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Determine a basis for the solution set of the homogeneous system:

$$\begin{align*} x_1 +x_2 +x_3 &=0\\ 3x_1+3x_2+x_3 &=0\\ 4x_1+4x_2+2x_3&=0 \end{align*}$$

Then the augmented matrix is:

$$ \left[\begin{array}{ccc|c} 1 & 1 & 1 &0\\ 3 & 3 & 1 &0\\ 4 & 4 & 2 &0\\ \end{array}\right] $$

Reduced Row Echelon Form $\to$

$$ \left[\begin{array}{ccc|c} 1 & 1 & 0 &0\\ 0 & 0 & 1 &0\\ 0 & 0 & 0 &0\\ \end{array}\right] $$

I already looked at this example but it didn't help much. I am wondering can someone help to find basis (choosing some parameter for variables $x_1,x_2,x_3$) from RREF.

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The first equation in the Reduced Row Echelon Form tells you that we need $x_1+x_2=0$ and the second equation says $x_3=0$. So If we take $x_2=t$ for $t\in\mathbb{R}$ then we must have $x_1=-x_2=-t$ and $x_3=0$.

Thus we have a 1-dimensional solution space determined by the vector

$(x_1,x_2,x_3)=(-1,1,0)$.

This follows from the above discussion since taking $x_2=t$ corresponds to scalar multiplication by t on $(-1,1,0)$.

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The equations to be satisfied are $x_1+ x_2+ x_3= 0$ $3x_1+ 3x_2+ x_3= 0$ $4x_1+ 4x_2+ x_3= 0$

The first thing I notice is that if we multiply the first equation by 3 and subtract the second equation we have $2x_3= 0$ so any solution to this system of equations must have $x_3= 0$. Setting $x_3= 0$ in these equations, we have $x_1+ x_2= 0$, $3x_1+ 3x_2= 0$, $4x_1+ 4x_2= 0$

all of which are equivalent to $x_1+ x_2= 0$ or $x_2= -x_1$. Any solution to this system of equations is for the form $(x_1, x_2, x_3)= (x_1, -x_1, 0)= x_1(1, -1, 0)$.

That is a one dimensional subspace of $R^3$ with basis $\{(1, -1, 0)\}$. (Of course, $\{(-1, 1, 0)\}$ is equivalent.)

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