Solving multiple Ax=b to find inverse of A

Question

Let's say I have $3$ systems $$Ax_1 = \begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}, Ax_2 = \begin{bmatrix}0 \\ 1\\ 0 \end{bmatrix}, Ax_3 = \begin{bmatrix}0 \\ 0\\ 1 \end{bmatrix}$$

If I solve for $x_1, x_2, x_3$, it gives me $A^{-1}$ but I'm not sure why this happens. Could someone explain why this is the case?

Answer 1

Let's consider the matrix:

$$[Ax_1 \ Ax_2 \ Ax_3]=I\implies A [x_1 \ x_2 \ x_3]=I\implies [x_1\ x_2 \ x_3]=A^{-1}$$

Answer 2

Hint: Look at the following matrix

$$B=\begin{pmatrix}\vert&\vert&\vert\\x_{1}&x_{2}&x_{3}\\\vert&\vert&\vert\end{pmatrix}$$

and calculate $AB=\:?$.

Answer 3

Let $B=A^{-1}$. Then\begin{align}Ax_1=\begin{bmatrix}1\\0\\0\end{bmatrix}&\iff BAx_1=B\begin{bmatrix}1\\0\\0\end{bmatrix}\\&\iff x_1=B\begin{bmatrix}1\\0\\0\end{bmatrix}=\text{first column of }B.\end{align}The same argument applies to the other equations.