I have been given the following set of linear equations:
$-x_1-2x_2+3x_4=6$
$x_1+2x_2-2x_4=5$
$2x_1+4x_2+x_3-4x_4=-8$
Part of the question asks me to write this as an augmented matrix and bring it into reduced row echelon form: (See my answer below)
\begin{pmatrix} 1 & 2 & 0 & 0 & 27\\ 0 & 0 & 1 & 0 & -18\\ 0 & 0 & 0 & 1 & 11 \end{pmatrix}
I was then asked to find the solutions set of the system of linear equations: (See my answer below)
$x_4 = 11$
$x_3 = -18$
$x_1 = 27 - 2x_2$
(Note: I am unsure if my answer for $x_1$ is correct?).
The final part of the question (which is what this post is about) has asked me to find the solution set of the corresponding homogeneous system of linear equations. To my understanding, I simply make each equation equal to 0 and solve from there. Hence:
\begin{pmatrix} 1 & 2 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}
Leaving me with:
$x_1 = -2x_2$
$x_3 = 0$
$x_4 = 0$
Is this correct, or I have gone about this the wrong way?