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Math Help - Inverse Matrices Problem

  1. #1
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    Inverse Matrices Problem

    Hi
    The following question, can someone tell me if this is the right method of doing it.
    If, X =\begin {bmatrix}x \\ y \\ z \\\end {bmatrix} and  D = \begin {bmatrix} 0\\ 0\\ c\\ \end{bmatrix}, find conditions on a, b and c so the system of equations represented by CX = D has an infinite number of solutions.

    previous question asked to find det C = \begin{bmatrix}1 & 5 & 1 \\ 1 & 6 & -1 \\ 2 & a & b \\ \end{bmatrix}

    det C = b +2a-22

    i made b=0 to find a which equals 11 therefore det C =0 and c=0 when i made the equation X=C^{-1}D

    P.S
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  2. #2
    Member HappyJoe's Avatar
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    You need to remember the following fact:

    A matrix C has an inverse, if and only if \text{det}(C)\neq 0.

    You are choosing a and b, such that \text{det}(C)=0, in which case the matrix C is not invertible, so the expression " C^{-1}" does not make sense in this case.
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  3. #3
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    so how would you suggest i approach this question.
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  4. #4
    Member HappyJoe's Avatar
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    I would do it in the "old-fashioned" way. To solve the equation CX=D for X, you row-reduce the augmented matrix

    \begin{bmatrix}1 & 5 & 1 & 0\\ 1 & 6 & -1 & 0 \\ 2 & a & b & c \\ \end{bmatrix}.

    For some values of a, b, and c, the last row may end up looking like

    \begin{bmatrix}0 & 0 & 0 & 1\\ \end{bmatrix},

    in which case there are no solutions. On the other hand, the last row could also end up looking like

    \begin{bmatrix}0 & 0 & 0 & 0\\ \end{bmatrix},

    in which case you need to consider, if the last variable (the z-variable) can be a free variable, thus giving rise to an infinite number of solutions.
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