Multiplication of inverse and non-inverse matrices

This is correct. For $AB$ to be invertible $A,B$ must by invertible as well.

Another way to see this is with determinants. The determinant of a matrix $A$ is denoted by $|A|$ and one can prove that $A$ is invertible iff $|A|\neq0$.

We can also prove that $$ |AB|=|A|\cdot|B| $$

which gives us the required result since for the multiplication of two numbers to be non-zero so must both of the numbers be non-zero

Case 1: $B$ is non-invertible.

There exists a non-zero vector $x$ such that $Bx=0$. It implies that $ABx=0$. So, $AB$ is non-invertible.

Case 2: $A$ is non-invertible and $B$ is invertible

There exists a non-zero vector $x$ such that $Ax=0$.

Since $B$ is invertible, take $y=B^{-1}x$ Clearly, $y$ is non-zero vector. Then, it follows,

$ABy=AB(B^{-1}x)=A(Ix)=Ax=0$. So, $AB$ is non-invertible.