True or flase? If the product of two matrices is invertible then each matrices in the product is invertible.

It was said to be true but I didn't understood the argument: a matrix A \begin{pmatrix} 1 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 1 \end{pmatrix}

is non-invertible, its determinant equals $0$, furthermore it is equal to the product of diagonal coefficients of $A$, $A$ not being triangular.

Okay but $A*A$ isn't invertible neither, it gives:

\begin{pmatrix} 3 & 2 & 3\\ 2 & 2 & 2\\ 3 & 2 & 3 \end{pmatrix}

So I don't undertand the point, it should have demonstrate that $A*A$ was invertible and that $A$ wasn't.

  • $\begingroup$ True${}{}{}{}{}{}{}$ , and both matrices in your question are non-invertible, so what is odd here? $\endgroup$
    – DonAntonio
    Mar 30, 2016 at 22:11
  • $\begingroup$ $\det(AB) = \det(A)\det(B)$. If either matrix is singular, then it has a zero determinant. Hence, the determinant of the product is 0. $\endgroup$ Mar 30, 2016 at 22:14
  • 1
    $\begingroup$ You do need that both matrices in the product are square-- you can have a 2x3 $A$ and a 3x2 $B$, so neither is invertible, with $AB$ an invertible 2x2 (e.g. start with 2x2 identity matrices and add a column (resp. row) of $0$'s to get $A$ (reps.$B$) ). $\endgroup$
    – Ned
    Mar 30, 2016 at 22:36

2 Answers 2


We can say that also without using determinants. Let $f_A$ and $f_B$ be the linear applications associated with our matrices and let $g=f_{AB}=f_A\circ f_B$. If $AB$ is invertible, $g$ is a isomorphism, hence it is both injective and surjective, hence $f_A$ is surjective and $f_B$ is injective.

But that is the same as stating that both $A$ and $B$ are invertible matrices.


There is an additional property of matrices which is how I think of this. $$\det(AB)=\det(A)\det(B)$$ This means that if the product $AB$ has det $0$, then necessarily at least one of $A$ and $B$ does.


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