I gotta solve this System of linear equations dependent on $a$:
$$x+y+z=1$$ $$-2x+2y+az=3$$ $$-ax+y+2z=2$$
I'm transforming this to a matrix.
$$ M_1 = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1\\ -2 & 2 & a & 3\\ -a & 1 & 2 & 2 \end{array}\right] $$
I'm switching x with y column and sort.
$$ M_2 = \left[\begin{array}{ccc|c} 1 & 1 & 1 &1\\ 1 & -a & 2 &3\\ 2 & -2 & a &2 \end{array}\right] $$
Solving this:
$$ M_3 = \left[\begin{array}{ccc|c} 1 & 1 & 1&1\\ 0 & -a-1 & 1&2 \\ 0 & 0 & a+2 &5 \end{array}\right] $$
Therefore I'm getting $$a= 3$$
Putting it into the $M1$ yields no result, as I end up with.
$$ M = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1\\ 0 & 4 & 5 & 1\\ 0 & 4 & 5 & 5 \end{array}\right] $$
Is there an error in my calculation?
I have severely edited my original answer.
Your first error occurs when forming $M2$. You did a combination of swapping the second and third row (which you are allowed to do) with swapping the first and second column (which you are not allowed to do).
Each row of the matrices $M1$, $M2$, and $M3$ represents an equation in $x, y,$ and $z$. By swapping rows, you just change the order in which the equations are listed; this is perfectly find because it does not change the solutions to the system. Swapping rows however corresponds to interchanging variables; this is not allowed because it effects the solutions of the system. For example, the solutions of $x + 2y = 0$ are not the same as the solutions of $y + 2x = 0$.
Your next error occurs when forming $M3$. You have done the row reduction incorrectly.
Your final error occurs when you conclude that $a = 3$. The third row of $M3$ represents the equation $(a + 2)z = 5$. That does not imply that $a = 3$. However, by imposing a restriction on $a$, you can find $z$ in terms of $a$.
I advise you try again. Start with $M1$ (which is the correct augmented matrix for the system) and row reduce it, making sure to only use the allowed row operations.