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Math Help - Determine values of the constant k for which vectors are linearly independent?

  1. #1
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    Determine values of the constant k for which vectors are linearly independent?

    Problem:

    Determine all values of the constant k for which the given set of vectors is linearly independent in R4.

    {(1,0,1,k),(-1,0,k,1),(2,0,1,3)}

    Attempt:

    So far, I have the system of equations:

    x-y-2z=0
    x+ky+z=0
    kx+y+3z=0


    Now, the determinant of the coefficient matrix yields 2(k+1)(2-k).

    Here is my question: How does the determinant of the coefficient matrix allow me to deduce that "the system has only the trivial solution, and the vectors are linearly dependent if and only if k=-1 or 2"?

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by divinelogos View Post
    Problem:

    Determine all values of the constant k for which the given set of vectors is linearly independent in R4.

    {(1,0,1,k),(-1,0,k,1),(2,0,1,3)}

    Attempt:

    So far, I have the system of equations:

    x-y-2z=0
    x+ky+z=0
    kx+y+3z=0


    Now, the determinant of the coefficient matrix yields 2(k+1)(2-k).

    Here is my question: How does the determinant of the coefficient matrix allow me to deduce that "the system has only the trivial solution, and the vectors are linearly dependent if and only if k=-1 or 2"?

    Thanks for the help!
    Take the determinant and find the values for k which make the determinant not equal to 0
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Take the determinant and find the values for k which make the determinant not equal to 0
    That's what I did. Can you answer the question though?
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    Quote Originally Posted by divinelogos View Post
    That's what I did. Can you answer the question though?
    What does it mean if the determinant is 0?
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    Quote Originally Posted by divinelogos View Post
    Now, the determinant of the coefficient matrix yields 2(k+1)(2-k).

    Here is my question: How does the determinant of the coefficient matrix allow me to deduce that "the system has only the trivial solution, and the vectors are linearly dependent if and only if k=-1 or 2"?
    I got 2k(k + 1) for the determinant.

    I'll let someone else answer the other question. I keep getting definitions mixed up as I am typing. Briefly if the determinant is 0 then there are no coefficients x, y, z that solve the system. Since x, y, z do not exist then the vectors are linearly independent. If the determinant is non-zero then there is a solution for x, y, z, so coefficients exist to make the linear combination sum to 0. Thus the vectors are linearly dependent.

    -Dan
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    What does it mean if the determinant is 0?
    I will answer my own question then. If the determinant is zero, then one or more of the column vectors can be written as a linear combination of one or more column vectors; therefore, the column space is lin. dep.
    Last edited by dwsmith; March 20th 2011 at 11:27 AM. Reason: as not an
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  7. #7
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    Another way of looking at it- because the right side of the system of equations is all "0"s, x= y= z= 0 is an obvious solution to the system of equations. If the determinant of the coefficients is 0, there is NOT a unique solution so there exist non-zero solutions, contradicting the definition of "indepdendent".
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