I have not tried induction yet. What I'm trying to prove is actually: If [MATH]B = -A[/MATH] then [MATH]det(B) = (-1)^{n} \cdot detA[/MATH].Do you think that the problem has anything to do with the sign in from of the A?
What have you tried? Are you stuck somewhere? Have you even tried to see if statement is even true???? Have you considered doing it by induction? I'll ask again, have you tried to see if it is true?? Does the sign in front of the A really matter. Hint: It could be the case that matrix B = -A.
Are you saying that det(A) is always positive? Hmm, then wouldn't that make det(B) positive. But wait let(B) = (-1)ndet(A). And the right hand side changes from positive to negative based on (-1)n if we assume that set(A) is always positive. This scenario is all possible and if it is then determinant of any matrix is 0. Do you see this logic?I have not tried induction yet. What I'm trying to prove is actually: If [MATH]B = -A[/MATH] then [MATH]det(B) = (-1)^{n} \cdot detA[/MATH].
The issue here is if you do prove the above is true (and you should state that B=-A) how would that help prove your theorem which you posted.What I'm trying to prove is actually: If [MATH]B = -A[/MATH] then [MATH]det(B) = (-1)^{n} \cdot detA[/MATH].
If you want to do this by induction then go for it. But there is a simpler way.How can I prove that the determinant of the matrix [MATH]-A, \in M_{n}[/MATH], [MATH]det(-A)[/MATH] is positive when [MATH]n[/MATH] is even and negative when [MATH]n[/MATH] is odd?
My intent was to walk through the problem by demonstrating that if you change the signs on one row (or column) that we multiply the determinant by -1. So if you have a matrix with an odd rank then [math]det(-A) = (-1)^{odd power} det(A)[/math], from which the theorem follows.Dan,
OK what you are showing is true. I just don't agree that when you go from A to -A you just change the signs of 1 row. I do agree that if you have a matrix A and then you multiply any row (or column) by -1 then the det of both matrices will differ by a sign. That is absolutely true.
To me, -A is the matrix which you add to A to get the zero matrix. Am I wrong????
Dan, I think that you are being a little sloppy. I agree that every time you change the signs of one row that the determinant changes its sign. We both agree with this. But the problem does that say that det(A) = det(-A) if n ie even and det(A) = - det(-A) if n is odd. This only works if det(-A)>=0, not when det(-A)<0My intent was to walk through the problem by demonstrating that if you change the signs on one row (or column) that we multiply the determinant by -1. So if you have a matrix with an odd rank then [math]det(-A) = (-1)^{odd power} det(A)[/math], from which the theorem follows.
-Dan
Okay, you've completely lost me. Why does it matter if the determinant of A is greater than or less that 0? I'm willing to admit that I might be wrong but I have no idea where.Dan, I think that you are being a little sloppy. I agree that every time you change the signs of one row that the determinant changes its sign. We both agree with this. But the problem does that say that det(A) = det(-A) if n ie even and det(A) = - det(-A) if n is odd. This only works if det(-A)>=0, not when det(-A)<0
My bad. The original question should have been: show that If [MATH]B=−A[/MATH] then [MATH]det(B)=(−1)^{n}⋅detA.[/MATH]I might be missing something completely obvious but here goes.
You asked Where does the condition on the positivity or negativity of |-A| come in?
The OP asked how to show that det(−A) is positive when n is even and negative when n is odd.