Does this vector set span $\Bbb{R}^3$?

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?

• 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. – JMoravitz Sep 11 '18 at 22:17
• The "determinant of the first column" does not exist. – amsmath Sep 11 '18 at 22:18
• Can I not find the determinant in this way @amsmath? – Jwan622 Sep 11 '18 at 22:20
• @JMoravitz what words are redundant? – Jwan622 Sep 11 '18 at 22:20
• Well, at least I did not understand what you did there. – amsmath Sep 11 '18 at 22:20

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