If $A$ is invertible show that $\det(A) \neq 0$

If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$,
thus $\det(A) =0$.

I proved that way, but looks like too simple so I think maybe there is a trick that I missed.
Or I proved it right?

Modification:
($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$
($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)

Now I think I complete the proof.

Solution 1:

Perhaps the simplest way to prove this, and in two lines, is the following... but it depends on what definitions you have (all the time square matrices of the same order):

A matrix $\,A\,$ is invertible iff there exists a matrix $\,B\,$ s.t. $\;AB= BA=I\;$ iff when reduced (by rows or columns) no row (column) is all zeros iff $\,\det A\neq 0\,$ .

The last double implication follows from the fact that the determinant of an upper (lower) triangular matrix is the product of the entries in its main diagonal (and, of course, on the fact that interchanging rows only multiplies the matrix by $\,\pm 1\,$)

If your definitions cover the above there you have an easy proof of what you want.

Solution 2:

Your proof of the $\Rightarrow$ implication is right. For the other one, do you know about the Adjugate Matrix? To each $A\in M_{n\times n}(F)$ we can associate a matrix $adj(A)$ defined as the tranpose of the cofactor matrix of $A$ such that one always has $A\cdot adj(A)=\det (A) I_{n\times n}$. If $\det(A)\neq 0$ then $\det(A)$ is invertible and therefore you get $A^{-1}=(\det(A))^{-1}adj(A)$.

Notice that the result can be actually strengthened: given a unitary commutative ring $R$, a matrix $A\in M_{n\times n}(R)$ is invertibile if and only if its determinant is invertible in $R$. The proof of this fact goes exactly as the former one just by noticing that what we really used previously was the fact that any non zero element in a field is invertible, as I underlined.