Let $A$ a matrix $n\times n$ over $\mathbb{R}$. I'm trying to prove that A is invertible if and only if $\det(A)\neq0$ using the Cauchy-Binet theorem.
I know that the Cauchy -Binet theorem is $$\det(A B)=\det(A)\cdot \det(B)$$
But for now, I couldn't think of any solutions to solve the proof.
On
Suppose that $A$ is invertible. Then there is a matrix $B$ such that $AB=I$. The Cauchy-Binet theorem then implies that $$1=det(I)=det(AB)=det(A)det(B)$$ so that $det(A)\neq 0\neq det(B)$.
Conversely, suppose $det(A)\neq 0$. Then $B=\frac{1}{detA}adj(A)$ satisfies $AB=I$, so that $A$ is invertible. Here $adj(A)$ is the adjugate of the matrix $A$.
Edit: Given the comments on my answer, I'm including a link to a thread about left/right inverses:
If $A$ is invertible, then $AA^{-1}=I$ (identity matrix), so $$ 1=\det I=\det(AA^{-1})=\det A\det(A^{-1}) $$ and therefore $\det A\ne0$.
The converse doesn't follow from Binet's theorem, but rather from the fact that the determinant is multilinear and alternating on the columns of a matrix.
Fact 1. If $A$ has a zero column, then $\det A=\det A+\det A$, so $\det A=0$.
Fact 2. If $A$ has two identical columns, then $\det A=-\det A$, by swapping them, so $\det A=0$.
Fact 3. If $A$ is not invertible, then $\det A=0$.
Since we want to show that $\det A=0$, possibly with a column swap we can assume that the last column is a linear combination of the other $n-1$ columns. Say $A=[a_1\ \dots\ a_{n-1}\ a_n]$, with $$ a_n=c_1a_1+\dots+c_{n-1}a_{n-1} $$ Then, by multilinearity and facts 1 and 2, we have $$ 0=\det[a_1\ \dots\ a_{n-1}\ 0]= \det A -c_1\det[a_1\ \dots\ a_{n-1}\ a_1] -\dots -c_{n-1}\det[a_1\ \dots\ a_{n-1}\ a_{n-1}] $$