## anonymous 3 years ago Let V be the span of the vectors (1,3,-1,2),(-1,2,-1,-4). Express V as the solution set of a homogenous linear system.

1. anonymous

I'm not quite sure I understand the question...

2. anonymous

I was having the same issue. i dont even know where to start lol

3. anonymous

it's asking to express the general form I suppose?$A\vec x=\vec0$and we have$A=\left[\begin{matrix}1&-1\\3&2\\-1&-1\\2&-4\end{matrix}\right]$and$\vec x=\left[\begin{array}~x_1\\x_2\\x_3\\x_4\end{array}\right]$but what does it want us to write exactly?

4. anonymous

oh that matrix is sideways I think

5. anonymous

no, it's just that the $$\vec x$$ is 2 rows long$A\vec x=\vec0$and we have$A=\left[\begin{matrix}1&-1\\3&2\\-1&-1\\2&-4\end{matrix}\right]$and$\vec x=\left[\begin{array}~x_1\\x_2\end{array}\right]$but what does it want us to write?

6. anonymous

ummm i have no clue

7. anonymous

well, the solution set is supposed to be the vector space, so we need to write the set of all solution in set notation somehow it seems

8. anonymous

but it says the "homogeneous solution" so I guess let's first solve the above

9. anonymous

but I'm thinking maybe homogeneous doesn't mean =0

10. anonymous

ohhhh ya i think it does mean =0

11. anonymous

so then I think that the span is the set$V=\{\vec x:A\vec x=\vec0\}$where$A=\left[\begin{matrix}1&-1\\3&2\\-1&-1\\2&-4\end{matrix}\right]$

12. anonymous

not sure how exactly they want you to state it, but the set of all vectors that solve that system seems to be the vector space they are talking about

13. anonymous

so its just saying to express the V as the solution set of a homogenious system?

14. anonymous

usually the span of those vectors is the set of all linear combinations of those vectors, so you if not for the word "homogenous" I would say the span is just$S=\{c_1\langle1,3,-1,2\rangle\},c_2\langle-1,2,-1,4\rangle\}$

15. anonymous

typo, I meant $S=\{c_1\langle1,3,-1,2\rangle,c_2\langle-1,2,-1,4\rangle\}$

16. anonymous

hard to know what they want you to write exactly imo

17. anonymous

correct i definitely see that

18. anonymous

but I don't think what I just wrote is "homogenous"