Show that $\forall n,k\geq 2$, there are invertible non diagonal matrices $\in M_{n}(\mathbb{R})$ such that $A_{1}^{-1}+A_{2}^{-1}+\cdots +A_{k}^{-1}=(A_{1}+A_{2}+\cdots +A_{k})^{-1}$.
I only need one example of $2 \times 2$ matrix with this property, because then I can solve it easily using a block matrix.
You wish to have
$$A^{-1}+B^{-1}=(A+B)^{-1}$$ the same as $$(A+B)(A^{-1}+B^{-1})=I+AB^{-1}+BA^{-1}+I=I$$ or $$AB^{-1}+BA^{-1}=-I$$
Let $C=AB^{-1}$ then the equation becomes
$$C+C^{-1}=-I$$ multiplying and rearranging, $$C^2+C+I=0$$
Now the characteristic polynomial of $C=\begin{pmatrix} a&b\\c&d\end{pmatrix}$ is $$x^2-tr(C)x+\det(C)=0$$ So you have
$$a+d=-1$$ and $$ad-bc=1$$
For example $$C=\begin{pmatrix} 1&-3\\1&-2\end{pmatrix}$$ Now chose ANY invertible $B$ and let $A=CB$.
For those of us who are incapable of taking an idea and using it, who need everything spelled out for them, here is how to generalise. Given $A$ and $B$ as above, let $X=A+B$ then $X^{-1}=A^{-1}+B^{-1}$ and so to get three matrices you have the equation. $$(X+C)=X^{-1}+C^{-1}$$ then as before you can solve for $CX^{-1}=D$ and let $C=DX$.
