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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?

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  • $\begingroup$ You'll get $x_2+2x_3+3x_4=6$ and $0=-8$ in the second case. The system has a solution iff $a_1+a_3=2a_2$. $\endgroup$ – Michael Hoppe Feb 25 '18 at 13:30
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RREF yields $$\begin{matrix} 1&0&-1&2&a_2-1_1\\ 0&1&2&3&a_1\\ 0&0&0&0&a_1-2a_2+a_3 \end{matrix}$$ Now iff $a_1+a_3=2a_2$ the system has a solution, obviously $$\begin{pmatrix}a_2-2a_1\\a_1\\0\\0\end{pmatrix}+s\begin{pmatrix}-1\\2\\-1\\0\end{pmatrix} +t\begin{pmatrix}2\\3\\0\\-1\end{pmatrix}.$$ So in both cases you've missed the right solution.

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