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Simplify: (a+2b+3c, 4a+5b+6c, 7a+8b+9c)

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data
    "Simplify: (a+2b+3c, 4a+5b+6c, 7a+8b+9c)."
    All of these are vectors ([tex]a=\vec{a}[/tex], etc.)

    2. Relevant equations
    I know that set of vectors [tex]\vec{x}_1+\cdots+\vec{x}_n \in V[/tex] is linearly independent
    if linear combination [tex]\alpha_1\vec{x}_1+\cdots+\alpha_n\vec{x}_n[/tex] equals [tex]\vec{0}[/tex]
    iff all scalars [tex]\alpha_1\,\cdots,\alpha_n[/tex] equal 0.

    3. The attempt at a solution
    (I don't understand it, although it's (partially) in my notebook)

    (a+2b+3c, 4a+5b+6c, 7a+8b+9c) =
    = (a,4a,7a) + (a,4a,8b) + (a,4a,9c) +
    + (a,5b,7a) + (a,5b,8b) + (a,5b,7a) +
    + (a,6c,8b) + ____0___ + ___0____ +
    + (2b,4a,9c) + (2b,6c,7a) +
    + (3c,4a,8b) + (3c,5b,7a) + =
    = 48(a,b,c) 72(a,b,c) + 84(a,b,c) + 128(a,b,c) - 105(a,b,c) + 45(a,b,c)
    = (a,b,c)(45+48-72+84+128)

    The whole first 2 lines
    " = (a,4a,7a) + (a,4a,8b) + (a,4a,9c) +
    + (a,5b,7a) + (a,5b,8b) + (a,5b,7a) + "

    were removed, because these vectors were lin. dependent.

    So, only vectors as [tex](\alpha\vec{a}, \beta\vec{b}, \gamma\vec{c})[/tex] count as lin. independent. But, if you add some of the last "simplified" ones ...+(2b,4a,9c) + (2b,6c,7a) + (3c,4a,8b) + (3c,5b,7a) and add them, you don't get the original one?!

    And, how is (a,6c,8b) same as 48(a,b,c)? Or where did 48 come from?



    Addendum: How do you make strikethrough block of text? I've tried with HTML's <s> and <strike> and <del>, also found nothing with [tex]LaTeX[/tex].
     
  2. jcsd
  3. Jun 8, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What do you mean by "simplified"? What I would do is write it as a "linear combination" : (a+2b+3c, 4a+5b+6c, 7a+8b+9c)= a(1, 4, 7)+ b(2, 5, 8)+ c(3, 6, 9).
     
  4. Jun 8, 2009 #3
    Instructions say (word-for-word):
    PS: Where's the button for tex tags?
     
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