Inverse of matrix sum

Question

I found on the Wikipedia page "Determinant" the following property: For any invertible $m \times m$ matrix $X$, $\det(X + AB) = \det(X) \det(I_m + BX^{-1}A)$.

Is this true? If so, how is this proved?

Answer 1

Two properties you can use to get there: $$det(AB)=det(A)det(B)$$ $$det(AB)=det(BA)$$

Answer 2

To prove equality $$ \det(X+AB)=\det(X)\det(I+BX^{-1}A), \tag1$$ for any invertible matrix $X$ and arbitrary matrices $A, B$, one needs, beside the well known equality $$\det(AB)=\det(A)\det(B)=\det(BA), \tag2$$ also the Sylvester's determinant theorem which says that $$\det(I+AB)=\det(I+BA). \tag3$$ With (2) and (3) the proof of (1) is easy: $$\det(X+AB)=\det(X)\det(I+X^{-1}AB)=\det(X)\det(I+BX^{-1}A). $$

Note that (1) can be proved without the Sylvester's determinant theorem if $A$ or $B$ is invertible. For instance, if $A$ is invertible, then $$ \det(X+AB)=\det(A)\det(A^{-1}X+B)=\det(A)\det(A^{-1}+BX^{-1})\det(X)= $$ $$=\det(X)\det(A^{-1}+BX^{-1})\det(A)=\det(X)\det(I+BX^{-1}A). $$