If $A,B$ are invertible so $AB$ is invertible

Question

I thought about the following proofs but I am not sure about them

1. there is $C,D$ so that $AC=CA=I$ and $BD=DB=I \rightarrow CABD=I \rightarrow$ due to associativity roles is no matrix $E$ so that $EAB=ABE=I$

2. let look at $A=\begin{pmatrix} 1 & 0\\ 3 & 3 \end{pmatrix}$, $B=\begin{pmatrix} 0 & 3\\ 2 & -3 \end{pmatrix}$, $AB=\begin{pmatrix} 0 & 3\\ 6 & 0 \end{pmatrix}$ so $A,B$ is invertible but $AB$ is not, but using Matlab $AB$ does have an inverse.

Answer 1

The product of two invertible matrices is always invertible. If $A^{-1}$ and $B^{-1}$ both exist, then $(AB)^{-1} = B^{-1}A^{-1}$.

This can be verified by pre and post multiplication.

Premultiplication:

$B^{-1}A^{-1}AB = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I $

The postmultiplication should be obvious.

The matrix product in your example is indeed invertible as it has a nonzero determinant.

Answer 2

After you let $C$ be the inverse of $A$ and $D$ be the inverse of $B$, we have the following:

$DCAB = DIB = DB = I$ and $ABDC = AIC = AC = I$.

Therefore, $AB$ is invertible, with inverse $DC = B^{-1}A^{-1}$.

Answer 3

In your example $AB$ is invertible since it has non-zero determinant. Notice that since there is $C$ such that $AC=CA=I$ and $D$ so that $BD=DB=I$ then

$(DC)(AB)=D(CA)B=DB=I$

$(AB)(DC)=A(BD)C=AC=I$