<s>[INST] If $A^2=A,\,B^2=B,\,C^2=C$ and $A+B+C=I,$ prove $AB=BC=AC=\mathbb{O}$. 
Let $A,\,B,\,C$ be square matrices such that $A^2=A,\,B^2=B,\,C^2=C$ and $A+B+C=I.$
Prove that $AB=BC=CA=\mathbb{O}.$

Attempt. We have $A(B+C)=A(I-A)=A-A^2=\mathbb{O}$ and similarly we get
$B(A+C)=C(A+B)=\mathbb{O}$, but this is where i get so far.
Thanks in advance for the help.
 [/INST] Roland's answer gives a great and simple linear algebra argument.
Here is a purely algebraic development:
\begin{align*}
(A+B+C)^2 & = A^2 + B^2 + C^2 + AB + AC + BC + BA + CA + CB \\
& = I  + A(B+C) + BC + (B+C)A + CB \\
& = I + A(I - A) + (I-A)A + BC + CB \\
& = I + A - A^2 + A - A^2 + BC + CB \\
& = I + BC + CB \\
I & = I + BC + CB \\
0 & = BC + CB \\
\end{align*}
By permuting the roles of $A,B,C$, we get $BC + CB = AC + CA = AB + BA = 0$.
This would allow us to conclude if we knew that $AB = BA$. It is not one of the hypotheses, but at this point we can deduce it:
\begin{align*}
AB + BA & = 0 \\
AB & = - BA \\
A^2B & = - ABA \\
AB & = - ABA \\
AB & = BA^2 \\
AB & = BA \\
\end{align*}
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