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I have a set $S=\{(-3,2,4,1), (0,1,5,-4), (2,-1,-1,5)\}$.

How do I extend this set to be a basis for $\mathbb{R}^{4}$?

What I've already tried is letting that fourth vector be $(a, b, c, d)$ and then I put all the vectors into augmented matrix form (as a homogenous linear equation) and reduced it to row echelon form. I got the following matrix:

$$\begin{bmatrix}1 & -4 & 5 & d\\0 & 9 & -11 & b-2d\\0 & 0 & 2 & -b+ \frac{3}{7}c+\frac{2}{7}d\\0 & 0 & 0 & \frac{3}{7}a+\frac{15}{14}b-\frac{3}{14}c\end{bmatrix}$$

I also know that the final vector must not be a linear combination of the previous vectors so that the set remains linearly independent, with a trivial solution.

But I'm not sure how I would to proceed after this step. Or perhaps this method is entirely wrong?

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  • $\begingroup$ You will have a basis if you can find a,b,c,d so that the matrix you have reduces to the identity. $\endgroup$ – DaveNine Oct 21 '18 at 16:41
  • $\begingroup$ You are almost done. Choose $a,b,c$ such that $\frac 37 a + \frac{15}{14} b - \frac 3{14} c\neq 0$. This will ensure that your matrix is of full rank, which precisely means that your vectors are linearly independent. $\endgroup$ – Ennar Oct 21 '18 at 16:57
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It's much simpler than that. Just pick $a$, $b$, $c$ and $d$ such that$$\begin{vmatrix}-3&0&2&a\\2&1&-1&b\\4&5&-1&c\\1&-4&5&d\end{vmatrix}\neq0.$$

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  • $\begingroup$ So any vector that gave me a non zero determinant would work? $\endgroup$ – Vaishnavi MadhuBhuvaneshwaran Oct 21 '18 at 23:21
  • $\begingroup$ Yes. And if the determinant is $0$, then it doesn't work. $\endgroup$ – José Carlos Santos Oct 21 '18 at 23:24
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The method is fine.

All you have to do now is to select some values for your $a,b,c,d$ which makes the determinant non- zero

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Most vectors would work to extend a (linearly independent) set of three vectors to a basis. What you could do is basically pick your favourite vector (like $(1,0,0,0)$ or $(1,e,\pi,\sqrt2)$, or almost anything else), and then you just have to make sure that you haven't been unlucky in your choice.

If you have been unlucky, and the vector you chose by some miracle happened to be linearly dependent of the other three, then pick another one and try again.

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