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Math Help - Diagonalisation of a matrix

  1. #1
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    Diagonalisation of a matrix

    anyone help me on this.

    the eigenvectors are 2,-1,3.

    This is B,
    1 1 1
    2 1 2
    3 2 4

    This is B-1,
    0 2 -1
    2 -1 0
    -1 -1 1

    I cant seem to workout how you manage to get from B to B-1.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by naisy18 View Post
    anyone help me on this.

    the eigenvectors are 2,-1,3.

    This is B,
    1 1 1
    2 1 2
    3 2 4

    This is B-1,
    0 2 -1
    2 -1 0
    -1 -1 1

    I cant seem to workout how you manage to get from B to B-1.
    1. Use Cramer's rule

    2. Augment B with I_{3 \times 3} then reduce the first three columns to
    I_{3 \times 3} using row operations, and then what is left in the
    augmentation part of the matrix will be B^{-1}

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    1. Use Cramer's rule

    2. Augment B with I_{3 \times 3} then reduce the first three columns to
    I_{3 \times 3} using row operations, and then what is left in the
    augmentation part of the matrix will be B^{-1}

    RonL
    Im still not too sure what your saying to be honest.

    If you or any other poster has time could you show me where you would start and how you would start it, as im still pretty stuck. cheers
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  4. #4
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    Naisy, what are you doing in this material? You simply MUST be able to find the inverse of a simple 3x3 matrix.

    Cpt. Black used some terms with which you simply must be familiar. If you are not, you must look them up and get familiar. It will not serve you at all to remain absent this information.
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  5. #5
    Lord of certain Rings
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    Can't we use the given eigenvalues to make the computation a tad easier
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  6. #6
    Eater of Worlds
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    Here's a way to find an inverse, but it's rather cumbersome. May not be taught much these days. I don't know. It's comes from the adjoint.

    \begin{bmatrix}1&1&1\\2&1&2\\3&2&4\end{bmatrix}

    Using the cofactors:

    C_{11}=det(\begin{bmatrix}1&2\\2&4\end{bmatrix})=0

    C_{12}= -det(\begin{bmatrix}2&2\\3&4\end{bmatrix})=-2

    C_{13}=det(\begin{bmatrix}2&1\\3&2\end{bmatrix})=1

    C_{21}= -det(\begin{bmatrix}1&1\\2&4\end{bmatrix})=-2

    C_{22}= det(\begin{bmatrix}1&1\\3&4\end{bmatrix})=1

    C_{23}= -det(\begin{bmatrix}1&1\\3&2\end{bmatrix})=1

    C_{31} = det(\begin{bmatrix}1&1\\1&2\end{bmatrix})=1

    C_{32} = -det(\begin{bmatrix}1&1\\2&2\end{bmatrix})=0

    C_{33} = det(\begin{bmatrix}1&1\\2&1\end{bmatrix}) = -1

    From these we build a matrix:

    \begin{bmatrix}0&-2&1\\-2&1&1\\1&0&-1\end{bmatrix}

    Take the transpose and we get:

    \begin{bmatrix}0&-2&1\\-2&1&0\\1&1&-1\end{bmatrix}

    Now, finally, B^{-1}=\frac{1}{det(B)}\cdot{adj(B)}

    det(B)=-1

    So, we have B^{-1}=\begin{bmatrix}0&2&-1\\2&-1&0\\-1&-1&1\end{bmatrix}
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by galactus View Post
    Here's a way to find an inverse, but it's rather cumbersome. May not be taught much these days. I don't know. It's comes from the adjoint.
    Actually this is how I originally learned it (while learning engineering).
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