TITLE: Multiplication of inverse and non-inverse matrices
QUESTION [6 upvotes]: I have thought about the combination of multiplication product of invertible and non-invertible matrices:

invertible $\cdot$ invertible = invertible

non-invertible $\cdot$non-invertible = non-invertible

non-invertible $\cdot$invertible = non-invertible

invertible $\cdot$non-invertible = non-invertible
Is it right? I have thought about from the point of view that non-invertible matrix is row equivalent to a matrix with a zero row there for multiple it from both right and left will produce a matrix with a zero row and a zero column respectfully, and opposite goes for invertible matrices

REPLY [3 votes]: 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.