Cfrgtkky

If A and B are square, nonsingular matrices and X is a square matrix, then
$ (A + XBX^T)^{−1} = A^{−1} − A^{−1}X(B^{−1} + X^TA^{−1}X)^{−1}X^TA^{−1} $

This equation so complex, I can't see the key variable to transform and solve it. Plz help me!

First premultiply RHS by $A+XBX^T$, you will get identity matrix.

Then postmultiply RHS by $A+XBX^T$, you will get identity matrix again.

Therefore, it will be the inverse of this matrix by definition.

I am showing the procedure for postmultiplication for instance,

$$(A^{-1}-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1})(A+XBX^T)$$
$Longrightarrow I+A^{-1}XBX^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XBX^T$
$Longrightarrow I+A^{-1}X[B-(B^{-1}+X^TA^{-1}X)^{-1}-(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XB]X^T$

Now, let $B^{-1}+X^TA^{-1}X = M$,

In order to prove that this expression is I, we have to show that,
$$B-M^{-1}-M^{-1}X^TA^{-1}XB = 0$$
$$Longleftrightarrow B = M^{-1}(I+X^TA^{-1}XB)$$
$$Longleftrightarrow MB = I+X^TA^{-1}XB$$
$$Longleftrightarrow M = (I+X^TA^{-1}XB)B^{-1}$$
$$Longleftrightarrow M = B^{-1}+X^TA^{-1}X$$

Hence, this turns out to be I and similarly the other one will be I(After quite lengthy calculation).
Hope it is helpful.