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Linear independent and combination

  1. Mar 26, 2014 #1
    1. The problem statement, all variables and given/known data
    let r be an element of R
    ..... 1................. 1 ...................r^2.....................3 + 2r
    u =( 1 )........v = ( 4 )...........w = (1 )............b = ( 5 + 12r)
    .......0..................1...................r^2 ....................... 2r


    (sorry don't know how to type matrices)

    1. For which values of r is the set {u, v, w} linearly independent?

    2. For which values of r is the vector b a linear combination of u. v, w?

    3. For which of these values of r can b be written as a linear combination of u, v and w in more than one way

    2. Relevant equations

    (the matrices)

    3. The attempt at a solution

    So for 1, I reduced the matrix ( u v w ) to become something like this:

    _1_1_r^2
    (_1_4_1)
    0_1_r^2

    __1_1_r^2
    → (0_1_r^2)
    __0_0_1-4r^2


    so for it to be linearly independent 1 - [itex] 4r^2 [/itex] =/= 0

    so r =/= [itex]\pm 1/2[/itex]

    for part b)

    We want something like:

    c1(u) + c2(v) + c3(w) = b

    I reduced everything and got:


    c1 = 3
    c2 = 2r - [itex] r^2 [/itex] (2/(1 - 2r))
    c3 = [itex]\frac{2(1 + 2r)}{(1 + 2r)(1 - 2r)}[/itex]

    Is it alright to say that for it to be a linear comb. r can't = [itex]\pm1/2[/itex]? (Is it correct that I didn't cancel out the 1 + 2r ?)

    and I don't really get part C
     
  2. jcsd
  3. Mar 26, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Using the "tex" and "/tex" tags (in "[" and "]")
    "u= \ begin{pmatrix}1 \\ 1 \\ 0 \ end{pmatrix} v= \ begin{pmatrix}1 \\ 4 \\ 1 \ end{pmatrix} w= \ begin{pmatrix}r^2 \\ 1 \\ r^2\ end{pmatrix} b= \ begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\ end{pmatrix}" (without the spaces) will give

    [tex]u= \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix} v= \begin{pmatrix}1 \\ 4 \\ 1 \end{pmatrix} w= \begin{pmatrix}r^2 \\ 1 \\ r^2\end{pmatrix} b= \begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}[/tex]


    The basic definition of "linear independent" says that four vectors u, v, w, and b, will be linearly independent if and only if the only way we can have pu+ qv+ sw+ tb= 0 is if p= q= s= t= 0. Here, for any numbers, p, q, s, and t
    [tex]pu+ qv+ sw+ tb= \begin{pmatrix}p \\ p \\ 0 \end{pmatrix}+ \begin{pmatrix}q \\ 4q \\ q \end{pmatrix}+ \begin{pmatrix}sr^2 \\ s \\ sr^2\end{pmatrix}+ \begin{pmatrix}t(3+ 2r) \\ t(5+ 12r) \\ t(2r)\end{pmatrix}= \begin{pmatrix}p+ q+ r^2+ 3t+ 2tr \\ p+ 4q+ s+ 5t+12rt \\ q+ sr^2+ 2tr\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}[/tex]

    so we have the equations [itex]p+ q+ sr^2+ 3t+ 2tr= 0[/itex], [itex]p+ 4q+ s+ 4t+ 12rt= 0[/itex], and [itex]q+ sr^2+ 2tr= 0[/itex] which we must solve for p, q, s, and t ("r" is just a parameter). Since there is no "p" in the last equation we can subtract the first equation from the second to eliminate p from those and have [itex]3q+ s(1- r^2)+ t+ 10rt= 0[/itex] and [itex]q+ sr^2+ 2tr= 0[/itex]. Multiply that last equation by 3 and subtract from the previous equation to eliminate q and we have [itex]s(1- 2r^2)+ t(1+ 8r)= 0[/itex]. If either [itex]1- 2r^2= 0[/itex] or [itex]1+ 8r=0[/itex] that will be true without s and t being equal to 0 so those vectors will NOT be "linearly independent". For any other value of r, they will be which also answers (b).

    "Writing b as a linear combination of u, v, and w" means writing
    [tex]\begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}= x\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}+ y\begin{bmatrix}1 \\ 4 \\ 1 \end{bmatrix}+ z\begin{bmatrix}r^2 \\ 1 \\ r^2\end{bmatrix}= \begin{bmatrix}x+ y+ zr^2 \\ x+ 4y+ z \\ y+ zr^2\end{bmatrix}[/tex]
    for some numbers x, y, and z.

    That is the same as the equations [itex]3+ 2r= x+ y+ zr^2[/itex], [itex]5+ 12r= x+ 4y+ z[/itex], and [itex]2r= y+ zr^2[/itex].

    The question is "for what values of r do those equations have more than one solution for x, y and z?"
     
  4. Mar 27, 2014 #3
    Hi there, but how do I find the ones that have more than one way of writing it?

    Do I solve the matrices for [itex]3+ 2r= x+ y+ zr^2[/itex], [itex]5+ 12r= x+ 4y+ z[/itex], and [itex]2r= y+ zr^2[/itex] and make it to have a free parameter so that there are infinite solutions?
     
    Last edited: Mar 27, 2014
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