Let A be a real 101x101 matrix.
If A is anti-symmetric, then |A^1001 + A^1003|= 0.
Why is this true? This was the correct answer to a multiple-choice question.
note that 101 is odd.
therefore, det(A) = det(-A^{t}) = (-1)^{101}det(A^{t}) = -det(A) (since det(A) = det(A^{t})),
hence det(A) = 0.
now A^{1001} + A^{1003} = (A^{1001})(I + A^{2}),
thus det(A^{1001} + A^{1003}) = det(A^{1001})det(I + A^{2})
= (det(A))^{1001}(det(I + A^{2})) = (0)(det(I + A^{2})) = 0