How can I prove that if $A$ is a singular square matrix, then $\det A = 0$ without using Binet's Theorem? Well, I need to prove that if $A$ is a $n \times n$ matrix and it is singular, then $\det A = 0$, in order to show Binet's Theorem $\det(AB) = \det(A)\cdot\det(B)$ in the case when both $A$ and $B$ are singular square matrices.
I'll thank any help with this issue!

I would use two proof

*

*Permutation change the sign of the determinant

*Multiplying a row by a constant multiply the determinant by this constant

Then, you can just say if you have a singular matrix S, 2 row $r_i$ and $r_j$ are linearly dependent. You can multiply $r_i$ by a constant C to get a new matrix S' where $r_i$ = $r_j$.
We know
\begin{equation} 
det(S') = C det(S)
\end{equation}
Then, you can swap $r_i$ and $r_j$ but you have the same matrix has before so
\begin{equation}
det(S') = -det(S') = -C det(S) = C det(S)
\end{equation}
The equation det(S) = -det(S) means det(S) is 0.
