Suppose $A,B$ are square matrices of size $n\times n$. Can $AB$ be invertible, even though both $A$ and $B$ are singular (not invertible)?
And if not, does it follow that if $A_1 \times A_2 \times … \times A_n$ is the invertible product of square matrices, then all factors $A_i$ are invertible?
Best Answer
$$\det(A_1\cdot A_2\cdot\ldots\cdot A_n)\neq0\quad\Leftrightarrow\quad \det A_1\cdot\det A_2\cdot\ldots\cdot\det A_n\neq0\quad\Leftrightarrow\quad\det A_k\neq 0\quad \forall k$$