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Math Help - Matrix Arithmetic

  1. #1
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    Matrix Arithmetic

    Solve each of the following matrix equations for X:
    a) AX=C
    b) AX+D=C
    Where A= ⎡2 1 2⎤, C=⎡2⎤and D= ⎡-1⎤
    ⎢0 1 6⎥ ⎢0⎥ ⎢0 ⎥
    ⎣2 1 0⎦ ⎣2⎦ ⎣3 ⎦
    Thanks in advance.
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  2. #2
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    Quote Originally Posted by mandarep View Post
    Solve each of the following matrix equations for X:
    a) AX=C
    b) AX+D=C
    Where A= ⎡2 1 2⎤, C=⎡2⎤and D= ⎡-1⎤
    ⎢0 1 6⎥ ⎢0⎥ ⎢0 ⎥
    ⎣2 1 0⎦ ⎣2⎦ ⎣3 ⎦
    Thanks in advance.
    a) X = A^{-1} C

    b) X = A^{-1} (C-D)

    To get the inverse matrix you either have to follow the procedure that will be outlined in your textbook or class notes, or use a CAS. Since A is 3x3 matrix the former approach will require some small amount of time and effort on your part.
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  3. #3
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    Quote Originally Posted by mandarep View Post
    Solve each of the following matrix equations for X:
    a) AX=C
    b) AX+D=C
    Where A= ⎡2 1 2⎤, C=⎡2⎤and D= ⎡-1⎤
    ⎢0 1 6⎥ ⎢0⎥ ⎢0 ⎥
    ⎣2 1 0⎦ ⎣2⎦ ⎣3 ⎦
    Thanks in advance.
    A^{-1}AX=A^{-1}C\Rightarrow X=A^{-1}C

    \displaystyle X=((x_{ij}))=\sum_{k=1}^n (a_{ik}c_{kj})

    \displaystyle A^{-1}=\frac{1}{\text{det}(A)}A

    \text{det}(A)=-4

    \displaystyle -\frac{1}{4}\begin{bmatrix}2&1&2\\0&1&6\\2&1&0\end{  bmatrix}\begin{bmatrix}2\\0\\2\end{bmatrix}=C_{3x1  }

    \displaystyle X=((x_{11}))=-\frac{1}{4}\sum_{k=1}^n (a_{i1}c_{1j})=4+0+4=-2

    \displaystyle X=((x_{21}))=-\frac{1}{4}\sum_{k=1}^n (a_{i2}c_{2j})=0+0+12=-3

    \displaystyle X=((x_{31}))=-\frac{1}{4}\sum_{k=1}^n (a_{i3}c_{13})=\cdots
    Last edited by dwsmith; January 24th 2011 at 06:30 PM. Reason: Changed x to c
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  4. #4
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    Thanks so much guys.
    I thought that was the method I would have to use.
    You're both brilliant.
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  5. #5
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    Alternatively, set up your matrix equations

    \displaystyle \left[\begin{matrix}2&2&1\\0&1&6\\2&1&0\end{matrix}\righ  t]\left[\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right] = \left[\begin{matrix}2\\0\\2\end{matrix}\right]

    and go through row operations to create an upper triangular matrix, which you will then be able to solve the system with.


    \displaystyle R_3 - R_1 \to R_3

    \displaystyle \left[\begin{matrix}2&\phantom{-}2&\phantom{-}1\\0&\phantom{-}1&\phantom{-}6\\0&-1&-1\end{matrix}\right]\left[\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right] = \left[\begin{matrix}2\\0\\0\end{matrix}\right]


    \displaystyle R_3 + R_2 \to R_2

    \displaystyle \left[\begin{matrix}2&2&1\\0&1&6\\0&0&5\end{matrix}\righ  t]\left[\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right] = \left[\begin{matrix}2\\0\\0\end{matrix}\right].


    So \displaystyle 5x_3 = 0 \implies x_3 = 0


    \displaystyle x_2 + 6x_3 = 0 \implies x_2 = 0


    \displaystyle 2x_1 + 2x_2 + x_3 = 2 \implies x_1 = 1.


    So the solution is \displaystyle \mathbf{x} = \left[\begin{matrix}1\\0\\0\end{matrix}\right]
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  6. #6
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    Thank you for all your help.
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