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

  1. May 4, 2008 #1
    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.
     
  2. jcsd
  3. May 4, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

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

    what that tells you is that solving Ax= 0 is the same as solving
    [tex]\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][/tex]

    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.
     
  4. May 4, 2008 #3
    Thanks for your help HallsofIvy, I misinterpreted the solution.
     
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