Determinant of $ (-A) $ in dependance of determinant $ ( A) $ by $ n \times n $ Matrices I would like to know what is $det( -A) $  dependance of $ det( A) $ when $A$ is a $n \times n$ Matrix.
Well after a lot of trying out many examples which included looking into $ n \times  n $ when $ n = 2, 3,4, 5 $  I came to the conclusion that $ det A = - det (-A) $ when n is odd like $3, 5 $ etc. And that $ det A = det( -A)$  when n is even such that $  n= 2, 4 $ etc. What do you think?
Thanks
 A: Your observation is correct, and the fact stems directly from the well known property of determinants, which is that the determinant is a multilinear function. This means that if you multiply one column (or row) of a matrix by some factor $\alpha$, the entire determinant also changes by the factor $\alpha$.
If it is not immediately clear to you why this property means that $\det(-A)=\det(A)$ for even-sized matrices and $\det(-A)=-\det(A)$ for odd-sized matrices, then it is very good practice for you to actually work it out!
A: The best way to understand is by example. Consider a simple case of $2\times 2$ matrix.
$\det\begin{pmatrix} ka & kb \\ c & d \end{pmatrix} =k\times\det\begin{pmatrix} a & b \\ c & d \end{pmatrix}  $
$\begin{align} \det\begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix} &=k\times\det\begin{pmatrix} a & b \\ kc & kd \end{pmatrix} \\&=k^2\times\det\begin{pmatrix} a & b \\ c & d \end{pmatrix}  \end{align} $
Determinant is linear and homogenous in each row and each column separately .
$det(kA) =k^n det(A) $
When we multiply a matrix by a scalar, we multiply all entries by that scalar. i.e
$k•A=k•(a_{ij})_{n \times n}=(ka_{ij})_{n\times n}$
Hence, $det(kA) =k^{\text {(no of rows) }} det(A) $
