Prove that a given matrix $B$ is invertible. It is given that $ A,X ,A-AX $ are invertible and $ (A-AX)^{-1}=X^{-1}B$ .
Prove that $B$ is invertible ?
My working is given below, since $ A-AX ,X^{-1} $ are both invertible, $B $ is invertible.
$ X^{-1}B=(A-AX)^{-1}$
$B= X(A-AX)^{-1}$
$B^{-1}=(X(A-AX)^{-1})^{-1}$
$B^{-1} = (A-AX)X^{-1}$
Is the proof correct ?
 A: Yes, since we can write $B$ as the product of invertible matrices, it is also invertible.
A: I'll just elaborate @Lee's comment. How do you know that product of invertible matrices is invertible? So we'll go by the definition of invertible matrices: Consider square matrices $A$ and $B$. If $AB=BA=I$, the identity matrix, then we say $A=B^{-1}$ or equivalently, $B=A^{-1}$
In your case, 
$B= X(A-AX)^{-1}.$  Let $P=(A-AX)X^{-1}$. Now show that: 
$BP=PB=I$, identity matrix. 
Clearly, $PB=(A-AX)(X^{-1}X)(A-AX)^{-1}=(A-AX)(I)(A-AX)^{-1}=I$ 
Similarly, $BP=I$ and hence by definition of invertible matrices $P=B^{-1}$ and hence $B$ is invertible.
A: In general if you have two invertible matrix $C$ and $D$, the inverse of $CD$ is $(CD)^{-1}=D^{-1}C^{-1}$ therefore your proof could be endend at line 2 when you write:
$$
B=X(A-AX)^{-1}
$$
because you have written $B$ as the product of two invertible matrices, which is an invertible matrix, therefore you have already proved that $B$ is invertible.
The other two lines are correct aswell, in those lines you find $B^{-1}$, but you already knew from line 2 that $B$ was invertible
