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Math Help - Linear Algebra matrix with variables

  1. #1
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    Linear Algebra matrix with variables

    I have no clue where to go with this matrix:


    1 0 k = 1
    k 1 0 = 1
    0 k 1 = 1



    The problem says to find all values of k, if any, for which the system of linear equations has no solution.


    x + kz = 1
    kx + y = 1
    ky + z = 1
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Linear Algebra matrix with variables

    take the determinant of A =  \begin{bmatrix} 1 & 0 & k \\ k & 1 & 0 \\ 0 & k & 1 \end{bmatrix} to get Det(A) = k^3 + 1 , we want the determinant equal to zero. (Why? Can you anwer this?) so  k = -1 . so the matrix  Ax = b either has zero solutions or infinite solutions. Now how do we know which? Well note that  \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} are linearly. independent (the 3rd vector is not independent) so clearly if  \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} has a solution at all.  x = \begin{bmatrix} x_1 \\ -1 \end{bmatrix} which means  x_2 = -2 , but clearly  \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -2 \\ -1 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} \not =\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}  so. for k = -1, no soln. exist.
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  3. #3
    MHF Contributor

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    Re: Linear Algebra matrix with variables

    if k = -1, our system becomes:

    x - z = 1 --> x > z.
    y - x = 1 --> y > x > z.
    z - y = 1 --> z > y > x > z, so z > z, which is impossible.
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