Determinant of Skew-Symmetric Matrix: Is it Zero for Odd n?

[bmb2009]

Homework Statement

A square (nn) matrix is called skew-symmetric (or antisymmetric) if AT =
-A. Prove that if A is skew-symmetric and n is odd, then detA = 0. Is this true
for even n?

Homework Equations

Det(A) = Det(AT) where AT= the transpose of matrix A

The Attempt at a Solution

I started to try and say that since AT=-A then Det(AT) = Det(-A) so Det(A) = Det(-A) b/c the law that Det(A)=Det(AT) but I didnt know where to go from here.. specifically what impact does the odd number of n have to do with anything.. Any help

 

[bmb2009]

and I assume that I need to say that Det(-A) = -Det(A) for the odd number on n so then I could conclude 2Det(A)=0 but i don't see why: Det(-A) = -Det(A) is true for odd number n matrices

 

[micromass]

You can easily calculate det(-I), no?? (where I is the identity matrix)
Then you can write det(-A)= det((-I)A).