I am trying to prove the alternating property of the determinant. From what I have seen as far as now everything resolves around the idea that if I have a matrix and begin to switch columns/rows then the amount of switches required to return to the original matrix will determine my sign.
Example: Let this matrix be A and be the original matrix
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
Now switch the colums. \begin{pmatrix} b & a \\ d & c \end{pmatrix}
The derminant of this new matrix will be (-1)(D|A|) because it takes one column switch to go to original. This expression is generalized by $(-1)^{n};$ n = the amount of switches.
My question is where does this $(-1)^{n}$ come from. What is its logic/intuition behind it. I'm not looking for an answer that says if its odd its negative even positive. I'd like to genuinely understand. Please give me a basic answer that I can eventually expand on.
Thank you for your time
A matrix represents a linear transformation of your space (in this case $\mathbb{R}^2$), and a linear transformation can distort space. The determinant tells you how much the area of some given shape in the space will change after the transformation.
For example, the matrix $\begin{bmatrix}2&0\\0&2\end{bmatrix}$ represents a scaling in both directions by 2. The determinant of this matrix is 4 because any given shape, when scaled in both directions by 2, will have an area 4 times as large.
The determinant being negative means the transformation also "flips" shapes around. Switching the columns of a matrix will have the effect of switching the final positions of the corresponding basis vectors; this will create the effect of "flipping" the space.