# Stuck with matrix equation

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.

• Pre multiply by $(A+I)^{-1}$. Your equation is $(A+I)X=C-B \implies X=(A+I)^{-1}(C-B )$. – Yadati Kiran Nov 21 '18 at 17:21
• I don't understand how you got to $(A+I)X=(C−B)$ – Arcturus Nov 21 '18 at 17:29
• 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 $X$. – Yadati Kiran Nov 21 '18 at 17:32
• 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. – Arcturus Nov 21 '18 at 17:34
• $AX+B=C−X$. 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. $(A+I)^{-1}(A+I)X=(A+I)^{-1}(C-B)\implies X=(A+I)^{-1}(C-B)$ – Yadati Kiran Nov 21 '18 at 17:39

$$X=(A+I)^{-1}(C-B )=\begin{bmatrix}\frac14 &\frac18 &\frac{-11}{8}\\\frac{-1}{2} &\frac14 &\frac14\\0 &0 &1\end{bmatrix}\begin{bmatrix}-2 &1\\2 &0\\-1 &1\end{bmatrix}=\begin{bmatrix}\frac98 &\frac{-9}{8}\\\frac54 &\frac{-1}{4}\\-1 &1\end{bmatrix}$$

We get two independent solutions for $$X$$.