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Math Help - Determinants Question What did I do Wrong?

  1. #1
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    Determinants Question What did I do Wrong?

    Does anyone know why this is wrong? I listed step by step what I did.

    Find the following determinants by cofactor expansion along a convenient row or columns

    |5 -1 5 0 -3 |
    |4 0 3 -2 2 |
    |3 2 1 0 3 |
    |7 0 -3 0 4 |
    |2 -1 4 3 -2 |

    First, I took R5 ->2R5+3R2
    I then got
    |-5 -1 5 -0 -3|
    |4 0 4 -2 2|
    |3 2 1 0 3 |
    |7 0 -3 0 4 |
    |16-2 17 0 2 |

    I then got -2 as a cofactor and used that as a pivot so i have -2 * the matrix below.
    |-5 -1 5 -0 -3|
    |3 2 1 0 3 |
    |7 0 -3 0 4 |
    |16-2 17 0 2 |

    I then performed the following row operations:
    R1-> 2R1 + R2
    R4 -> R4 + R2 and I get
    |-7 0 11 -3|
    |3 2 1 3|
    |7 0 -3 4|
    |19 0 18 5|

    I used the row 3,2,1,3 as the pivot and I have the resulting matrix
    -4 cofactor times
    |-7 11 -3|
    |7 -3 4|
    |19 18 5|

    If I just take the value of the determinant 3 x 3 I get the correct answer, but shouldn't i multiply the matrix by -4? Intuitively I feel like it should be multiplied by -5. Any thoughts?
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  2. #2
    Senior Member Danneedshelp's Avatar
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    let A= \begin{bmatrix}5&-1&5&0&-3\\4&0&3&-2&2\\3&2&1&0&3\\7&0&-3&0&4\\2&-1&4&3&-2\end{bmatrix} \frac{3}{2}R_{2}+R_{5}\rightarrow{R_{5}} \begin{bmatrix}5&-1&5&0&-3\\4&0&3&-2&2\\3&2&1&0&3\\7&0&-3&0&4\\8&-1&\frac{17}{2}&0&1\end{bmatrix}

    As you can see, we can expand from the -2 (row 2 column 4)...

    (-1)^{6}(-2) \begin{bmatrix}5&-1&5&-3\\3&2&1&3\\7&0&-3&4\\8&-1&\frac{17}{2}&1\end{bmatrix}

    Now we have a 4X4 matrix to reduce...

    \begin{bmatrix}5&-1&5&-3\\3&2&1&3\\7&0&-3&4\\8&-1&\frac{17}{2}&1\end{bmatrix} (-1)R_{1}+R_{4}\rightarrow{R_{4}} \begin{bmatrix}5&-1&5&-3\\3&2&1&3\\7&0&-3&4\\3&0&\frac{7}{2}&4\end{bmatrix} 2R_{1}+R_{2}\rightarrow{R_{2}} \begin{bmatrix}5&-1&5&-3\\13&0&11&-3\\7&0&-3&4\\3&0&\frac{7}{2}&4\end{bmatrix}

    Now, expand from -1 (row 1 column 2)...

    We have...

    (-1)^{3}(-1)\begin{bmatrix}13&11&-3\\7&-3&4\\3&\frac{7}{2}&4\end{bmatrix}

    Then...

    det(A)= \begin{bmatrix}13&11&-3\\7&-3&4\\3&\frac{7}{2}&4\end{bmatrix} \begin{matrix}13&11\\7&-3\\3&\frac{7}{2}\end{matrix}

    Thus, det(A)=(-2)[(-156)+(132)+(\frac{-147}{2})-(27)-(182)-(308)]=1229

    (the -2 comes from our two previous cofactors)

    Which agrees with my Ti-89!
    Last edited by Danneedshelp; July 30th 2009 at 12:57 PM.
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  3. #3
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    Hmm I don't think this is correct according to the answer in the back and my calculator the answer should be -511 but I can't seem to figure out why
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Your calculator is wrong, Danneedshelp's answer is the correct one.
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Did you post the correct matrix? Maybe you forgot a sign or something. When I enter the orginal matrix you posted into my 89, I get 1229, everytime.
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  6. #6
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    I think I figured it out for the first term it should be -5 and not positive 5 which is why there is a difference. Thanks for the help. God I jst hate it when I am off my a negative sign drives me crazy
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