Question

For the equation, 4x - 2y + z = 0, write the subspace of sulutions in R(3) as the span of two vectors in R(3).

Answer 1

when u solve the equation u get:
\[ \large x=\frac{1}{2}y-\frac{1}{4}z. \]

Answer 2

then u construct a "solution" vector:
\[ \large \begin{pmatrix}
x\\ y \\ z
\end{pmatrix}=\begin{pmatrix}
1/2y-1/4z\\ y \\ z
\end{pmatrix} \]

Answer 3

now split the vector into two vectors, one for each parameter.

Answer 4

so y(1/2, 1, 0) and z(-1/4, 0, 1).... which is just what i would do after reduced echelon in any other matrix, which is what I thought, but the solution in the book states this...
\[span \left\{ (1, 0, -4)^{t}, (0, 1, 2)^{t}\right\}\]
I dont understand how they got that.

Answer 5

u have
\[ \large y(1/2,1,0)^t+z(-1/4,0,1) \]
replace \(y=2\alpha\) and \(z=-4\beta\) and see what u get.

Answer 6

\[\alpha(1, 2, 0)^{t}, \beta(1, 0, -4)^{t}\]

Answer 7

but thats not the same thing is it?

Answer 8

your book must have a typo.
the second vector u got is the same as the first in the solution.

Answer 9

okay I will work the next one and see what I get... so is writing \[span \left\{ (1, 2, 0)^{t}, (1, 0, -4)^{t} \right\}\]
the same thing as writing it
\[x=y \left(\begin{matrix}1 \\ 2\\ 0\end{matrix}\right) + z \left(\begin{matrix}1 \\ 0\\-4\end{matrix}\right)\]

Answer 10

and also would it of made any difference if I solved for y or z instead of x in the beginning, even though the vectors would be different it would still be the right answer wouldn't it? is there any particular reason you chose x to start with.

Answer 11

it is not the same thing. actually the answer to the problem is
\[ \large \text{span}\{(1,2,0)^t,(1,0,-4)^t\}. \]
and what u wrote last is wrong. don't forget \(x\) is NOT

Answer 12

a vector

Answer 13

regarding which variable u solve in terms of the others. it really makes no difference.

of course the vectors u obtain will be different.

Answer 14

thank you.

Answer 15

u r welcome