Relationship between null-space and determinant

This is the key insight to this proof: If we have $Ax=b$, then by multiplying both sides by $A^{-1}$ on the left, we get $x=A^{-1}b$. Thus, if $A$ is invertible, then $Ax=b$ has only one solution.

Now, let's consider $Ax=\mathbf{0}$. If $A$ has a non-trivial null space, that, by definition, means that there are multiple possible solutions for this equation. However, if $A$ is invertible, then $Ax=\mathbf{0}$ only has one solution, as shown in the first paragraph. Thus, $A$ can not be invertible and have a non-trivial null space at the same time, so if it has a non-trivial null space, it must be not invertible.