# "Invertible Matrix" ⇔ "Non-zero determinant" - SEMATH INFO -

A matrix is invertible if and only if its determinant is non-zero, i.e.,
.

### Proof

First we will prove
.
If $A$ is an invertible matirx. there exists an inverse matrix $A^{-1}$ satisfying
. The determinant of the left-hand side of this equation can be represented as the product of the determinants, i.e.,
, because, in general, the determinant of the product is equal to the product of the determinants. On the other hand, the determinant of the right-hand side of the equation (the identity matrix) is $1$. We thus see that
. Therefore $|A|$ must not be zero, that is,
Next we will prove
.
It is known that the product of a square matrix and its adjugate matrix is equal to the product of the identity matrix and the determinant, i.e.,
where $\tilde{A}$ is the adjugate matrix of $A$.
If $|A|\neq 0$, we see that
. This equation means that the matrix $\frac{1}{| A |} \tilde{A}$ is the inverse of $A$. Therefore, the inverse matrix of $A$ exists, that is, $A$ is invertible.
The following statement is obtained from the above.
.