Why is $\det⁡(-A)=(-1)^n\det(A)$? Why is $\det⁡(-A)=(-1)^n\det(A)$?
 A: Do you know that $\det (AB)=\det A\cdot \det B$ for any two square matrices $A,B$?
If so, notice that $-A=(-I)A$ where $-I$ is the matrix with $-1$ on the diagonal and $0$ elsewhere. Clearly, $\det(-I)=(-1)^n$.
More generaly, $\alpha A=(\alpha I)A$ and hence $\det(\alpha A)=\alpha^n\det A$ for any scalar $\alpha$.
A: An elementary definition of the determinant of the matrix $A=(a_{i,j})$ is given by the expression
$$
\det(A)=\sum_{\pi}s(\pi)a_{1,\pi(1)}\cdots a_{n,\pi(n)}
$$
where the sum is extended over all permutations of the set $\{1,...,n\}$ and $s(\pi)=\pm1$ is the sign of the permutation $\pi$.
Since each summand is the product of exactly $n$ of the entries of the matrix, if you reverse all the signs each summand--and so the whole sum--will be altered by a factor of $(-1)^n$.
A: Lots of ways to see this, mathematical and geometrical. Here are some mathsy ways first...


*

*Determinants are linear in every row or column. Since you have $n$ of them which you multiply by $-1$ you get this result.

*More explicitly, writing out a typical definition with an alternating tensor $\epsilon$, there are $n$ $A$  matrix elements.

*You could verify this by computing $\det (-I)$ as Dennis did.

*For diagonalizable matrices, determinants are products of eigenvalues, and you're changing $n$ of them in sign.


Most interesting are geometrical approaches. For instance, as a change of basis for $A$ invertible (the other case being trivial due to vanishing determinants), the matrix determinant has an interpretation as the change in volume form. A negative Jacobian factor corresponds to inversion or handedness changes.
In this way of looking at it, we see the result is equivalent to the observation that in even numbers of dimensions, changing the sign of everything can be achieved just by spinning everything around, whereas in odd numbers of dimensions it cannot be smoothy achieved without pushing a basis vector through zero (i.e. a non invertible matrix) where the $\det$ is zero.
