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Math Help - Finding a value for A so that a system of linear equations is inconsistent

  1. #1
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    Angry Finding a value for A so that a system of linear equations is inconsistent

    Consider the system of linear equations
    x − 3z = −3
    −2x − Ay + z = 2
    x + 2y + Az = 1


    Find the respective values of A such that the system is inconsistent..

    I've tried rewriting this as an augemented matrix, then doing row operations to get the bottom row as [0 0 {(A+3)/2 - 5/A} | 2-(4/A)]
    so for it to be inconsistent i know that it has to look like [0 0 0 | K] for any real k
    so (A+3)/2 - 5/A should equal zero, but 2-4A shouldn't because that would make the system have infinitely many solutions, not NO solutions.

    but i can't find any values of A that solve both these equations!

    Please help, I'm so confused
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Finding a value for A so that a system of linear equations is inconsistent

    Note that the system of linear equations does not have a solution only if the determinant of  \begin{bmatrix}1 & 0 & -3 \\ -2 & -A & 1 \\ 1 & 2 & A \end{bmatrix} is 0, the determinant of this matrix is  -A^2 - 3A + 10 which is 0 at (using quadratic formula) A = -5, A = 2

    so to see plug in A = -5, and to verify if the system is inconsistent, first note that By simple examination i see that -3 * \begin{bmatrix}1 \\ -2 \\ 1 \end{bmatrix} +  -1 * \begin{bmatrix}0 \\ 5 \\ 2 \end{bmatrix} = \begin{bmatrix}-3 \\ 1 \\ -5 \end{bmatrix} so the last column vector is linearly dependent, so we can throw it out.

    Now  \begin{bmatrix}1 & 0 & -3 \\ -2 & 5 & 1 \\ 1 & 2 & -5 \end{bmatrix}  x = \begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix} is aking to solving
     \begin{bmatrix}1 & 0  \\ -2 & 5 \\ 1 & 2  \end{bmatrix}  x = \begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix}
    so by observation we note that a solution \begin{bmatrix}x \\ y \end{bmatrix} of this system must have x = -3, which constraints y = \frac{-4}{5} so our soln, if we have one is \begin{bmatrix}-3 \\ -\frac{4}{5} \end{bmatrix} but
    \begin{bmatrix}1 & 0  \\ -2 & 5 \\ 1 & 2  \end{bmatrix} \begin{bmatrix}-3 \\ -\frac{4}{5} \end{bmatrix} = \begin{bmatrix}-3 \\ 2 \\ \frac{-23}{5} \end{bmatrix} which does not equal \begin{bmatrix}-3 \\ 2 \\ 1 \end{bmatrix} Thus for A = -5, this system is inconsistent (has no solutions).

    Now for A = 2, using the above analysis, it has an infinitely many soln.

    So only for A = -5, the system of lin eqn is inconsistent.
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