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Math Help - Matrix Inversion

  1. #1
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    Matrix Inversion

    I'm trying to solve the following system of equations by matrix inversion, however when I substitute back my final solutions for x,y,z into the equations, one of them doesn't work. I checked everything a couple of times but I can't find the mistake.

    Q: Solve the system of equations by matrix inversion if possible:

    3x-2y+z=1
    x+y+z=0
    2x-y+2z=1

    and here is the solution: ImageShack Album - 6 images
    )))))))
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  2. #2
    Behold, the power of SARDINES!
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    You inverse for the matrix is correct, but I think there are simpler and more efficient ways to invert a matrix. With that said I think you just made multiplication error
    when you multiplied the column vector by the inverse. It should give
    \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\<br />
0 & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{2} & -\frac{1}{6} & \frac{5}{6}<br />
\end{bmatrix} \begin{bmatrix} 1 \\ 0 \\1\end{bmatrix}=\begin{bmatrix} 0 \\ -\frac{1}{3} \\ \frac{1}{3}\end{bmatrix}
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    You inverse for the matrix is correct, but I think there are simpler and more efficient ways to invert a matrix. With that said I think you just made multiplication error
    when you multiplied the column vector by the inverse. It should give
    \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\<br />
0 & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{2} & -\frac{1}{6} & \frac{5}{6}<br />
\end{bmatrix} \begin{bmatrix} 1 \\ 0 \\1\end{bmatrix}=\begin{bmatrix} 0 \\ -\frac{1}{3} \\ \frac{1}{3}\end{bmatrix}
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