Determinant of $ (-A) $ in dependance of determinant $ ( A) $ by $ n \times n $ Matrices [closed]

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!

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) $