity matrix and get something like: \begin{equation} X=(A+2X)A^{-1} \end{equation} I would like some tips on how to solve this, thanks.

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  • $\begingroup$ $2X = X\times 2\times I$, which you can use to factor out the $X$ $\endgroup$ – DanielV Feb 17 '15 at 19:20
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Advance as

$$ XA=A+2X \implies X(A-2I) =A \implies X = A(A-2I)^{-1}. $$

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