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Math Help - Two quick questions

  1. #1
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    Two quick questions

    Question 1
    If I have matrix v1 = (1,-1,2) v2 = (4,-6,7) and v3 (-1,1,k)

    Am I right in saying that if k = -2, v1,v2,v3 is linearly dependent, because v3 = -1(v1)?
    And following that, if I had to write one vector as a linear combination of the other two, would it be correct to say v1 = 0(v2) - 1(v3)

    Question 2
    The following matrix is row equivalent to A. Find the general solution to Ax = 0
    1 -4 -2 0 3 -5
    0 0 1 0 2 -1
    0 0 0 0 0 1
    0 0 0 0 0 0

    I get that the basic variables are x1 and x3, since they are pivots. But is x6 a basic variable or is it free? I'm confused because 1x6 = 0 and 0x6 = 0.
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  2. #2
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    Re: Two quick questions

    Question 1: yes
    Question 2:
    What you end up getting after bringing that matrix on RREF is
    A=\begin{pmatrix}1&-4&0&0&7&0\\ 0&0&1&0&2&0\\ 0&0&0&0&0&1\\ 0&0&0&0&0&0 \end{pmatrix},
    we see that the free variables are  x_2,x_4,x_5 and the lead variables are  x_1,x_3,x_6 since they are the pivots.
    You ask if  x_6 are free, it's not because the only value it can take is 0 since it is a pivot in the third row.
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  3. #3
    MHF Contributor

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    Re: Two quick questions

    \begin{pmatrix}1 & -4 & -2 & 0 & 3 & - 5 \\ 0 & 0 & 1 & 0 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\  0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
    is equivalent to the three equations
    x_1- 4x_2- 2x_3+ 3x_5- 5x_6= 0
    x_3+ 2x_5- x_6= 0
    x_6= 0

    From the last, x_6= 0 so that x_3+ 2x_5= 0 and x_3= -2x_5. Putting that into the first equation, x_1- 4x_2- 2(-2x_5)+ 3x_5= x_1- 4x_2+ 7x_5= 0. Then the best we can do is say x_1= 4x_2- 7x_5 and take x_2= r, x_4= s and x_5= t as parameters, x_1= 4r- 7t, x_2= r, x_3= -2t, x_4= s, x_5= t, x_6= 0.
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