Question

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.

Answer 1

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

Answer 2

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

Answer 3

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?

Answer 4

oh that matrix is sideways I think

Answer 5

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?

Answer 6

ummm i have no clue

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

ohhhh ya i think it does mean =0

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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\[S=\{c_1\langle1,3,-1,2\rangle\},c_2\langle-1,2,-1,4\rangle\}\]

Answer 15

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

Answer 16

hard to know what they want you to write exactly imo

Answer 17

correct i definitely see that

Answer 18

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