"Dividing" matrix by a vector

Question

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 $\mathbf{A}$.

Answer 1

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.

Answer 2

Write the coefficient of $x_j$ in $i^{th}$ row of $x$ as elements $a_{ij}$ in c.

$\mathbf{c} = \left[\begin{array}{r} 2x_1-4x_2-x_3-3x_4+2x_5\\ -x_1+2x_2+x_3+x_5\\ x_1-2x_2-x_3-3x_4-x_5\\ -x_1+4x_2-x_3+5x_5 \end{array}\right]$

$\mathbf{A} = \left[\begin{matrix} 2& -4& -1& -3& 2\\ -1& 2& 1& 0& 1\\ 1& -2& -1& -3& -1\\ -1& 4& -1& 0& 5 \end{matrix}\right]$