Consider the linear system below: $$\begin{array}{ccccccc} x_1&-&2x_2&+&&&x_4&=&1\\ 2x_1& -& 5x_2& -& 2x_3& +& k^2x_4 &= &-2\\ &&x_2&+&2x_3&-&x_4&=&4 \end{array}$$
For $k = \sqrt{3}$, describe the solution set of the system.
Attempt at a solution
I tried to take its RREF, this is the matrix I acquired.
$$ \begin{matrix} 1 & 0 & 4 & -1 & 9 \\ 0 & 1 & 2 & -1 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end{matrix} $$
The values at the right-hand coloumn are the solutions, however I have no clue where to move on from here..
From your reduced matrix: $x_1+4x_3-x_4=9\\ x_2+2x_3-x_4=4$ where $x_4$ and $x_3$ are free. You therefore get $\\x_1=-4s+t+9\\ x_2=-2s+t+4$
Can you get the solution set from here?