linear algebra
I'm reading this text and I'm a bit confused about how they determined the determinant of this matrix:
The factoring out of the -7 and the -3 is because of rule 2 in theoreum 3.3 right?
What is the intuition behind these rules? Aren't row operations supposed to keep matrices equal? Aren't matrices that result from the elementary row operations supposed to be equivalent? So why would that change the determinants if you say interchange two rows?
We are changing the system of equations, what we are not changing is the set of solutions to the system of equations.
The following figure shows the equations \begin{align*} -x+2y &= 2 \\ x-y &= 0 \end{align*} and what happens after we add the first equation to the second, i.e., the system of equations \begin{align*} -x+2y &= 2 \\ y &= 2. \end{align*}
We can see that the system of equations has indeed changed, but the set of solutions (in this case $\{(2,2)\}$) is preserved.
To prove this: If $\mathbf{x}$ satisfies equations $E_1$ and $E_2$, then it satisfies equations $E_1$ and $E_1+E_2$. Conversely, if $\mathbf{x}$ satisfies equation $E_1$ and $E_1+E_2$, then it satisfies equations $E_1$ and $(E_1+E_2)-E_1$ (which is $E_2$).
We conclude that $\mathbf{x}$ satisfies equations $E_1$ and $E_2$ if and only if $\mathbf{x}$ satisfies equations $E_1$ and $E_1+E_2$.
Thus the set of solutions is preserved.
In your example, if you interchange two first colums you'll get $$ \left[\matrix{ x_1 & x_2 & x_3 \\ 2 & 1 & 3 \\ 4 & 4 & 2 \\ 1 & 1 & 4}\right]\quad\sim\quad \left[\matrix{ x_2 & x_1 & x_3 \\ 1 & 2 & 3 \\ 4 & 4 & 2 \\ 1 & 1 & 4}\right]. $$ What happened in the top most row is that the roles of $x_1$ and $x_2$ have been interchanged. When you get a solution to the modified system as $(2,-2,4)$, you must not forget to switch the variables back to the original ones.
Similar thing happens if you take, say, the first column minus the second column $$ \left[\matrix{ x_1 & x_2 & x_3 \\ 2 & 1 & 3 \\ 4 & 4 & 2 \\ 1 & 1 & 4}\right]\quad\sim\quad \left[\matrix{ \tilde x_1 & \tilde x_2 & \tilde x_3 \\ 1 & 1 & 3 \\ 0 & 4 & 2 \\ 0 & 1 & 4}\right]. $$ To see the relation between the old and the new variables, we write $$ A_1x_1+A_2x_2+A_3x_3=(A_1-A_2)\tilde x_1+A_2\tilde x_2+A_3\tilde x_3= A_1\tilde x_1+A_2(\tilde x_2-\tilde x_1)+A_3\tilde x_3. $$ hence, $x_1=\tilde x_1$, $x_2=\tilde x_2-\tilde x_1$ and $x_3=\tilde x_3$, i.e. your variables has changed again. It means that
working with columns you are working with variables while working with rows you are working with equations.
In general, elementary row operations are equivalent to left multiplication by an elementary matrix transformation $$ Ax=b\quad\implies\quad \underbrace{LA}_{\tilde A}x=\underbrace{Lb}_{\tilde b}, $$ while column operation corresponds to right multiplication by an elementary matrix. Because this multiplication has to be done for $A$ and matrix multiplication is not commuting, we have to compensate this action by the inverse matrix $$ Ax=b\quad\implies\quad \underbrace{AR}_{\tilde A}\underbrace{R^{-1}x}_{\tilde x}=b. $$
Best Answer
For a simple intuition and check of the Properties of the determinant let consider
$$\begin{vmatrix}1&0\\0&1\end{vmatrix}=1$$
but for interchanging rows/colums
$$\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1$$
multiply a row/column by a scalar
$$\begin{vmatrix}2&0\\0&1\end{vmatrix}=2$$
add first row to second row
$$\begin{vmatrix}1&0\\1&1\end{vmatrix}=1$$