[I did notice similar questions were asked here before, but I couldn't find a satisfactory answer for me to grasp as a beginner, so I chose to post this question]
I'm just starting to teach myself linear algebra with Linear Algebra and Group Theory. The book starts with the concept of determinant with the definition of even & odd permutations by giving the example of a {3 by 3 array} with the following equation:
$$\begin{vmatrix} a_{11}\ \ a_{12}\ \ ...\ \ a_{1k}\ \ ...\ \ a_{1n}\\a_{21}\ \ a_{22}\ \ ...\ \ a_{2k}\ \ ...\ \ a_{2n}\\...\ \ ...\ \ ...\ \ ...\ \ ...\ \ ...\\a_{n1}\ \ a_{n2}\ \ ...\ \ a_{nk}\ \ ...\ \ a_{nn}\end{vmatrix}=\displaystyle\sum_{(p_1, p_2, ..., p_n)}(-1)^{[p_1, p_2, ..., p_n]}a_{1p_1}a_{2p_2}...a_{np_n}$$
where $[p_1, p_2, ..., p_n] $ is the number of inversions of permutation $p_1, p_2, ..., p_n$
So because there is no justification given in the book for the above equation for {$n$ by $n$ array} and the concept of determinant feels a bit odd at the first place, I tried to investigate it myself:
[Step 1]: I started with a {2 by 2 array} first, by taking a system of two equations in two unknowns:
$$(eq1):\ a_{11}x_1+a_{12}x_2=b_1\\(eq2):\ a_{21}x_1+a_{22}x_2=b_2$$ $$A=\begin{Vmatrix} a_{11}\ \ a_{12}\\a_{21}\ \ a_{22}\end{Vmatrix}$$with the determinant $detA$
[Step 2]: I then tried to eliminate {$x_2$ in $eq1$} and {$x_1$ in $eq2$} by doing the following:
$$(eq1 \cdot a_{22})-(eq2 \cdot a_{12})=(a_{11}a_{22}-a_{12}a_{21})x_1\\(eq2 \cdot a_{11})-(eq1 \cdot a_{21})=(a_{11}a_{22}-a_{12}a_{21})x_2$$
[Step 3]: I noticed that the two coefficients above both give me the determinant of the array, so I then postulated the following statement:
the determinant is {the coefficient of unknown $x_k$ in $k$th row} after eliminating other unknowns from the $k$th row by {multiplication} and {subtracting other rows} in the array.
that is to say, if I have a $n$th order array $$N=\begin{Vmatrix} a_{11}\ \ a_{12}\ \ ...\ \ a_{1k}\ \ ...\ \ a_{1n}\\a_{21}\ \ a_{22}\ \ ...\ \ a_{2k}\ \ ...\ \ a_{2n}\\...\ \ ...\ \ ...\ \ ...\ \ ...\ \ ...\\a_{n1}\ \ a_{n2}\ \ ...\ \ a_{nk}\ \ ...\ \ a_{nn}\end{Vmatrix}$$ I can eventually transfer $N$ into $$\begin{Vmatrix} detN & 0 & ... & 0 & ... & 0\\0 & detN & ... & 0 & ... & 0\\... & ... & ... & ... & ... & ...\\0 & 0 & ... & 0 & ... & detN\end{Vmatrix}$$
[Step 4]: I tested my statement with a {3 by 3 array} and it seems to work. And the idea of odd & even permutation seems to become more intuitive as it has to do with the order of subtracting depending on the row of the unknowns.
So here comes my questions
- if my guess is right, how do I construct the permuatation formula at the beginning of the question for {$n$ by $n$ array} without defining determinant by using a set of formalistic operation at the first place ?
- I've seen multiple answers talking about the geometric intuition of determinant (and I roughly get the idea). How does the intuition of permutation connects, or transfers into the geometric intuition ?
[Note: I have never studied abstract algebra, so answers without using notations in abstract algebra will be much appreciated :)]
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EDIT: I think I figured out my question 2 (the geometric intuition).... correct me if I am wrong
So again using a {2 by 2 array} as an example:
[Step 1]: Assume again I have the following equations and array
$$(eq1):\ a_{11}x_1+a_{12}x_2=b_1\\(eq2):\ a_{21}x_1+a_{22}x_2=b_2$$
$$A=\begin{vmatrix} a_{11}\ \ a_{12}\\a_{21}\ \ a_{22}\end{vmatrix}$$ with the determinant $detA$
[Step 2]: I can immediately transfer the array into
$$\begin{Vmatrix} detA & 0\\0 & detA\end{Vmatrix}$$
[Step 3]: Becasue the above array is the coefficient of $x_1$ and $x_2$, I can write the unknowns down as a vector
$$\begin{bmatrix} x_1\\x_2\end{bmatrix}$$ which makes the {$detA$ array} a linear transformation when multiply with this vector