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Math Help - matrix equation

  1. #1
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    matrix equation

    Colomn vector
    Last edited by Tom123; November 7th 2007 at 06:37 AM.
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  2. #2
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    Quote Originally Posted by Tom123 View Post
    Colomn vector x is defined as x = (x1 x2 x3 )^t

    Solve the matrix equation Ax=b for x1 x2 x3 for:

    a = 3 and b = (18 16 62 )^t

    a = 5 and b = (4 6 24 )^t
    I don't see "a" in the matrix equation, only "A" and these can't be the same since the given a's are scalars, not matrices.

    -Dan
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    Oh sorry... A=

    2 0 a
    2 2 1
    8 a+3 4
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tom123 View Post
    Colomn vector x is defined as x = (x1 x2 x3 )^t

    Solve the matrix equation Ax=b for x1 x2 x3 for:

    a = 3 and b = (18 16 62 )^t

    a = 5 and b = (4 6 24 )^t
    Quote Originally Posted by Tom123 View Post
    Oh sorry... A=

    2 0 a
    2 2 1
    8 a+3 4
    a = 3

    b = \left ( \begin{matrix} 18 \\ 16 \\ 62 \end{matrix} \right )

    Solve
    \left ( \begin{matrix} 2 & 0 & 3 \\ 2 & 2 & 1 \\ 8 & 6 & 4 \end{matrix} \right ) \cdot \left ( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right ) = \left ( \begin{matrix} 18 \\ 16 \\ 62 \end{matrix} \right )

    To be precise there are exactly a zillion ways to solve this. I'll assume you don't know any of the fancier ones, and do the matrix multiplication. We get three equations:
    2x_1 + 3x_3 = 18
    2x_1 + 2x_2 + x_3 = 16
    8x_1 + 6x_2 + 4x_3 = 62

    So solve the first equation for x_3:
    x_3 = 6 - \frac{2}{3}x_1
    and sub that into the remaining two equations:
    2x_1 + 2x_2 + \left ( 6 - \frac{2}{3}x_1 \right ) = 16 \implies \frac{4}{3}x_1 + 2x_2 = 10
    and
    8x_1 + 6x_2 + 4 \left ( 6 - \frac{2}{3}x_1 \right ) = 62 \implies \frac{16}{3}x_1 + 6x_2 = 38

    Again, solve the top equation, this time for x_2:
    x_2 = 5 - \frac{2}{3}x_1
    and sub that into the remaining equation:
    \frac{16}{3}x_1 + 6 \left ( 5 - \frac{2}{3}x_1 \right ) = 38 \implies \frac{4}{3}x_1 = 8

    Now solve for x_1:
    x_1 = \frac{3}{4} \cdot 8 = 6

    Thus
    x_2 = 5 - \frac{2}{3} \cdot 6 = 1
    and
    x_3 = 6 - \frac{2}{3} \cdot 6 =  2

    So finally
    \left ( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right ) = \left ( \begin{matrix} 6 \\ 1 \\ 2 \end{matrix} \right )

    You do the other one.

    -Dan
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