I'm having trouble understanding the proof of Theorem 1 on this page:

Pauls Online Notes : Linear Algebra - Finding Inverse Matrices
I don't understand how they proved the first conditional $\displaystyle (a)\Rightarrow (b)$

If A in invertible, then this means that $\displaystyle A^{-1}$ exists and that $\displaystyle AA^{-1}=I=A^{-1}A$.

It seems that they assumed that $\displaystyle AX=0$, selected a solution $\displaystyle x_0$ to write $\displaystyle Ax_0=0$, and then mulitplied both sides by $\displaystyle A^{-1}$ so that

$\displaystyle A^{-1}Ax_0=A^{-1}0$

$\displaystyle Ix_0=I0$

$\displaystyle x_0=0$

I don't understand the logic here. It just looks like circular reasoning to me. Here is a quote from the proof:

So it seems that the equation that they are multiplying by $\displaystyle A^{-1}$ is:

$\displaystyle Ax_0=0$

But where does that come from? How is it that substituting$\displaystyle x_0$ into $\displaystyle AX$ gives $\displaystyle Ax_0=0$. If I look at it this way:

$\displaystyle \left(\begin{array}{ccc}a_{11}&a_{12}&...a_{1n}\\a _{21}&a_{22}&...a_{2n}\\.&.&.\\a_{n1}&a_{n2}&...a_ {nn}\end{array}\right)$$\displaystyle \left(\begin{array}{c}x_1\\x_2\\.\\x_n\end{array}\ right)$=$\displaystyle \left(\begin{array}{c}b_1\\b_2\\.\\b_n\end{array}\ right)$

To get the right side to be equal to zero, they must have taken $\displaystyle x_0=\left(\begin{array}{c}0\\0\\.\\0\end{array}\ri ght)$, but this isn't ANY solution. Can anyone help clarify this?