Prove: If $A$ is singular, so is $AB$.
Contrapositive: If $AB$ is invertible then $A$ is invertible.
If $AB$ is invertible then there exist a matrix $C$ such that $(AB)C=I$.
by associativity, $A(BC)=I$.
right multiplying by a vector $b∈R^n$ we get $A(BCb)=b$.
now we can write $BCb$ as a vector $x∈R^n$ and so
for all $b∈R^n$ the equation $Ax=b$ has a solution.
the above is true IFF
the columns of A span $R^n$
IFF the columns of A are linearly independent
IFF the homogenous system of equations $Ax=0$ has only the trivial solution
IFF matrix $A$ is row equivalent to $I$
because $A$ is row equivalent to $I$, $A$ is invertible.
so by contraposition if $A$ is singular, so is $AB$.
Best Answer
I think you've overthought the problem. If $AB$ is invertible, then $\exists C (= (AB)^{-1})$ such that $(AB)C=I$. But then $A(BC)=I$ so $A^{-1}=BC$ and $A$ is invertible.
Taking the contrapositive, if $A$ is singular (i.e., not invertible), then $AB$ can't be invertible (i.e., must be singular).