# How do I extend a set so that it becomes a basis for $\mathbb{R}^{4}$?

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?

• You will have a basis if you can find a,b,c,d so that the matrix you have reduces to the identity. – DaveNine Oct 21 '18 at 16:41
• 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. – Ennar Oct 21 '18 at 16:57

It's much simpler than that. Just pick $$a$$, $$b$$, $$c$$ and $$d$$ such that$$\begin{vmatrix}-3&0&2&a\\2&1&-1&b\\4&5&-1&c\\1&-4&5&d\end{vmatrix}\neq0.$$
• Yes. And if the determinant is $0$, then it doesn't work. – José Carlos Santos Oct 21 '18 at 23:24
All you have to do now is to select some values for your $$a,b,c,d$$ which makes the determinant non- zero
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.