# Describing span(x) geometrically

1. Mar 24, 2013

### Cottontails

1. The problem statement, all variables and given/known data
There is a vector space with real entries, in ℝ3 with the subset X = $$\begin{pmatrix} 2\\ -1\\ -3 \end{pmatrix}\\ , \begin{pmatrix} 4\\ 0\\ 1 \end{pmatrix}\\ , \begin{pmatrix} 0\\ 2\\ 7 \end{pmatrix}$$
and you have to describe span(x) geometrically.

2. Relevant equations
In answering this question, I found the span of the subset through $$a \begin{pmatrix} 2\\ -1\\ -3 \end{pmatrix}\\ + b \begin{pmatrix} 4\\ 0\\ 1 \end{pmatrix}\\ + c\begin{pmatrix} 0\\ 2\\ 7 \end{pmatrix}= \begin{pmatrix} x\\ y\\ z \end{pmatrix}$$

3. The attempt at a solution
This formed the matrix:
$$\begin{pmatrix} 2 & 4 & 0 & | & x\\ -1 & 0 & 2 & | & y\\ -3 & 1 & 7 & | & z \end{pmatrix}$$
Using row operations, I then made the matrix into reduced row echelon form and it was non-trivial with the final result being:
$$\begin{pmatrix} 1 & 2 & 0 & | & x/2\\ 0 & 1 & 1 & | & x/4+y/2\\ 0 & 0 & 0 & | & -x/4-7y/2+z \end{pmatrix}$$
So, we can then interpret the span geometrically as the plane in ℝ3 with the equation $-1/4x - 7y/2 + z=0$
Is this right?

Last edited: Mar 24, 2013
2. Mar 24, 2013

### vela

Staff Emeritus
You made a mistake somewhere. Those three vectors are linearly independent, so you shouldn't end up with a row of zeros. But if that were the correct matrix, then yes, the span would be the plane you said.

3. Mar 24, 2013

### Cottontails

Sorry, I didn't realise but I put the first entry as 3 instead of -3 and now I have corrected it. Is that correct now?

4. Mar 24, 2013

### vela

Staff Emeritus
Yes, that matches what I get.