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Math Help - MATRIX question help

  1. #1
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    MATRIX question help

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  2. #2
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    The matrix product is just a matter of definition:
    A = \left ( \begin{matrix} 3 & 1 & -2 \\ 2 & 1 & 0 \end{matrix} \right )

    B = \left ( \begin{matrix} 1 & 2 \\ -1 & 1 \\ 2 & -1 \end{matrix} \right )

    So
    AB = \left ( \begin{matrix} 3 & 1 & -2 \\ 2 & 1 & 0 \end{matrix} \right ) <br />
\left ( \begin{matrix} 1 & 2 \\ -1 & 1 \\ 2 & -1 \end{matrix} \right ) = \left ( \begin{matrix} 3 \cdot 1 + 1 \cdot -1 + -2 \cdot 2 & 3 \cdot 2 + 1 \cdot 1 + -2 \cdot -1 \\ 2 \cdot 1 + 1 \cdot -1 + 0 \cdot 2 & 2 \cdot 2 + 1 \cdot 1 + 0 \cdot -1 \end{matrix} \right ) = \left ( \begin{matrix} -2 & 9 \\ 1 & 5 \end{matrix} \right )
    See the pattern?

    You do BA. I get
    BA = \left ( \begin{matrix} 7 & 3 & -2 \\ -1 & 0 & 2 \\ 4 & -1 & -4 \end{matrix} \right )

    -Dan
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  3. #3
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    Hello, Amanda!

    (b) Write the system: . \begin{array}{ccc}2u + 5v & = & 16 \\ u+2v & = & 7\end{array} . in matrix form: . Au \,=\,B

    Use the inverse matrix method to solve the system for u and v.
    We have: . \begin{pmatrix}2 & 5 \\ 1 & 2\end{pmatrix}\begin{pmatrix}u \\ v\end{pmatrix} \;=\;\begin{pmatrix}16 \\ 7\end{pmatrix} .[1]


    I assume we are to find the inverse of matrix A . . . ack!


    We have: . \begin{vmatrix}\;2 & 5 & | & 1 & 0\; \\ \;1 & 2 & | & 0 & 1\;\end{vmatrix}

    . . \begin{array}{c}R_1\!-\!R_2 \\ \\ \end{array}\;\begin{vmatrix}\;1 & 3 & | & 1 & \text{-}1 \;\\ \;1 & 2 & | & 0 & 1\;\end{vmatrix}

    . . \begin{array}{c} \\ R_2\!-\!R_1\end{array}\;\begin{vmatrix}\;1 & 3 & | & 1 & \text{-}1\; \\ \;0 & \text{-}1 & | & \text{-}1 & 2\;\end{vmatrix}

    . . \begin{array}{c}R_1\!-\!3R_3 \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & | & \text{-}2 & 5\; \\ \;0 & 1 & | & 1 & \text{-}2\;\end{vmatrix}

    Hence: . A^{-1}\:=\:\begin{pmatrix}\text{-}2 & 5 \\ 1 & \text{-}2\end{pmatrix}


    Left-multiply both sides of [1] by A^{-1}

    . . \begin{pmatrix}\text{-}2 & 5 \\ 1 & \text{-}2\end{pmatrix}\begin{pmatrix}2 & 5 \\ 1 & 2\end{pmatrix}\begin{pmatrix}u \\ v\end{pmatrix} \;=\;\begin{pmatrix}\text{-}2 & 5 \\ 1 & \text{-}2\end{pmatrix}\begin{pmatrix}16\\7\end{pmatrix}

    . . \begin{pmatrix}\text{-}4+5 & \text{-}10 + 10 \\2-2 & 5 - 4\end{pmatrix}\begin{pmatrix}u \\ v\end{pmatrix} \;=\;\begin{pmatrix}\text{-}32 + 35 \\ 16-14\end{pmatrix}

    . . \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}u \\ v\end{pmatrix} \;= \;\begin{pmatrix}3 \\ 2\end{pmatrix}\quad\Rightarrow\quad\begin{pmatrix}  u\\v\end{pmatrix}\;=\;\begin{pmatrix}3\\2\end{pmat  rix}


    Therefore: . u = 3,\;v = 2
    . .
    (I need a nap!)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Any other method (Elimination, Substitution, Cramer's Rule,
    . . Augmented Matrix) would have been much shorter.

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