Find a set of vectors Find a set of vectors $\{u,v\}$ in $\mathbb{R}^4$ that spans the solution set of the equations:
Vector $u = [\,]$, vector $v = [\,]$.
\begin{eqnarray}
w-x-2y-4z&=&0\\5w+2x+y+3z&=&0
\end{eqnarray}
I have attempted this problem several times. I put it in reduced form which is 
$\begin{bmatrix}
1&&0&&-2&&-4\\0&&1&&\frac{11}{7}&&\frac{23}{7}
\end{bmatrix}$
I am mostly confused on how the variables correspond to the numbers. Then this leaves my answer at: 
$$ u=[2,-11/7,1,0]\qquad v=[4,-23/7,0,1]$$
 A: Although the approach seems to be correct, there appear to be some errors in the row-reduction which should be as follows:
$\begin{bmatrix}
1&&-1&&-2&&-4\\5&&2&&1&&3
\end{bmatrix}$
Perform operation $-5R_1+R_2\Rightarrow R_2$ to get
$\begin{bmatrix}
1&&-1&&-2&&-4\\0&&7&&11&&23
\end{bmatrix}$
Perform operation $\frac{1}{7}R_2\Rightarrow R_2$ to get
$\begin{bmatrix}
1&&-1&&-2&&-4\\0&&1&&\frac{11}{7}&&\frac{23}{7}
\end{bmatrix}$
Perform operation $R_2+R_1\Rightarrow R_1$ to get
$\begin{bmatrix}
1&&0&&-\frac{3}{7}&&-\frac{5}{7}\\0&&1&&\frac{11}{7}&&\frac{23}{7}
\end{bmatrix}$
This will give
$$ u=\left[-\frac{3}{7},\frac{11}{7},1,0\right]\qquad v=\left[-\frac{5}{7},\frac{23}{7},0,1\right]$$
But if you wish to avoid fractions, you can use any constant multiple of these two vectors, such as
$$ u=\left[-3,11,7,0\right]\qquad v=\left[-5,23,0,7\right]$$
A: Row reduction correspond to and can be directly applied to the equations themselves. 
The system of equations is equivalent to the one you found at the end of row reduction (unless you miscalculated), 
$$\matrix{w-2y-4z&=&0\\x+\frac{11}7y+\frac{23}7z&=&0}$$
And, to find a solution, just rearrange them, and arbitrarily choose values for $y$ and $z$. (A standard choice is taking $y=1,z=0$ and $y=0,z=1$, but you're also free to take e.g. $y=7,z=0$ and $y=0,z=7$.)
$$\matrix{w&=&2y+4z\\x&=&-\frac{11}7y-\frac{23}7z}$$
This seems just coinciding with your solution. 
A: w- x- 2y+ 4z= 0 and 5w+ 2x+ y+ 3z= 0.  From the first equation we get w= x+ 2y- 4z.  Putting that into the second equation gives 5(x+ 2y- 4z)+ 2x+ y+ 3z= 7x+ 11y- 17z= 0 and from that, z= (7/17)x+ (11/17)y.  Putting that back into w= x+ 2y- 4z we get w= x+ 2y- (28/17)x- (44/17)y= (-11/17)x- (6/17)y.  So every vector in this space is of the form (w, x, y, z)= ((-11/17)x- (6/17)y, x, y, (7/17)x+ (11/17)y)= ((11/17)x, x, 0, (7/17)x)+ (-(6/17)y, 0, y, (11/17)y)= (11/17, 1, 0, 7/17)x+ (-6/17, 0, 1, 11/17)y.
So a basis is {((11/17, 1, 0, 7/17), (-6/17, 0, 1, 11/17)).  Of course we can multiply each of those by 17 without changing the space they span.  That would give as a basis {(11, 1, 0, 7), (-6, 0, 1, 11)}.
