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Math Help - [SOLVED] Inverse matrix with adj(A)

  1. #1
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    [SOLVED] Inverse matrix with adj(A)

    So I'm trying to find A^-1 (inverse) matrix. The theorem I'm supposed to use for this is of a matrix using its Adjoint.

    A^-1 = [1 / det(A)] * [adj(A)]

    The matrix used is A =

    2 0 0
    8 1 0
    -5 3 6

    I expanded the matrix using cofactors and obtained

    c11=6, c12=48, c13=29
    c21=0, c22=12, c23=6
    c31=0, c32=0, c33=2

    so the matrix cofactor is

    6 48 29
    0 12 6
    0 0 2

    and the adjoint of A is

    6 0 0
    48 12 0
    29 6 2

    now what do I do? I'm stuck here.
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  2. #2
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    Quote Originally Posted by thekrown View Post
    So I'm trying to find A^-1 (inverse) matrix. The theorem I'm supposed to use for this is of a matrix using its Adjoint.

    A^-1 = [1 / det(A)] * [adj(A)]

    The matrix used is A =

    2 0 0
    8 1 0
    -5 3 6

    I expanded the matrix using cofactors and obtained

    c11=6, c12=48, c13=29
    c21=0, c22=12, c23=6
    c31=0, c32=0, c33=2

    so the matrix cofactor is

    6 48 29
    0 12 6
    0 0 2

    and the adjoint of A is

    6 0 0
    48 12 0
    29 6 2

    now what do I do? I'm stuck here.
    now you only need to compute the deteminant of A.

    \det A = 6 \left|\begin{matrix} 2 & 0 \\ 8 & 1 \end{matrix}\right| - 0 \cdot \left|\begin{matrix} 2 & 0 \\ -5 & 3 \end{matrix}\right| + 0 \cdot \left|\begin{matrix} 8 & 1 \\ -5 & 3 \end{matrix}\right| = 12

    and finally,

    A^{-1} = \frac{1}{12}  \left(\begin{matrix} 6 & 0 & 0 \\ 48 & 12 & 0 \\ 29 & 6 & 2\end{matrix}\right)
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  3. #3
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    Alright so the end result should then be

    A^-1 = 1/12 * 6 (det of the adj(A))

    and so for my final answer I have A^-1 = 1/2, yes?
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  4. #4
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    A(-1) is a matrix

    No, the inverse of a matrix is not a value, like it's determinant.
    The inverse of a matrix A = adjoint(A)/det(A).
    It's not det(adj(A))*(adj(A)) [which is what I think you are suggesting]

    Anyway, I don't see how the determinant of adj(A)=6. It's 144 . (12)
    The answer dedust posted is the inverse of A.
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  5. #5
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    Okay so I understand my mistake. The answer is 1/det(A) * ajd(A) (where ads(A) is the matrix starting with line 6 0 0).

    However, nahduma really confused me. Just so I can correct my mistake of trying to find the det of such a matrix:

    6 0 0
    48 12 0
    29 6 2

    I originally ended up with

    (-1)^1+1 6det

    12 0
    6 2

    + (-1)^1+2 0 det (matrix doesn't matter here is mult by zero).

    + (-1)^1+3 0 det (matrix doesn't matter here it's mult by zero).

    So with this I ended up with 6 multiplied by ad-bc which is 12x2 - 6x0. We get 6x24 = 144.

    Originally I forgot to count the ad-bc part and ended up with just 6.

    Now back to nahduma's post, did you just get 12^2 by simplifying 144 or did you use a method which gave you 12^2?
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  6. #6
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    det(adjoint)

    the determinant of the cofactor matrix of a n x n matrix A is det(A)^(n-1)
    since the adjoint is just the transpose of the cofactor matrix, its determinant is also det(A)^(n-1).

    For a 3x3 matrix, its det(A)^2. in this case, 144.
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