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Math Help - Negative Determinent of a matrix

  1. #1
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    Negative Determinent of a matrix

    First post, so bear with me lol.

    Solve the following system of equations using matrix inversion method:

    10x + 14y = 818
    8x + 11y = 646

    Of course, first order of business is to calculate the determinent, which works out to -2 (10 x 11 - 14 x 8 = -2...unless i calculated it wrong.)

    My question is where do I go from here that I have a negative determinent, assuming i calculated it right?

    Cheers

    Ibrox
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  2. #2
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    Hello,
    Quote Originally Posted by ibrox View Post
    First post, so bear with me lol.

    Solve the following system of equations using matrix inversion method:

    10x + 14y = 818
    8x + 11y = 646

    Of course, first order of business is to calculate the determinent, which works out to -2 (10 x 11 - 14 x 8 = -2...unless i calculated it wrong.)

    My question is where do I go from here that I have a negative determinent, assuming i calculated it right?

    Cheers

    Ibrox
    Welcome to the forums

    You have \begin{pmatrix} 10 & 14 \\ 8 & 11 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 818 \\ 646 \end{pmatrix}
    You calculated the determinant of the matrix A=\begin{pmatrix} 10 & 14 \\ 8 & 11 \end{pmatrix}

    Now compute the inverse of A (which exists since the determinant is not 0) : The inverse of a matrix B=\begin{pmatrix} a&b \\ c&d \end{pmatrix} is \frac{1}{\text{det}(B)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

    Apply this formula here, to get A^{-1}


    You had A \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 818 \\ 646 \end{pmatrix}
    So multiply to the left by A^{-1} :

    A^{-1}A  \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}= A^{-1} \begin{pmatrix} 818 \\ 646 \end{pmatrix}
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  3. #3
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    Quote Originally Posted by ibrox View Post
    First post, so bear with me lol.

    Solve the following system of equations using matrix inversion method:

    10x + 14y = 818
    8x + 11y = 646

    Of course, first order of business is to calculate the determinent, which works out to -2 (10 x 11 - 14 x 8 = -2...unless i calculated it wrong.)

    My question is where do I go from here that I have a negative determinent, assuming i calculated it right?

    Cheers

    Ibrox
    The determinant is used in finding the inverse of the coefficient matrix .....

    6.3 - The Inverse of a Square Matrix
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  4. #4
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    Okay, I get how to calculate a matrix, its just you proceed as normal...thanks guys.
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  5. #5
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    Sorry, one more thing:

    The part b to this question is "Suppose now that both right hand-side values are increased by an amount "delta (i.e. triangle..dont have symbol lol). What value of delta causes x to = 35 and what is the value of y at this point?

    Here's what i did:

    x = [23] + [-5.5 7] [delta]
    y = [42] [ 4 -5] [ 0 ]

    Which of course gives:

    23 - 5.5delta = 35
    42 + 4delta

    Then:
    -5.5 delta = 12
    delta = -2.18

    Am I correct? If not, where did I go wrong and where do I go from here?

    Thanks again.

    Ibrox
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  6. #6
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    Hello, ibrox!

    I interpreted the set-up differently . . .


    Suppose now that both right hand-side values are increased by an amount \Delta.
    What value of \Delta makes x = 35
    and what is the value of y at this point?

    I think the new equations are: . \begin{array}{ccc}10x + 14y &=& 818 + \Delta \\ 8x +11y &=& 646 + \Delta\end{array}

    We have: . \begin{pmatrix}10 & 14 \\ 8 & 11 \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} \;=\; \begin{pmatrix}818+\Delta \\ 646+\Delta \end{pmatrix} .[1]

    The inverse matrix is still: . A^{-1} \:=\:\begin{pmatrix}\text{-}\frac{11}{2} & 7 \\ 4 & \text{-}5\end{pmatrix}


    Multiply both sides of [1] by A^{-1}\!:\quad \begin{pmatrix}\text{-}\frac{11}{2} & 7 \\ 4 & \text{-}5 \end{pmatrix}\begin{pmatrix}10&14\\8&11\end{pmatri  x}\begin{pmatrix}x\\y\end{pmatrix} \;=\; \begin{pmatrix}\text{-}\frac{11}{2} & 7 \\ 4 & \text{-}5 \end{pmatrix}\begin{pmatrix}818+\Delta \\ 646+\Delta\end{pmatrix}

    And we get:. \begin{pmatrix}x\\y\end{pmatrix} \;=\;\begin{pmatrix}(\text{-}4499-\frac{11}{2}\Delta) + (4522+7\Delta) \\ (3272 + 4\Delta) + (\text{-}3230 - 5\Delta) \end{pmatrix} \;=\;\begin{pmatrix}23 + \frac{3}{2}\Delta \\ 42 - \Delta \end{pmatrix}


    If x = 35, we have: . 23+\tfrac{3}{2}\Delta \:=\:35 \quad\Rightarrow\quad \boxed{\Delta \,=\,8}

    Then: . y \:=\:42 - 8 \quad\Rightarrow\quad\boxed{ y \,=\,34}

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