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Matrices problem, K != ? so that its a linear transformation R^3->R^2?

  1. Nov 26, 2005 #1
    Hello everyone, i'm confused on this problem:
    It says:
    A linear transformation T:R^3->R^2 whose matrix is
    2 -4 -3
    -3 6 0+k

    is onto if k != ?
    != meaning, not equal.

    So my thinking was, For it to be a transformation into R^2, doesn't that mean k isn't suppose mean that the column
    -3
    0+k
    has to some how come out to be
    0
    0
    so your only left with
    2 -4
    -3 6

    so that would be in R^2 right?
    i'm confused, am i thinking wrong? or is there some trick to easily find what k isn't suppose to be? Thanks!
     
  2. jcsd
  3. Nov 26, 2005 #2

    HallsofIvy

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    T will be "onto" R2 as long as, after row reduction, the second row is not [0 0 0]. What value of k would make the second row [0 0 0]?
     
  4. Nov 26, 2005 #3
    THanks Halls, but I'm slighly confused,
    I row reduced and got:
    2 -4 -3
    0 0 2k

    so 2k = 0?
    so no value of k will equal 0, unless its 0, which isn't right.
    So i thought maybe i have to write it out like:
    2x - 4y -3k = 0
    0 + 0 +2k = 0
    k = 0;
    2x -4y = 0;
    so if x = 2, and y = 1 then u'll get 0 = 0, but thats for the top row....hm..what am i not getting? THanks!
     
  5. Nov 26, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You started with
    [2 -4 -3]
    [-3 6 0+k]
    which I take to mean
    [2 -4 -3]
    [-3 6 k]

    To "row-reduce" add 3/2 the first row to the second row.
    3/2(2)= 3 and adding that to -3 gives 0
    3/2(-4)= -6 and adding that to 6 gives 0
    3/2(-3)= -9/2 and adding that to k give k- 9/2. that will equal 0 when
    k= 9/2.
     
  6. Nov 27, 2005 #5
    ahhh i c were I messed up! thanks Ivy it worked great!! :biggrin:
     
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