When is the Determinant of a Square Matrix Equal to Its Negative?

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Homework Statement



Suppose A is a square matrix of size n. When is det(-A) = -det(A)?

Homework Equations



N/A

The Attempt at a Solution



My approach to the problem is to simply multiply the size n identity matrix by -1, then multiplied by A. For example: det((-1)*IdentityMatrix[n]*A) = det((-1)*IdentityMatrix[n])*det(A). At this point I could answer the original question by saying this is true when n is odd. But I get the impression I am overlooking some other very obvious answer or condition, and am wondering if anyone can think of a different approach. Thanks.
 
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you reasoning seems sound to me...
 
Looks good to me too.

Remember that for scalar c and matrix A, det(cA)=cndet(A) where n is the size of A. Plugging in c = -1 gives your answer.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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