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So if I have a matrix

$\left[\begin{array}{ccc|c} 1 & 3 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{array}\right]$

The solution is:

$\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right]$ = $\left[\begin{array}{c} -5 \\ 0 \\ 3 \end{array}\right]$ + $x_2$$\left[\begin{array}{c} -3 \\ 1 \\ 0 \end{array}\right]$

Obviously I see how the first transformed into the second, but what is the exact process from getting the solution from the given matrix? Why was the first column just thrown out? Thanks

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Here's the process I use:

Step one: Row reduce your matrix until it is in RREF. You've already done that in this case. Good job!

$$\left[\begin{array}{ccc|c} 1 & 3 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{array}\right]$$

Step two: Determine your pivots. Pivots are the first nonzero numbers in each row before the vertical line (if there is one). There are two in this matrix.

$$\require{enclose}\left[\begin{array}{ccc|c} \enclose{circle}1 & 3 & 0 & -5 \\ 0 & 0 & \enclose{circle}1 & 3 \end{array}\right]$$

Step three: All variables corresponding to columns without pivots are free variables. Set them equal to arbitrary letters. In this case, only the second column doesn't have a pivot so the next thing I'd write down is

$$\begin{matrix}\text{Let } x_2 = t, & t\in \Bbb R\end{matrix}$$

Step four: Solve the system of equations for your pivot variables in terms of your free variables:

$$\begin{cases} x_1 + 3x_2 = -5 \\ x_3 = 3\end{cases} \iff \begin{cases} x_1 + 3t = -5 \\ x_3 = 3\end{cases} \iff \begin{cases} x_1 = -3t -5 \\ x_3 = 3\end{cases}$$

Step five: Write down the solution.

$$\begin{matrix}\pmatrix{x_1 \\ x_2 \\ x_3} = \pmatrix{-3t-5 \\ t \\ 3} = t\pmatrix{-3 \\ 1 \\ 0} + \pmatrix{-5 \\ 0 \\ 3}, & t\in \Bbb R\end{matrix}$$

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  • $\begingroup$ Thanks for the detail, much appreciated $\endgroup$ – Brian Jones Oct 6 '15 at 1:04
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    $\begingroup$ I figured it out, was a silly question. (Studying all day will do that to you haha) Was asking why the bottom 3 was on the side without the t, and obviously because it doesn't have a t to take out of it. $\endgroup$ – Brian Jones Oct 6 '15 at 1:06
  • $\begingroup$ No problem. :-) $\endgroup$ – user137731 Oct 6 '15 at 1:07

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