Given augmented matrix [A|0]: ->RREF ->x=?

[RogerDodgr]

Given augmented matrix [A|0]: -->RREF -->x=?

I started with:
|0 1 0|0|
|0-3 1|0|
|0 0 2|0|
--->RREF--->
|0 1 0|0|
|0 0 1|0|
|0 0 0|0|
The RREF is correct according to solution manual, but
the solution manual says that the vector x =
|1|
|0|
|0|

I think I understand why x_3 and x_2 =0, But I don't understand why x_1 =1.

 

[HallsofIvy]

After have the "RREF" what do you do with it?

what that tells you is that solving Ax= 0 is the same as solving
\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c}x \\ y \\ z\end{array}\right]= \left[\begin{array}{c}0 \\ 0 \\ 0\end{array}\right]

which is the same as solving the equations 0x+ 1y+ 0z= 0, 0x+ 0y+ 1z= 0, 0x+ 0y+ 0z= 0.
Obviously the first equation tells you that y must be 0. The second equation tells you that z= 0. The third equation doesn't tell you anything!

So how do you know that x= 1? You don't! In fact, x can be anything and <x, 0, 0> would still satisfy that equation. The solution is any multiple of <1, 0, 0>. The "solution space" is the vector space spanned by <1, 0, 0>. In fact, you could use <x, 0, 0> for any x but '1' happens to be a nice easy number.

 

[RogerDodgr]

Thanks for your help HallsofIvy, I misinterpreted the solution.