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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

gbox
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    You should replace "inverse" with "invertible". Then it is all right. – MooS Feb 06 '15 at 14:27
  • Yes, the product $AB$, where either $A$ or $B$ is not invertible, is not invertible. – egreg Feb 06 '15 at 14:28
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    For the sake of a proof, you might want to check that $rank(AB) \leq rank(A)$ and $rank(AB) \leq rank(B)$. – MooS Feb 06 '15 at 14:30
  • You seem to be thinking of square matrices, all of a common size. If we were to consider arbitrary matrices, even limited to ones with a defined product, the outcomes would not be so binary. – hardmath Dec 28 '21 at 18:32

2 Answers2

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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

Belgi
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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.

Madhan Kumar
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