Question

To find out if a vector is in the span of a set of vectors, i would place all the vectors in a matrix, row reduce... and what should i get to tell me if they are in the span?

Answer 1

The question is vector (4,11,12) in the span of u=(1,2,3) and v=(2,5,6).

\[\left[\begin{matrix}1 & 2 & 4\\ 2 & 5 & 11\\ 3 & 6 & 12\end{matrix}\right]\]

Answer 2

so what should I get when I row reduce? if I get the identity?
@helder_edwin

Answer 3

don't forget that the matrix you wrote corresponds to a linear system, so it is an augmentged matrix. so you should have written something like this
\[ \large \left(\begin{array}{cc|c}
1 & 2 & 4\\ 2 & 5 & 11\\ 3 & 6 & 12
\end{array}\right) \]

Answer 4

so when you row-reduce the matrix u will get two pivots.

the vector (4,11,12) will be in the spam of the other two nes if the last row turns completely cero when row-reducing

Answer 5

\[\left[\begin{matrix}1 & 0 & -2 \\ 0 & 1 & 3\\ 0 & 0 & 0\end{matrix}\right]\]

Answer 6

so why is it that the last row has to be zero? what does that tell me?... i see that i can check by multiplying the first by -2 and the second by 3 and i will get (4,11,12)... what if i got that last 2 rows of zeros... couldnt it still be possible that the first vector times a scalar plus the second still give me a vector im looking for?

Answer 7

first, u have a three by two linear system, so for it to be consistent (have a solution), since u cannot have more than two pivots the last row(s) has(ve) to turn zero.

second, the row-reduced matrix u got tells u that the vector (4,11,12) is indeed in the span of the other two, and the way to obtain said vector is as u pointed out to multiply the first one by -2 and the second one by 3 and then add.

third. should u have lost two rows, then the system would still be consistent. but the second vector should have been a scalar multiple of the first one and just as the third one should have. so for instance, you could have had
\[ \large \left(\begin{array}{cc|c}
1 & 2 & 3\\ 2 & 4 & 6\\ -1 & -2 & -3
\end{array}\right) \]

Answer 8

this last matrix would be row-reduced into
\[ \large (1\quad 2\ | \ 3) \]

Answer 9

so x=3-2y and you would have infinitely many solutions.

Answer 10

ohhh okay... thank you... so now if i wanted to find an equation for the plane of those two vectors u and v... i would take the cross product... and once i did that i got (-7,0,1) so does that mean the equation of the plane is -7x+z=0 or is it 7x=z, or is there something i am missing?

Answer 11

no it is fine. both equations are the same.

Answer 12

so then that is right then... that is the equation then?