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Math Help - Help with matrices

  1. #1
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    Help with matrices

    HI I'm really close to the answer, but do not know where I went wrong..

    Consider the two equations 4x+8y=1 and 2x-ay=11 ( One is on top of the other) so in matrix form it is
    \begin{bmatrix} 4 & 8 \\2 & -a \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 1 \\11 \end{bmatrix}

    For what values of a does the system have a unique solution?

    Arranging in augmented matrix and using row operations, I reduced the matrix to
    \begin{bmatrix} 4 & 8 \\0 & 2a+8 \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 1 \\-22 \end{bmatrix}

    I had to invert the matrix premultiplied to the matrix containing x and y, so I had to get the det, I got 8a+32

    \begin{bmatrix} x \\y \end{bmatrix} = \frac{1}{8a+32}\begin{bmatrix} 2a+8 & -8 \\0 & 4 \end{bmatrix}\begin{bmatrix} 1\\-22\end{bmatrix}
    Then I multiplied the determinant into the inversed matrix, simplified, was left with
    \begin{bmatrix}\frac{a+4}{4a+16} &\frac{-1}{a+4} \\0 &\frac{1}{2a+8} \end{bmatrix}\begin{bmatrix} 1 \\-22 \end{bmatrix}
    multiplying them, I get
    \begin{bmatrix}\frac{a+4}{4a+16}+\frac{22}{a+4}\\0  +\frac{-22}{2a+8}\end{bmatrix}
    So my answer is x = \frac{a+92}{4a+16}.. and y = \frac{-22}{2a+8}
    but it's wrong.. can someone show me where I went wrong?

    Will really appreciate it, thank you so much!
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  2. #2
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    Re: Help with matrices

    Hello, Tutu!

    First of all, you didn't answer the question.


    \text{Given: }\:\begin{bmatrix} 4 & 8 \\2 & \text{-}a \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 1 \\11 \end{bmatrix}

    \text{For what values of }{\color{red}a}\text{ does the system have a unique solution?}

    The system does not have a unique solution if its determinant equals zero.

    . . D \:=\:\begin{vmatrix}4&8 \\ 2&\text{-}a \end{vmatrix} \;=\;(4)(\text{-}a) - (8)(2) \;=\;\text{-}4a - 16

    . . If D = 0\!:\;\text{-}4a - 16 \:=\:0 \quad\Rightarrow\quad a \:=\:\text{-}4


    \text{The system has a unique solution for all }a \ne \text{-}4.
    . . . . a \,\in\,(\text{-}\infty,\text{-}4) \cup (\text{-}4,\infty)
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