[Math] Find the general solution of this matrix

gaussian eliminationlinear algebramatricessystems of equations

I am trying to find the general solution to this system of equations using an augmented matrix, and then using the gauss elimination, but I cant seem to get it into row echelon form no matter what i try. The system is:
$$x-2y-z+3w=0$$
$$-2x+4y+5z-5w=3$$
$$3x-6y-6z+8w=-3$$
If someone could please give me some pointers on where to go, I would be extremely grateful. Thank you in advance. Matt

Best Answer

First take notice that you have a 3 by 4 coefficients matrix, meaning that you have more variables than equations, therefore there is a general solution or no solution.

We have

$$ \left( \begin{array}{cccc|c} 1 & -2 & -1 & 3& 0 \\ -2 & 4 & 5 & -5 &3 \\ 3 & -6 & -6 & 8 & -3 \\ \end{array} \right) $$

We will do the following elementary row operations $2R_1+R_2\to R_2$ and $-3R_1+R_3\to R_3$ to get

$$ \left( \begin{array}{cccc|c} 1 & -2 & -1 & 3& 0 \\ 0 & 0 & 3 & 1 &3 \\ 0 & 0 & -3 & -1 & -3 \\ \end{array} \right) $$

Next $R_2+R_3\to R_3$ and we got

$$ \left( \begin{array}{cccc|c} 1 & -2 & -1 & 3& 0 \\ 0 & 0 & 3 & 1 &3 \\ 0 & 0 & 0 & 0 & 0\\ \end{array} \right) $$

Which has written in the comments, the third row is a linear combination of the first and second rows

And we got the row echelon form

going back to the system of linear eqautions

\begin{array}{lcl} x -2y -z +3w = 0 \\ 3z+w = 3 \end{array}

Let set $w=t$ and from the second equation we get

$$3z=3-t\iff z=1-\frac{t}{3}$$

Back substituting to the first equation we get

$$x-2y-(1-\frac{t}{3})+t=0\iff x=2y-\frac{4}{3}t+1$$ setting $y=s$ we get

$$\begin{pmatrix} 2s-\frac{4}{3}t+1 \\ s\\ 1-\frac{t}{3}\\ t \end{pmatrix}=t \begin{pmatrix} \frac{4}{3} \\ 0\\ -\frac{1}{3}\\ 1 \end{pmatrix}+s \begin{pmatrix} 2 \\ 1\\ 0\\ 0 \end{pmatrix}+ \begin{pmatrix} 1 \\ 0\\ 1\\ 0 \end{pmatrix}$$

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