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Math Help - a question involving the determinant...

  1. #1
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    a question involving the determinant...

    given: A =
    |a b c|
    |d e f| det(A) = 5
    |g h i|

    B =
    |a u c|
    |d v f| det(B) = -3
    |g w i|

    FIND:
    |-3a c 2b - u|
    |-3d f 2e - v|
    |-3g i 2h - w|

    -it's pretty much "B" except with the column of A in it...not sure what to do about that
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  2. #2
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    Re: a question involving the determinant...

    Given
    aei + bfg + cdh - ceg - bdi - afh = 5
    avi + ufg + cdw - cvg - udi - afw = -3

    Find
    (-3a)(f)(2h-w) + c(2e-v)(-3g) + (2b-u)(-3d)(i) - (2b - u)f(-3g) - c(-3d)(2h - w) - (-3a)(2e - v)(i) = ?
    -6afh + 3afw - 6ceg + 3cvg - 6bdi + 3udi + 6bfg - 3ufg + 6cdh - 3cdw + 6aei - 3avi
    [6aei + 6bfg + 6cdh - 6ceg - 6bdi - 6afh] + [-3avi - 3ufg - 3cdw + 3cvg + 3udi + 3afw]
    6(aei + bfg + cdh - ceg - bdi - afh) + (-3)(avi + ufg + cdw - cvg - udi - afw)
    6(5) + (-3)(-3)
    30 + 9
    39
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  3. #3
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    Re: a question involving the determinant...

    Hello, mathlover700!

    I hope I interpreted the problem correctly . . .


    \text{Given: }\:A = \begin{vmatrix} a&b&c \\ d&e&f \\ g&h&i \end{vmatrix},\;\text{det}(A) = 5

    B =\begin{vmatrix}a&u&c \\ d&v&f \\ g&w&i\end{vmatrix},\;\text{det}(B) = -3

    \text{Find the determinant of: }\:\begin{vmatrix}-3a & c & 2b-u \\ -3d & f & 2e-v \\ -3g & i & 2h-w \end{vmatrix}

    2A \;=\;2\cdot\begin{vmatrix}a&b&c\\d&e&f\\g&h&y\end{  vmatrix} \;=\;\begin{vmatrix}2a&2b&2c\\2d&2e&2f\\2g&2h&2i \end{vmatrix}

    . . \text{det}(2A)\,=\,2(5)\,=\,10


    2A-B \;=\;\begin{vmatrix}2a&2b&2c\\2d&2e&2f \\ 2g&2h&2y\end{vmatrix} - \begin{vmatrix}a&u&c\\d&v&f \\ g&w&i\end{vmatrix} \;=\;\begin{vmatrix}a&2b-y&c\\d&2e-v&f \\ g&2h-w&i \end{vmatrix}

    . . \text{det}(2A-B)\,=\,10 - (\text{-}3)\,=\,13

    Multiply -3 times column-1: . \begin{vmatrix}\text{-}3a&2b-u&c \\ \text{-}3d&2e-v&f \\ \text{-}3g&2h-w&i\end{vmatrix}
    . . \text{det}\,=\,-3(13)\,=\,-39

    Switch column-2 and column-3: . \begin{vmatrix}\text{-}3a & c & 2b-u \\ \text{-}3d & f & 2e-v \\ \text{-}3g & i & 2h-w \end{vmatrix}
    . . \text{det} \,=\,-1(-39) \,=\,+39



    Hey, I agree with Mathguy25 !
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  4. #4
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    Re: a question involving the determinant...

    Soroban, wow. My newbieness is showing. I attempted to approach the problem that way using the properties of manipulating columns but found no clear solution. Then I resorted to the basic way I know: brute force. Well done, sir. I'm impressed.
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  5. #5
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    Re: a question involving the determinant...

    Hello, mathguy25!

    And I am impressed by your dedication and stamina
    . . evident in your algebraic solution.

    Well done to you, sir!
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  6. #6
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    Re: a question involving the determinant...

    a question involving the determinant...-10-4-13.png
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  7. #7
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    Re: a question involving the determinant...

    thank you mathguy25 and thank you soroban for your help! I appreciate both solutions!
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  8. #8
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    Re: a question involving the determinant...

    Quote Originally Posted by Soroban View Post
    . . \text{det}(2A)\,=\,2(5)\,=\,10

    ...

    . . \text{det}(2A-B)\,=\,10 - (\text{-}3)\,=\,13
    I hope I'm interpreting your solution correctly, but I think there are two errors here that cancel each other out. \text{det}(2A)=2^n\text{det}A, where n is the dimension of A. This is because each column is multiplied by 2, and each of those multiplications multiplies the determinant by 2. In this case, \text{det}(2A)=8\text{det}A=40.

    Also, \text{det}(2A-B)=\text{det}2A-\text{det}B in this case*, but not in general.

    What's happening here is that \text{det}[A_i(2c_i+v)]=\text{det}[A_i(2c_i)]+\text{det}[A_i(v)], where the notation A_i(v) means the matrix that's the same as A except the ith column is replaced with the column vector v, and c_i represents the ith column of A, so \text{det}(2A-B) does indeed equal 13 and the rest of the proof works fine.

    *edit: whoops, no it doesn't. It would if det(2A)=2det(A) is what I should have said.
    Last edited by LoblawsLawBlog; April 16th 2013 at 02:13 PM.
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