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Math Help - show that the vectors are linearly dependent

  1. #1
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    show that the vectors are linearly dependent

    Show that the vectors u [-1 1] v [2 1] w[3 3] are linearly dependent.

    So, linearly dependent means that there is a solution for A,B,C that doesn't involve all zero's right?

    So, -a + 2b + 3c = 0 and a + b + 3c = 0

    I just solve these linear system of equations to show that the vectors are linearly dependent? Or am I completely wrong?

    Thank you
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  2. #2
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    Correct.

    The system of equations

    \begin{aligned}<br />
-a+2b+3c&=0\\<br />
a+b+3c&=0<br />
\end{aligned}

    can be solved by linear combination. Hint: Multiple solutions will work.
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  3. #3
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    Uhh, it has been a while since I solved a system of equations, especially with 3 variables.. is there a good method for doing this besides just "looking at it". I must be doing something wrong because I keep getting zero for all of the variables when I try to solve it..
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  4. #4
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    Quote Originally Posted by zodiacbrave View Post
    Show that the vectors u [-1 1] v [2 1] w[3 3] are linearly dependent.

    So, linearly dependent means that there is a solution for A,B,C that doesn't involve all zero's right?

    So, -a + 2b + 3c = 0 and a + b + 3c = 0

    I just solve these linear system of equations to show that the vectors are linearly dependent? Or am I completely wrong?

    Thank you

    You could also note that since

    u,v,w \in \mathbb{R}^2 (It's basis has exactly 2 vectors)

    You have 3 vectors in a 2D space so they must be linearly dependant
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  5. #5
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    Quote Originally Posted by zodiacbrave View Post
    Show that the vectors u [-1 1] v [2 1] w[3 3] are linearly dependent.

    So, linearly dependent means that there is a solution for A,B,C that doesn't involve all zero's right?

    So, -a + 2b + 3c = 0 and a + b + 3c = 0

    I just solve these linear system of equations to show that the vectors are linearly dependent? Or am I completely wrong?

    Thank you
    u,v,w being linearly dependent over the field F means that there exist scalars a_1,a_2,a_3 \in F, not all 0, such that a_1v + a_2u + a_3w = 0
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  6. #6
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    Quote Originally Posted by zodiacbrave View Post
    Uhh, it has been a while since I solved a system of equations, especially with 3 variables.. is there a good method for doing this besides just "looking at it". I must be doing something wrong because I keep getting zero for all of the variables when I try to solve it..
    Subtracting the first equation from the second gives us our first result:

    \begin{aligned}<br />
(a - (-a))+(b-2b)+(3c-3c)&=0\\<br />
2a-b&=0\\<br />
2a&=b.<br />
\end{aligned}<br />
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  7. #7
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    Right, so I get b = -2a then I substitute it back into one of the equations and find that a = c and then i put all of that into an equation and find that a = 0.. but from all of that i get a = 0, b = 0 and c = 0 which is all wrong..
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  8. #8
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    Quote Originally Posted by zodiacbrave View Post
    Right, so I get b = -2a then I substitute it back into one of the equations and find that a = c and then i put all of that into an equation and find that a = 0.. but from all of that i get a = 0, b = 0 and c = 0 which is all wrong..
    The correct equation is b=2a. Substituting, we obtain

    \begin{aligned}<br />
-a+2(2a)+3c&=0\\<br />
3a+3c&=0.\end{aligned}
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  9. #9
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    Okay,

    so if b = 2a, I subsitute it back into the equation and get 3a + 3c = 0

    then if I solve this for a, I get a = -c..

    No matter where I go with this, I end up with all zero answers. Is there something simple I am missing here? Maybe I need to look up system of linear equations?

    Thank you
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