

Suppose $A$ and $B$ are nonsingular. By theorem 6.2, a
matrix is nonsingular if and only if there exists an
inverse. In our case,
there exist $A^{-1}$ and $B^{-1}$ such that,
\[ A A^{-1} = A^{-1}A=B^{-1}B = B^{-1}B = I.\]
Then,
\[ AB(B^{-1}A^{-1}) = A(B B^{-1})A^{-1} = AIA^{-1} = AA^{-1}=I.\]
So $AB$ has an inverse, namely $B^{-1}A^{-1}$ and the
nonsingular.

Now suppose that $AB$ is nonsingular.
By theorem 6.1, a matrix has two possibilities only: it is
nonsingular and each equation $Ax=b$ has a unique solution for
$x$, or it is singular and the equation $Ax=b$ has either no
or an infinity of solutions, depending on the $b$.
It is enough to show that $Ax=0$ and $Bx=0$ have unique solutions.
Because $Ax=0$ has a solution, $x=0$, if it is singular it must
have an infinity of solutions $Ax=0$. If it has only one solution,
than it must be nonsingular. The same goes for $B$.

Suppose $Bx_1=Bx_2=0$. Then $ABx_1=ABx_2=0$ so $x_1=x_2=0$.
Thus $B$ is nonsingular. If $Ax_1=Ax_2=0$, then since $B$
is non-singular, there are unique solutions to $By_i=x_i$,
with $i=1,2.$ Hence $ABy_i=0$ and thus $y_1=y_2$, and
$x_1=By_1=By_2=x_2$. So $A$ is nonsingular.



