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Math Help - Determinant of this 5x5 matrix.

  1. #1
    s3a
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    Determinant of this 5x5 matrix.

    I tried this probably like 20 times (not exaggerating) and I keep making a mistake. I usually get -30 but the answer is 30. I even checked with software and it's 30. So, could someone please point out what I'm doing wrong?

    (My work is attached).

    Any help would be greatly appreciated!
    Thanks in advance!
    Attached Thumbnails Attached Thumbnails Determinant of this 5x5 matrix.-temp_q.jpeg  
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  2. #2
    A Plied Mathematician
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    Two errors:

    Third line, second matrix: should be a -4 multiplying the 2 x 2 determinant, because the sign associated with the expansion corresponding to -1 has itself a -1, thus keeping whatever sign you had before.

    Third line, first 2 x 2 determinant: (0)(-1) - (-3)(-3) = 0 - 9 = -9. You had a +9.

    The two errors were sign errors corresponding to the two different pieces, thus totally explaining the opposite sign.
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  3. #3
    MHF Contributor harish21's Avatar
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    when you are taking the det of the second 3x3 matrix in you second line, you should have:

    -2(2)(-(-1))\left(\begin{array}{cc}2&0\\1&3\end{array}\righ  t)=-4(6-0)=-24

    and \left|\begin{array}{cc}0&-3\\-3&1\end{array}\right| = 0(1)-(-3)(-3) = -9
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  4. #4
    s3a
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    I got it wrong again but I tried to see why I was wrong and I cannot see how the -1 gets an additional negative like: -(-1). I tried to explain myself on the new attached paper. If more explanation is needed, please tell me.
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  5. #5
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    Quote Originally Posted by s3a View Post
    I got it wrong again but I tried to see why I was wrong and I cannot see how the -1 gets an additional negative like: -(-1). I tried to explain myself on the new attached paper. If more explanation is needed, please tell me.

    Develop the determinant \left|\begin{array}{rrrrr}2&0&0&2&0\\2&0&3&0&0\\0&  2&0&0&-3\\0&0&0&-3&-1\\0&1&3&0&0\end{array}\right| as follows:

    first substract row 1 from row 2 and develop then by the first column:

    \left|\begin{array}{rrrrr}2&0&0&2&0\\0&0&3&-2&0\\0&2&0&0&-3\\0&0&0&-3&-1\\0&1&3&0&0\end{array}\right|= 2\cdot\left|\begin{array}{rrrr}0&3&-2&0\\2&0&0&-3\\0&0&-3&-1\\1&3&0&0\end{array}\right| . Now substract twice the 4th row from the 2nd one

    and, again, develop by the first column:

    =2\cdot\left|\begin{array}{rrrr}0&3&-2&0\\0&-6&0&-3\\0&0&-3&-1\\1&3&0&0\end{array}\right| =2\cdot(-1)\left|\begin{array}{rrrr}3&-2&0\\-6&0&-3\\0&-3&-1\end{array}\right| =-2\cdot(-27+12)=30

    Tonio
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  6. #6
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    Quote Originally Posted by s3a View Post
    I got it wrong again but I tried to see why I was wrong and I cannot see how the -1 gets an additional negative like: -(-1). I tried to explain myself on the new attached paper. If more explanation is needed, please tell me.
    The signs associated with the determinant: \left|\begin{array}{ccc}2&0&-3\\0&0&-1\\1&3&0\end{array}\right|

    are: \left|\begin{array}{ccc}+&-&+\\-&+&-\\+&-&+\end{array}\right|.

    The second row is: - + -
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