Linear system $Ax=b$ is inconsistent implies $\det(A)=0$

Question

Consider a matrix equation $Ax=b$.

If $\det(A)=0$, it follows that the equation has a non-unique solution or it does not have solutions at all.

I have proven that if the rows/columns of $A$ are linearly dependent, then $\det(A)=0$ and $Ax=0$ does not have a unique solution. Therefore the original equation $Ax=b$ does not have a unique solution.

However what I haven't been able to prove is that if the system of linear equations is inconsistent, then the determinant is zero. If the system is inconsistent, does it always imply that row operations on the matrix A lead to a zero row?

Answer 1

Let's prove the contrapositive. If $\det A\ne0$, $A^{-1}$ exists, so $Ax=b$ has exactly one solution, $x=A^{-1}b$, i.e. is not inconsistent.

Answer 2

If they wouldn't, then Elementary Row Operations would bring it to its Reduced Row Echelon Form which would have no zero rows. This can only happen if the RRE form is in fact $I_n$. This implies that $A$ is invertible so $\mathrm{det}(A) \neq 0$.