# Solve the system (3)

Solve the system

$x_1 + x_2 -3x_3 = -2$

$4x_1 + 3x_2 + 3x_3 = 2$

$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}\\\\\\\end{bmatrix} + \begin{bmatrix}\\\\\\\end{bmatrix} s$

Do I need to put this in RREF?

Or how should I go about doing this?

The most straight-forward way about solving this is to take the 2 equations and set them up as an Augmented Matrix and get RREF like so:

$\left[\begin{array}{ccc|c} 1 & 1 & -3 & -2\\ 4 & 3 & 3 &2 \end{array}\right] \longrightarrow RREF \longrightarrow \left[\begin{array}{ccc|c} 1 & 0 & 12 & 8\\ 0 & -1 & 15 & 10 \end{array}\right]$

Being that you have 2 leading variables the system of equations is dependent and write the system like so:

$x_1+12x_3=8$

and

$-x_2+15x_3=10$

let $x_3=s$ and rewrited the sytem as

$\left[\begin{array} \\x_1\\ x_2\\ x_3\\ \end{array}\right]= \left[\begin{array} \\ 8 \\ -10 \\ 0 \end{array}\right]+ \left[\begin{array} \\ -12\\ 15\\1 \end{array}\right]s$

and there you have it!

Almost anything works. Adding the two equations gives $x_1=-0.8x_2$. Substituting that in the first equation gives $3x_3=2+0.2x_2$. So we have $(x_1,x_2,x_3)=(0,0,\frac{2}{3})+\lambda(-\frac{4}{5},1,\frac{1}{15})$

• So I should try RREF then? – Shammy Jun 25 '16 at 19:36
• If you want to use matrices, that works too. But this case is really simple it is faster to do it directly as above. – almagest Jun 25 '16 at 19:38