Let $A,B,C\in M_{2}(C)$ such that $$A^2+B^2+C^2=AB+BC+CA$$
show that
$$\det{(A^2+B^2+C^2-BA-CB-AC)}=0$$
from:matrix indentity
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Use the determinant product formula $\det(AB)=\det(A)\det(B)$. – J.R. Mar 13 '15 at 11:20
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1@YourAdHere I don't know he can use this formula – Elaqqad Mar 13 '15 at 11:38
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Hint : if $A\in M_{2\times 2}(\mathbb{R})$ then : $$\det(A)=(tr(A)^2-tr(A^2))/2,$$ and we know that $$tr(A^2+B^2+C^2-BA-CB-AC)=tr(AB+BC+CA-BA-CB-AC)=0$$
Farhad
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