<s>[INST] "Dividing" matrix by a vector If I have $\mathbf{Ax}=\mathbf{c}$ where $\mathbf{x} = 
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]$ and $\mathbf{c} = \left[\begin{array}{r}
2x_1-4x_2-x_3-3x_4+2x_5\\
-x_1+2x_2+x_3+x_5\\
x_1-2x_2-x_3-3x_4-x_5\\
-x_1+4x_2-x_3+5x_5
\end{array}\right]$.
To solve the equation for $\mathbf{A}$, I would therefore like to isolate $\mathbf{A}$ - is there a way to do this? Like "dividing" by $\mathbf{a}$. I know that $\mathbf{A}$ is obviously the coefficient matrix but I would like some justification to actually show this like by mathematically solving the equation for $\mathbf{A}$. 
 [/INST] Notice that $$Ax = A \cdot x =\left[\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
a_{41} & \cdots & \cdots & \cdots & a_{45}
\end{array}\right]\cdot\left[\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}
\end{array}\right]=\left[\begin{array}{c}
a_{11} x_{1}+a_{12}x_{2}+a_{13}x_{3} +a_{14}x_{4} +a_{15}x_{5}\\
a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+a_{24}x_{4}+a_{25}x_{5}\\
a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+a_{34}x_{4}+a_{35}x_{5}\\
a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}+a_{45}x_{5}
\end{array}\right]$$ 
$$\left[\begin{array}{c}
a_{11} x_{1}+a_{12}x_{2}+a_{13}x_{3} +a_{14}x_{4} +a_{15}x_{5}\\
a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+a_{24}x_{4}+a_{25}x_{5}\\
a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+a_{34}x_{4}+a_{35}x_{5}\\
a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}+a_{45}x_{5}
\end{array}\right] =  \left[\begin{array}{r}
2x_1-4x_2-x_3-3x_4+2x_5\\
-x_1+2x_2+x_3+x_5\\
x_1-2x_2-x_3-3x_4-x_5\\
-x_1+4x_2-x_3+5x_5
\end{array}\right]$ $$ 
And you can continue from here.  
</s>