Solve the system of linear equations: $$\begin{cases} x_2+2x_3+3x_4=a_1 \\ x_1+2x_2+3x_3+4x_4=a_2 \\ 2x_1+3x_2+4x_3+5x_4=a_3 \end{cases} $$ where:
$1)$ $a_1 = a_2=a_3=0$ and $2)$ $a_1=12, a_2=30, a_3=40$
We are looking for all solutions. Write down the coefficient matrix and use Gauss elimination.
After doing some row operations, I got the following matrix: $$ \begin{pmatrix} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 0 \\ \end{pmatrix} $$
In case 1) the linear equations look like: $$\begin{cases} x_1-x_3-2x_4=0 \\ x_2+2x_3+3x_4=0 \\ \end{cases} $$ let $x_3 = s$ and $x_4=t$, therefore, $x_2 = -2t -3s$ and $x_1 = t + 2s$ These are all solutions.
In case 2) I have the same matrix after doing some row operations and the following linear equation system: $$\begin{cases} x_1-x_3-2x_4=0 \\ x_2+2x_3+3x_4=12 \\ \end{cases} $$ let $x_3 = t$ and $x_4 = s$, therefore, $x_1 = t + 2s$ and $x_2 = 12 - 2t - 3s$ and these appear to be all solutions.
Could you please confirm with me whether I am on the right track?