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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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up vote 2 down vote accepted

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


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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