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I'm trying to solve a matrix equation problem and I can't work out the correct form for the equation for it to be valid.

The matrices given are:

A= $\begin{bmatrix} 1 & -1 & 3\\ 4 & 1 & 5\\ 0 & 0 & 0\\ \end{bmatrix}$, B= $\begin{bmatrix} 1 & -1\\ 3 & 6\\ 1 & 0\\ \end{bmatrix}$, C= $\begin{bmatrix} -1 & 0\\ 5 & 6\\ 0 & 1\\ \end{bmatrix}$

The equation goes as follows: $AX + B = C - X$

I arrange it to: $X= (C - B)*(A+I)^{-1}$ via the following steps: $$AX + B = C - X$$ $$AX +X = C - B$$ $$X(A+I) = C - B /(A+I)^{-1}$$ $$X = (C - B) (A+I)^{-1}$$

But the problem is that the matrices $(C-B)$ and $(A+I)^{-1}$ can't be multiplied because they're not chained (the number of rows and collumns don't allow multiplication). I've been looking at this for over half an hour and can't figure out a different approach. Any help would be highly appreciated.

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  • $\begingroup$ Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B \implies X=(A+I)^{-1}(C-B )$. $\endgroup$ – Yadati Kiran Nov 21 '18 at 17:21
  • $\begingroup$ I don't understand how you got to $(A+I)X=(C−B)$ $\endgroup$ – Arcturus Nov 21 '18 at 17:29
  • $\begingroup$ See we have to be careful with matrix multplication. we have $AX+X$ so number of columns of $A$ must be equal to number of rows of $X$. $\endgroup$ – Yadati Kiran Nov 21 '18 at 17:32
  • $\begingroup$ Can you guide me step to step through how you got from $AX+B=C−X$ to $(A+I)X=(C−B)$? That's the only thing I don't understand. I can find the inverse of $A+I$ just fine. $\endgroup$ – Arcturus Nov 21 '18 at 17:34
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    $\begingroup$ $AX+B=C−X $. Adding addtitive inverses of $B$ and $-X$ on both sides we get $AX+(B-B)+X=C+(−X+X)-B\implies AX+I\cdot X=C-B\implies (A+I)X=C-B$. Assuming $(A+I)$ is invertible we premultiply both sides by $(A+I)^{-1}$ i.e. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)\implies X=(A+I)^{-1}(C-B)$ $\endgroup$ – Yadati Kiran Nov 21 '18 at 17:39
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$X=(A+I)^{-1}(C-B )=\begin{bmatrix}\frac14 &\frac18 &\frac{-11}{8}\\\frac{-1}{2} &\frac14 &\frac14\\0 &0 &1\end{bmatrix}\begin{bmatrix}-2 &1\\2 &0\\-1 &1\end{bmatrix}=\begin{bmatrix}\frac98 &\frac{-9}{8}\\\frac54 &\frac{-1}{4}\\-1 &1\end{bmatrix}$

We get two independent solutions for $X$.

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