# Det(A) = det(AT) = det(−A) = (−1)^n*det(A). Hence det(A) = 0 when n is odd Why?

RossH

## Homework Statement

det(A) = det(AT) = det(−A) = (−1)^n*det(A). Hence det(A) = 0 when n is odd
I understand all of this except for the last statement: =(-1)^n*det(A).
Why is this? Does it have to do with the cofactor expansion, and the "checkerboard" (as my professor puts it) of signs? I'm just not quite sure why that has to be. A quick explanation would be more than adequate. Thank you very much for your time. I didn't think that my post fit the rest of the template so I deleted it.

VeeEight
It follows from the fact that: if B is formed from multiplying a single row of A by scalar x, then det(B) = x(det(A)). This can be shown easily using the definition of the determinant - you know there is some bij = xaij (where i is the row you multiplied to), so you can factor out the x. The desired result comes from multiplying all rows of A by x.

Also, note that det(A) = det(AT) is true but det(A) = det(-A) only if the size of n is even, and your statement: "det(A) = 0 if n when n is odd" doesn't make sense. det(A)=0 tells you about the invertibility of A.

RossH
It follows from the fact that: if B is formed from multiplying a single row of A by scalar x, then det(B) = x(det(A)). This can be shown easily using the definition of the determinant - you know there is some bij = xaij (where i is the row you multiplied to), so you can factor out the x. The desired result comes from multiplying all rows of A by x.

Also, note that det(A) = det(AT) is true but det(A) = det(-A) only if the size of n is even, and your statement: "det(A) = 0 if n when n is odd" doesn't make sense. det(A)=0 tells you about the invertibility of A.

Wow I'm an idiot. I forgot to mention that I was talking about a skew symmetric matrix. So I guess it does make sense. Thank you for your help.

VeeEight
Okay, I thought you were talking about something specific but thought I'd throw in that last bit anyway to be safe. Cheers.