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Determine whether $S = \Big\{(1, 0, −1),(2, 1, 0),(0, 1, 1)\Big\}$ spans $\Bbb{R}^3$.

So I'm going to let $(u_1, u_2, u_3)$ be a random vector.

$$(u_1, u_2, u_3) = c_1(1,0,-1) + c_2(2,1,0) + c_3(0,1,1) \\ \hspace{0.5cm}= (c_1 + 2c_2, c_2 + c_3, -c_1 + c_3) $$

So it leads to the equations:

$$c_1 + 2c_2 = u_1$$ $$c_2 + c_3 = u_2$$ $$-c_1 + c_3 = u_3$$

$\Rightarrow$

$$\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 2 & 1 \end{bmatrix}$$

I know one way to check this: the determinant of this matrix is found by taking the determinant of the first column:

$$1 * 1 * -1 = -1$$

So the matrix has a unique solution so any vector in $\Bbb{R}^3$ can be written with those 3 vectors so these vectors span $\Bbb{R}^3$

Is this right?

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  • $\begingroup$ Yes, this is correct. You used quite a few more words than I would have in constructing the matrix., the columns are merely the vectors in question. $\endgroup$ – JMoravitz Sep 11 '18 at 22:17
  • $\begingroup$ The "determinant of the first column" does not exist. $\endgroup$ – amsmath Sep 11 '18 at 22:18
  • $\begingroup$ Can I not find the determinant in this way @amsmath? $\endgroup$ – Jwan622 Sep 11 '18 at 22:20
  • $\begingroup$ @JMoravitz what words are redundant? $\endgroup$ – Jwan622 Sep 11 '18 at 22:20
  • $\begingroup$ Well, at least I did not understand what you did there. $\endgroup$ – amsmath Sep 11 '18 at 22:20
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Yes, your solution is perfect. The determinant trick does it.

The other way to do it is to show that the vectors are linearly independent which again boils down to determinant being non-zero.

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Yes in that case the derivation by the determinant is the straightforward method.

As an alternative, particularly useful for larger matrices, we can use RREF

$$\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 2 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & -1 \end{bmatrix} $$

and note that since we have three pivots the given set is a linearly independent set of vectors.

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