"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.
.