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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..

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

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