prove that if A is idempotent, then det(A)=0 or det(A)=-1?

Answer 1

Minor error in the question: it should be "prove that det(A) = 0 or det(A) = 1". If det(A) = -1 A is definitely not idempotent!

det(AB) = det(A).det(B)
So det(A^2) = (det(A))^2.

If A is idempotent, A^2 = A. So det(A) = det(A)^2.
=> det(A) [det(A) - 1] = 0
=> det(A) = 0 or 1.

Answer 2

A is idempotent if A^2 = A
=> det (A^2)= det(A)
=> det(A) ^2 = det( A)
=> det(A) (det(A) -1) =0
=> det(A) =0 or det(A)-1 =0
so det(A)=0 or det(A) =1.

your statement is not quite correct! .