Prove that $I+\lambda A$,$I+\lambda B$,$I+\lambda C$ are invertible and A=B=C Given the matrices $A,B,C\in M_n(\mathbb R)$ which satisfy the condiditions:
$A+B+C+\lambda ΑΒ=0 \quad(1)$
$Α+Β+C+\lambda BC=0 \quad(2)$
$A+B+C+\lambda CA=0 \quad(3)$
for some $\lambda\neq0$.
(a) Prove that $I+\lambda Α$, $Ι+\lambda Β$, $I+\lambda C$ are invertible and $AB=BC=CA$.
(b) Prove that $A=B=C$.
For (a) i understand that the matrices are invertible, because we have non zero eigenvalue. In addition we have AB=BC=CA, because the (1)-(2) and the (2)-(3) give us these equalities.
I have been stucked in (b). I tried to use the equality $AB=BC=CA$ and the fact that $I+\lambda Α,Ι+\lambda Β,Ι+\lambda C$ are invertible, but i can't prove that $A=B=C$.
I thought that A, B and C are null matrix, but i don't find a way to prove that.
I want help for (b).
 A: For (a), the equations are equivalent to $-\lambda AB=-\lambda BC=-\lambda CA=A+B+C$. Since $\lambda$ is nonzero, we have $AB=BC=CA$.
Now suppose $(I+\lambda A)x=0$. Then $x=-\lambda Ax$. Hence $Cx=C(-\lambda Ax)=-\lambda CAx=(A+B+C)x$. Subtract $Cx$ from both sides and rearrange the remaining terms, we get $Bx=-Ax$. It follows that $Cx=-Ax$ too, because
$$
C\left(-\frac1\lambda x\right)=CAx=
ABx=-A^2x=-\frac1{\lambda^2}x.
$$
But then from $AB=BC$, we obtain $-\frac1{\lambda^2}x=-A^2x=ABx=BCx=A^2x=\frac1{\lambda^2}x$. Therefore $x=0$ and $I+\lambda A$ is invertible. By similar arguments, $I+\lambda B$ and $I+\lambda C$ are invertible too.
The statement in (b) is not true. For a counterexample, pick any three different scalars $a,b,c$ such that $a+b+c=0$ and consider
$$
A=\pmatrix{0&a\\ 0&0},\ B=\pmatrix{0&b\\ 0&0},\ C=\pmatrix{0&c\\ 0&0},\ \lambda=1.
$$
However, it is always true that $A^2=B^2=C^2$. This can be shown by considering $(A+B+C+\lambda AB)C=A(A+B+C+\lambda BC)$ and similar equalities.
A: This is not an answer, these are just ideas that came up into my head and I am not sure they'll help (I don't have much time now to try it) but try for example multiplying the equations by $I+\lambda A$ you'll get $A^2+AB+AC+\lambda AB=0$ 
Or maybe find the inverse of $I+\lambda A$, maybe it has a form that'll help.
