# 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 ?

• Small comment, you cannot take inverse if you don't know if it exist. But if we ignore it, then answer looks correct for me
– Lee
Jun 13, 2020 at 12:19

Yes, since we can write $$B$$ as the product of invertible matrices, it is also invertible.

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, $$B=A^{-1}$$

$$B= X(A-AX)^{-1}.$$ 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.
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