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Math Help - Systems of Linear Equations Problem

  1. #1
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    Systems of Linear Equations Problem

    I'm having trouble solving a linear equations system with an unknown coefficient. I am trying to solve it using the reduce row echelon form, but to no avail.

    x + y + 7z = -7
    2x + 3y + 17z = -16
    x + 2y + (k2 + 1)z = 3k

    [ 1 1 7 | -7 ]
    [ 2 3 17| -16]
    [ 1 2 k2+1| 3k ]

    Specifically, I am trying to simplify this equation to find:
    a) no solution "inconsistent"
    b) infinite number of solutions "consistent"
    c) one solution "unique"

    Right now I am just attempting a. I believe I have accidentally already solved c.

    So following the procedure...

    I subtract 2R1 from R2, giving in the second row:
    [ 0 1 3 | -2 ]

    I then subtract R1 from R3 giving in the third row:
    [ 0 1 k2-6 | 3k+7 ]

    Then subtracting R2 from R3 to give:
    [ 0 0 k2-9 | 3k+9 ]

    From here I am lost. My intuition tells me that 3rd row should simplify to something like this (x representing a non-zero value):
    [ 0 0 0 | x ]

    Any help would be greatly appreciated.
    Last edited by brettm; August 22nd 2013 at 10:02 AM.
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  2. #2
    Super Member ebaines's Avatar
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    Re: Systems of Linear Equations Problem

    From your last line simply divide through by k^2-9:

    [ 0 0 1 | (3k+9)/k^2-9]

    This tells you that z = (3k+9)/(k^2-9). As long as this value is defined you get a unique set of solutions for x, y, and z. For example if k=0 you get x=-1, y=1, and z = -1. If k= 4 you get x= -17, y= -11, z = 3. But notice that the (3k+0)/(k^2-9) is not defined for k= 3 or k=-3, so we need to see what happens for these two cases. For k=3 your last row becomes [0 0 0 | 15] which has no solution and for k = -3 it's [0 0 0| 0] which means the three original equations are not linearly indpendent and hence there are an infininite number of solutions.
    Thanks from brettm
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  3. #3
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    Re: Systems of Linear Equations Problem

    Hello,

    I can help you with the first one. For a system of equations to be consistent (with either one or infinite solutions) it has to be satisfied that

    \det\left(\bf A\right)\neq 0

    For this reason, calculating the determinant of the coefficient matrix, you can determine the values of k where the system is not defined:

    \det\left(\bf A\right) = \left|\begin{matrix}1 & 1 & 7 \\ 2 & 3 & 17 \\ 1 & 2 & k^2 + 1 \end{matrix}\right|=k^2 -9=0

    Thus,

    k=\pm 3 makes the determinant to be equal to zero, and thus, the system is not defined.
    Thanks from brettm
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  4. #4
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    Re: Systems of Linear Equations Problem

    Hello, brettm!

    \begin{array}{ccc} x + y +\qquad 7z &=& \text{-}7 \\ 2x + 3y +\qquad 17z &=& \text{-}16 \\ x + 2y + (k^2 + 1)z &=& 3k\end{array}

    Specifically, I am trying to simplify this equation to find:
    . . (a) no solution, "inconsistent"
    . . (b) infinite number of solutions, "consistent"
    . . (c) one solution, "unique"

    \text{We have: }\:\left[\begin{array}{ccc|c} 1&1&7 & \text{-}7 \\ 2&3&17 & \text{-}16 \\ 1&2& k^2\!+\!1 & 3k \end{array}\right]

    \begin{array}{c}\\ R_2-2R_1 \\ R_3-R_1 \end{array}\left[\begin{array}{ccc|c}1&1&7 & -7 \\ 0&1&3&-2 \\ 0&1&k^2\!-\!6 & 3k\!+\!7 \end{array}\right]

    \begin{array}{c}R_1-R_2 \\ \\ R_3-R_2 \end{array}\;\left[\begin{array}{ccc|c}1&0&4&\text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&k^2\!-\!9 & 3k\!+\!9 \end{array}\right]


    \text{If }k = 3\text{, we have: }\;\left[\begin{array}{ccc|c}1&0&4&\text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&0&18 \end{array}\right] \;\;\text{ . . . no solution}

    \text{If }k = \text{-}3\text{, we have: }\;\left[\begin{array}{ccc|c}1&0&4&\text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&0&0 \end{array}\right]\;\;\text{ . . . infinite solutions}


    \text{If }k \,\ne\,\pm3: \:
    \begin{array}{c} \\ \\ R_3 \div (k^2-9)\end{array}\;\left[\begin{array}{ccc|c} 1&0&4 & \text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&1&\frac{3}{k-3}\end{array}\right]

    . . . \begin{array}{c}R_1-4R_3 \\ R_2-3R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c} 1&0&0 & \frac{\text{-}5k+3}{k-3} \\ 0&1&0& \frac{\text{-}2k-3}{k-3} \\ 0&0&1& \frac{3}{k-3} \end{array}\right]\;\;\text{ . . . one solution}
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  5. #5
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    Re: Systems of Linear Equations Problem

    Thank you very much for your help everyone.

    I've been trying to simplify the unique solution to the equation(s), i.e.:

    \frac{-5k+3}{k-3}

    But I'm beginning to think that I am meant to just leave this answer in terms of k, is this correct? I'm confused because the question states "Determine the value of the constant k such that the system..."

    Also, what is the justification for arriving at k = \pm3 for no and infinite solutions? I am basically stating that for a system to have no solution, the bottom row of the matrix (excluding the augmented end) must have values of 0, with the augmented column having a non-zero value, 0x + 0y + 0z = 18 (not possible therefore there is no solution. So I determine the value of k when k^2 - 9 = 0; then substituting in both values into 3k + 9 seeing when it is a non-zero value, to find the answer for part a), no solutions. My justification for part b) is basically the same, just substituting in a value determined from k^2 - 9 = 0 that will give 3k + 9 = 0 in the augmented column.

    I am sure there has to be a better, more conventional way to explain this on my homework.
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  6. #6
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    Re: Systems of Linear Equations Problem

    Quote Originally Posted by berni1984 View Post
    Hello,

    I can help you with the first one. For a system of equations to be consistent (with either one or infinite solutions) it has to be satisfied that

    \det\left(\bf A\right)\neq 0

    For this reason, calculating the determinant of the coefficient matrix, you can determine the values of k where the system is not defined:

    \det\left(\bf A\right) = \left|\begin{matrix}1 & 1 & 7 \\ 2 & 3 & 17 \\ 1 & 2 & k^2 + 1 \end{matrix}\right|=k^2 -9=0

    Thus,

    k=\pm 3 makes the determinant to be equal to zero, and thus, the system is not defined.
    This is a better solution, but we should note that a zero determinant could mean that the system either has no solutions or INFINITELY MANY solutions. So we can state that if \displaystyle \begin{align*} k \neq \pm 3 \end{align*} there is unique solution, and then the OP will need to substitute in each of k = 3 and k = -3 and reduce the system to echelon form to determine if there are no solutions or infinitely many solutions.
    Thanks from brettm
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