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I have the homogeneous system of linear equations

$$ 2x_1 + 2x_2 + 4x_3 - 2x_4 = 0, $$ $$ x_1 + 2x_2 + x_3 + 2x_4 = 0, $$ $$ -x_1 + x_2 + 4x_3- 2x_4 = 0. $$

I row reduced to $$\begin{bmatrix}1 & 0 & 0 & -.625\\0 & 1 & 0 & 1.875\\0 & 0 & 1 & -1.125\end{bmatrix}$$

And came up with the general solution: $$\begin{bmatrix}x_1 \\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix}-.625 \\ 1.875 \\ -1.125 \\ t\end{bmatrix}$$ t = any real number

So I believe that the basis will be:
$$\begin{bmatrix}2\\1\\-1\end{bmatrix} + \begin{bmatrix}2\\2\\1\end{bmatrix} + \begin{bmatrix}4\\1\\4\end{bmatrix} + {\begin{bmatrix}-2\\2\\-2\end{bmatrix}}$$

Dimension = # of vectors in the basis so It would be 4.

My question is did I handle the free variable $x_4$ correctly in my answer? Or should I exclude the last vector, ${\begin{bmatrix}-2\\2\\-2\end{bmatrix}}$, because they are all free variables, which would make the Dimension = 3

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If the RREF is $$\begin{bmatrix}1 & 0 & 0 & -.625\\0 & 1 & 0 & 1.875\\0 & 0 & 1 & -1.125\end{bmatrix}$$

And came up with the general solution: $$\begin{bmatrix}x_1 \\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix}0.625t \\ -1.875t \\ 1.125t \\ t\end{bmatrix}=t\begin{bmatrix}0.625 \\ -1.875 \\ 1.125 \\ 1\end{bmatrix}$$

From here, can you state the dimension and a basis for the solution space now?

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  • $\begingroup$ The basis is all of the equations in the homogenous solution and the dimension would be 3? $\endgroup$ – Evan Kim Nov 16 '18 at 17:10
  • $\begingroup$ just checking, basis of what are we talking about? solution space? $\endgroup$ – Siong Thye Goh Nov 16 '18 at 17:12
  • $\begingroup$ yes, the solution space $\endgroup$ – Evan Kim Nov 16 '18 at 17:15
  • $\begingroup$ the solution is a multiple of the same vector, that vector forms a basis. $\endgroup$ – Siong Thye Goh Nov 16 '18 at 17:16
  • $\begingroup$ I'm sorry, you lost me there. multiple of what vector? Why is it the same vector? $\endgroup$ – Evan Kim Nov 16 '18 at 17:19

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