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Thread: matrices

  1. #1
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    matrices

    X is a 2 by 2 matric


    2 -3
    4 -1

    multiplies by unkonw matrix X

    =

    -12 -9
    1 3


    Find the matrix X
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  2. #2
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    Make the unknow matrix and label each part, eg.

    $\displaystyle \begin{bmatrix}
    {2}\;\; {-3}\\
    {4}\;\;{-1}

    \end{bmatrix} \cdot\begin{bmatrix}
    {a}\;\; {b}\\
    {c}\;\;{d}

    \end{bmatrix}=\begin{bmatrix}
    {-12}\;\; {-9}\\
    {1}\;\;\;{3}

    \end{bmatrix}$
    Perform the multiplication and solve for the variables, eg.
    $\displaystyle 2a-3c=-12$
    $\displaystyle 4a-c=1$
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  3. #3
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    Hello, DINOCALC09!

    $\displaystyle X$ is a 2x2 matrix such that: .$\displaystyle \begin{pmatrix}2 & \text{-}3 \\ 4 & \text{-}1\end{pmatrix}\,X \;=\;\begin{pmatrix}\text{-}12 & \text{-}9 \\ 1 & 3 \end{pmatrix}$

    Find the matrix $\displaystyle X$
    Let $\displaystyle X \:=\:\begin{pmatrix}w & x \\ y & z\end{pmatrix}$

    Then we have: .$\displaystyle \begin{pmatrix}2 & \text{-}3 \\ 4 & \text{-}1\end{pmatrix}\,\begin{pmatrix}w & x \\ y & z \end{pmatrix} \;=\;\begin{pmatrix}\text{-}12 & \text{-}9 \\ 1 & 3\end{pmatrix} $

    . . . . $\displaystyle \begin{pmatrix}2w-3y & 2x-3z \\ 4w - y & 4x-z\end{pmatrix} \;=\;\begin{pmatrix}-12 & -9 \\ 1 & 3\end{pmatrix}$


    We have two systems of equations: .$\displaystyle \begin{Bmatrix}2w - 3y & = & \text{-}12 \\ 4w - y &=&1\end{Bmatrix}\quad\begin{Bmatrix}2x-3z &=&\text{-}9 \\ 4x-z &=&3\end{Bmatrix} $

    Solve them . . .


    You should get: .$\displaystyle X \;=\;\begin{pmatrix}\frac{3}{2} & \frac{9}{5} \\ 5 & \frac{21}{5}\end{pmatrix} $
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  4. #4
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    Quote Originally Posted by DINOCALC09 View Post
    X is a 2 by 2 matric


    2 -3
    4 -1

    multiplies by unkonw matrix X

    =

    -12 -9
    1 3


    Find the matrix X
    Solve
    $\displaystyle \left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) X = \left ( \begin{matrix} -12 & -9 \\ 1 & 3 \end{matrix} \right )$

    Multiply both sides by the inverse of the coefficient matrix:
    $\displaystyle \left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) ^{-1} = \frac{1}{2 \cdot -1 - (-3) \cdot 4} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right )$

    So
    $\displaystyle \frac{1}{10} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right ) \left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) X = \frac{1}{10} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right ) \left ( \begin{matrix} -12 & -9 \\ 1 & 3 \end{matrix} \right )$

    $\displaystyle X = \frac{1}{10} \left ( \begin{matrix} 15 & 18 \\ 50 & 42 \end{matrix} \right )$

    -Dan
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