# Describe the solution set of the system

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..

• Hint: what system does your reduced row-echelon form represent? – Roger Burt Aug 21 '14 at 19:56
• @Roger Burt We must have posted at the same time. Didn't mean to get in your way. – Paul Sundheim Aug 21 '14 at 20:02
• @PaulSundheim It's all good. – Roger Burt Aug 21 '14 at 20:16

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 $\\x_1=-4s+t+9\\ x_2=-2s+t+4$