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My textbook doesn't contain any solution to the answer so I was wondering if my answer is right.

Let $$v_1 =\begin{bmatrix} 1\\ -3 \\ 4 \end{bmatrix}, v_2 = \begin{bmatrix} 6 &\\ 2\\ -1 \end{bmatrix}, v_3 = \begin{bmatrix} 2 &\\ -2\\ 3 \end{bmatrix}, \text { and } v_4 = \begin{bmatrix} -4 &\\ -8\\ 9 \end{bmatrix}$$

Find a basis for the subspace $W$ spanned by ${v_1,v_2,v_3,v_4}$?

What I did is that I reduced the matrix:

$$W = \begin{bmatrix} 1 & 6 & 2 &-4\\ -3 & 2 &-2 &-8\\ 4 &-1 & 3 & 9 \end{bmatrix}$$

then I reduced it rref --> $$W = \begin{bmatrix} 1 & 0 & 8 &-2\\ 0 & 1 & \frac{4}{20} &-1\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

therefore the basis for the subspace $W$ spanned by ${v_1,v_2,v_3,v_4}$. would be $B= {\begin{bmatrix} 1 \\ -3 \\ 4 \end{bmatrix},\begin{bmatrix} 6 \\ 2 \\ -1 \end{bmatrix} }$

Is this correct?

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    $\begingroup$ Yes, it seems correct. You can verify that indeed $v_3,v_4\in{\rm span}(v_1,v_2)$. $\endgroup$ – Berci Jan 27 '14 at 21:11
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    $\begingroup$ Yes it is correct. Minor mistakes in your row reduction but doesn't affect the final results. See here for full row reduction. $\endgroup$ – John Habert Jan 27 '14 at 21:16
  • $\begingroup$ O thank you, I didn't even notice that v3 is half of that of v4. Thanks XD. $\endgroup$ – user2551612 Jan 27 '14 at 21:16
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This question was answered in comments:

Yes, it seems correct. You can verify that indeed $v_3,v_4\in \text{span}(v_1,v_2)$. – Berci Jan 27 at 21:11

and

Yes it is correct. Minor mistakes in your row reduction but doesn't affect the final results. See here for full row reduction. – John Habert Jan 27 at 21:16

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