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Thread: Lin. Dep.

  1. #1
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    Lin. Dep.

    1.) Given the set with these two vectors:

    $\displaystyle \left\{ \left[ \begin {array}{c} a-1\\\noalign{\medskip}a+3
    \end {array} \right] , \left[ \begin {array}{c} 1\\\noalign{\medskip}a
    \end {array} \right] \right\} $

    I have to find the value(s) of a that makes it linearly depedent.

    WORK:

    Well, they are lin. dep. if they are multiples of each other. So I eye-balled it (not sure how else to do it) and saw that a = 3 will make it lin. dep because then you have the two rows are multiples of each other (scalar of 2)

    Is that the only value?

    2.) Like above..but with 1 more vector..given the set with these 3 vectors:

    $\displaystyle \left\{ \left[ \begin {array}{c} a+4\\\noalign{\medskip}3
    \\\noalign{\medskip}2\end {array} \right] , \left[ \begin {array}{c} a
    \\\noalign{\medskip}1\\\noalign{\medskip}0\end {array} \right] ,
    \left[ \begin {array}{c} 2\\\noalign{\medskip}1\\\noalign{\medskip}1
    \end {array} \right] \right\} $

    I have to find the value(s) of a that makes it linearly depedent.

    WORK:

    I can't eye-ball any a that will make these multiples of each other or will make a whole column all 0....I conclude that there is no value a for which this is linearly dependent??
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    1.) Given the set with these two vectors:

    $\displaystyle \left\{ \left[ \begin {array}{c} a-1\\\noalign{\medskip}a+3
    \end {array} \right] , \left[ \begin {array}{c} 1\\\noalign{\medskip}a
    \end {array} \right] \right\} $

    I have to find the value(s) of a that makes it linearly depedent.
    As you say, if they are linearly dependent they are multiples of each other. Thus we have the equation:
    $\displaystyle \left[ \begin {matrix} a-1 \\ a+3
    \end {matrix} \right] = c \left[ \begin {matrix}1 \\ a
    \end {matrix} \right] $
    where c is some constant.

    Thus we have that
    $\displaystyle a - 1 = c \cdot 1$
    $\displaystyle a + 3 = c \cdot a$

    Solving these simultaneously gives:
    $\displaystyle a = -1, 3$ (and $\displaystyle c = 2, -2$ respectively.)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    2.) Like above..but with 1 more vector..given the set with these 3 vectors:

    $\displaystyle \left\{ \left[ \begin {array}{c} a+4\\\noalign{\medskip}3
    \\\noalign{\medskip}2\end {array} \right] , \left[ \begin {array}{c} a
    \\\noalign{\medskip}1\\\noalign{\medskip}0\end {array} \right] ,
    \left[ \begin {array}{c} 2\\\noalign{\medskip}1\\\noalign{\medskip}1
    \end {array} \right] \right\} $

    I have to find the value(s) of a that makes it linearly depedent.
    Here we need to define linearly dependent a bit more carefully. If three vectors are linearly dependent we may take a linear combination of any two of them and produce a third. So one example of this is
    $\displaystyle \left[ \begin {matrix}a+4 \\ 3
    \\ 2 \end {matrix} \right] = c \left[ \begin {matrix} a
    \\ 1 \\ 0 \end {matrix} \right] +
    d \left[ \begin {matrix} 2\\ 1 \\ 1
    \end {matrix} \right] $
    where c and d are some constants.

    So we have the equations:
    $\displaystyle a + 4 = c \cdot a + d \cdot 2$
    $\displaystyle 3 = c \cdot 1 + d \cdot 1$
    $\displaystyle 2 = c \cdot 0 + d \cdot 1$

    I'll let you solve the system for a, c, and d.

    Note: You may set up the 2 other examples, reordering the vectors as you do, and you will get the same value for a, but obviously with different values of c and d for each.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    Here we need to define linearly dependent a bit more carefully. If three vectors are linearly dependent we may take a linear combination of any two of them and produce a third. So one example of this is
    $\displaystyle \left[ \begin {matrix}a+4 \\ 3
    \\ 2 \end {matrix} \right] = c \left[ \begin {matrix} a
    \\ 1 \\ 0 \end {matrix} \right] +
    d \left[ \begin {matrix} 2\\ 1 \\ 1
    \end {matrix} \right] $
    where c and d are some constants.

    So we have the equations:
    $\displaystyle a + 4 = c \cdot a + d \cdot 2$
    $\displaystyle 3 = c \cdot 1 + d \cdot 1$
    $\displaystyle 2 = c \cdot 0 + d \cdot 1$

    I'll let you solve the system for a, c, and d.

    Note: You may set up the 2 other examples, reordering the vectors as you do, and you will get the same value for a, but obviously with different values of c and d for each.

    -Dan

    Ohhh!! Wow, completed missed the other solution for #1.

    For #2, I can't find any a that makes it depedent. Whatever I try makes it independent

    I keep getting:

    $\displaystyle \left[ \begin {array}{ccc} 1&0&8\\\noalign{\medskip}0&1&10
    \\\noalign{\medskip}0&0&45\end {array} \right] $

    Is there no a that makes this lin dep? I'm going crazy.
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