Show that if $A^{ −1}$ exists and $x, y ∈ \Bbb R^n$ , then $(A + xy t )^{ −1}$ exists if and only if $y^t A^{ −1} x = −1$.

Question

My question is regarding #10, please see the image link below to view the full prompt.

I am unsure as to how to begin...

Any help would be appreciated. Thanks.

Full Prompt

Answer 1

Since $A$ is invertible, we have that $$\det(A+xy^t)=\det(A)\det(I+A^{-1}xy^t)=\det(A)(1+y^tAx).$$ Here I used exercise 9 in the last equation. The result follows.

Answer 2

You can find directly the inverse matrix. It has to have the form $$ (A+xy^t)^{-1}=A^{-1}+c·A^{-1}xy^TA^{-1}. $$ Now compute $c$ by evaluating the product $$ I=(A+xy^t)(A^{-1}+c·A^{-1}xy^TA^{-1}) $$