Let $A$ be an $n$ by$n$ matrix. Then $A$ is invertible if and only if $det(A)\ne 0$ .
27. Further properties of determinants
27. Further properties of determinants
Recall that $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$ is invertible if and only if its determinant $adbc$ is nonzero. We'll prove the analogue of this for $n$ by$n$ matrices.
We'll also prove the following:
$det(AB)=det(A)det(B)$ .
Proof of the invertibility criterion
Recall that $A$ is invertible if and only if its reduced echelon form (say ${A}_{red}$ ) is the identity matrix. To get from $A$ to ${A}_{red}$ , we used some row operations. The effect of these on the determinant is:

(type I) determinant unchanged,

(type II) determinant rescaled by a nonzero factor,

(type III) determinant switches sign.
Therefore $det(A)$ differs from $det({A}_{red})$ by a nonzero factor. In particular, $det(A)\ne 0$ if and only if $det({A}_{red})\ne 0$ .
If $A$ is invertible, $det({A}_{red})=det(I)=1$ , so $det(A)\ne 0$ .
If $A$ is not invertible, ${A}_{red}$ has a row of zeros (${A}_{red}$ is a square matrix in reduced echelon form, so if every row is nonzero then the leading entries have to go along the diagonal, and you get ${A}_{red}=I$ ; but we're assuming $A$ is not invertible). Since ${A}_{red}$ has a row of zeros, $det({A}_{red})=0$ , so $det(A)=0$ too.
Proof of multiplicativity of the determinant
We first prove $det(AB)=det(A)det(B)$ in the case where $A$ is an elementary matrix.

If $A={E}_{ij}(\lambda )$ then $AB$ is obtained from $B$ by ${R}_{i}\mapsto {R}_{i}+\lambda {R}_{j}$ , so $det(AB)=det(B)$ (such a row operation doesn't change the determinant). We also have $det(A)=1$ , so we can verify the formula $det(AB)=det(A)det(B)$ in this case just by calculating both sides.

If $A={E}_{i}(\lambda )$ then $AB$ is obtained from $B$ by ${R}_{i}\mapsto \lambda {R}_{i}$ , so $det(AB)=\lambda det(B)$ . We also have $det(A)=\lambda $ , so again we can verify the formula just by computing both sides.
This shows that $det(AB)=det(A)det(B)$ whenever $A$ is an elementary matrix.
By induction, we can now show that $det(AB)=det(A)det(B)$ whenever $A$ is a product of elementary matrices, in other words whenever $A$ is an invertible matrix.
If $A$ is not invertible, then $det(A)=0$ (by the previous theorem) so to verify the formula in this case, we need to prove that $det(AB)=0$ . Suppose for a contradiction that $det(AB)\ne 0$ . Then $AB$ is invertible. But if $AB$ is invertible then $B{(AB)}^{1}$ is an inverse for $A$ : $$A(B{(AB)}^{1})=AB{(AB)}^{1}=I.$$ This gives a contradiction, so we deduce that $det(AB)=0$ .
We have now checked the formula for all possible matrices $A$ , which proves the theorem.