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# Prove that the determinant of a skew symmetric matrix of even order is a perfect square

SHAIK AASIF AHAMED
7 years ago
Hello student,
see that for an [nXn] skew symmetric marix A, with n odd, we have [det A=0] , which is againa perfect square. if n is even, let [A=[aij] we have [aii=0] and [aij=-aji.] the proof is by induction on even numbers, but for a more general case, i.e. matrices with entriesin [{Q}.] for n = 2, it's clear. for [n4] : if [a1j=0] , for all j, then the determinant of A is 0 and sowe're done.
otherwise [a1j$\neq$0] for some [j > 1.] for simplicity suppose j = 2. then using columnoperations we can change the first row to [(0,a12, 0, 0, ... ,0).] similarly using row operations wecan change the first column to [(0, -a12, 0, 0, ... ,0).]