Q:

Prove the following: $[1-\lambda \operatorname{sum}(A^{-1})][1-\lambda \operatorname{sum}(B^{-1})]=1$

$\newcommand{\s}{\operatorname{sum}}$Problem: Let $E=[1]_{n\times n}$ and let $\s(X)$ be the sum of all elements of matrix $X$.
If $A$,$B$ $\in$ $M_{n \times n}$ $(\mathbb C)$ are regular matrices so that $A+B=\lambda E$, $\lambda \in \mathbb C$, prove the following:
$$[1-\lambda \s(A^{-1})][1-\lambda \s(B^{-1})]=1$$
We did not say much about sums of all elements. I did some research and found this topic, but I do not know how to use said information here.
This is what I've done so far:
$$[1-\lambda \s(A^{-1})][1-\lambda \s(B^{-1})]=1$$
$$1-\lambda \s(A^{-1})-\lambda \s(B^{-1})+\lambda^2\s(A^{-1}\s(B^{-1})=1$$
$$-\lambda \s(A^{-1})-\lambda \s(B^{-1})+\lambda^2\s(A^{-1})\s(B^{-1})=0$$
$$\lambda [-\s(A^{-1})-\s(B^{-1})+\lambda \s(A^{-1})\s(B^{-1})]=0$$
If $\lambda=0$ this is true, but if $\lambda \not= 0$ I have:
$$-\s(A^{-1})-\s(B^{-1})+\lambda \s(A^{-1})\s(B^{-1})=0$$
I am not sure how to make connection between $A$, $B$ and their inverse matrices.
I know that regular matrices $A$ and $B$ mean their inverse exists. Also $\det(A+B)=0$ which means $(A+B)^{-1}$ does not exist. 
Any help is greatly appreciated.

A:

Let us denote by $\sigma(M)$ the sum of the entries of matrix $M$. If we denote by $v^T=(1,1,\cdots, 1)$, then $\sigma(M)= v^TMv$. The way you defined it, $E=vv^T$. Now if $A+B=\lambda E$, you have
$$
\begin{aligned}
\lambda \sigma(A^{-1})\sigma(B^{-1})&= \lambda(v^TA^{-1}v)(v^TB^{-1}v)=
v^TA^{-1}(\lambda E) B^{-1}v\\
&=v^TA^{-1}(A+B)B^{-1}v =
v^TA^{-1}v+v^TB^{-1}v=\sigma(A^{-1})+\sigma(B^{-1})
\end{aligned}
$$

