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Thread: Determinants (do you notice any thing wrong??)

  1. #1
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    Red face Determinants (do you notice any thing wrong??)

    Hi guys,

    hoping somebody can check through my calculations for the determinant of A a 4 x 4 matrix.

    A = \begin{vmatrix}<br />
\-1 && 3 && 0 && 5 \\<br />
-2 && 0 && 4 && 0\\<br />
1 && 2 && -1 && 3\\<br />
1 && -2 && 1 && -1\\<br />
\end{vmatrix} = 1 \begin{vmatrix}<br />
\ 0 && 4 && 0 \\<br />
2 && -1 && 3\\<br />
-2 && 1 && -1\\<br />
\end{vmatrix} - 3 \begin{vmatrix}<br />
\ -2 && 4 && 0 \\<br />
1 && -1 && 3\\<br />
1 && 1 && -1\\<br />
\end{vmatrix} <br />

    <br />
+ 0 \begin{vmatrix}<br />
\ -2 && 0 && 0 \\<br />
1 && 2 && 3\\<br />
1 && -2 && -1\\<br />
\end{vmatrix} - 5 \begin{vmatrix}<br />
\ -2 && 0 && 4 \\<br />
1 && 2 && -1\\<br />
1 && -2 && 1\\<br />
\end{vmatrix}<br />

    <br />
= 1det(4)\begin{vmatrix}<br />
\ 2 && 3 \\<br />
-2 && -1\\<br />
\end{vmatrix}<br />
    <br />
 - 3 det(-2) \begin{vmatrix}<br />
\ -1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix} - det(4) \begin{vmatrix}<br />
\ 1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix}<br />
    <br />
 - 5det(-2) \begin{vmatrix}<br />
\ 2 && -1 \\<br />
-2 && 1\\<br />
\end{vmatrix} + det(4) \begin{vmatrix}<br />
\ 1 && 2 \\<br />
1 && -2\\<br />
\end{vmatrix}<br />

    =  1 (4(-2+6) - 3 (-2(1-3) - 4(-1-3))-5(-2(2-2)+4(-2-2))

    =  1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))

    =  1 (16) - 3 (4 + 16) +5(-16)

    If anyone noticed any wrong be greatly appreciate if they could correct it saying so.

    Thanks a bunch!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathsToday View Post
    Hi guys,

    hoping somebody can check through my calculations for the determinant of A a 4 x 4 matrix.

    A = \begin{vmatrix}<br />
\-1 && 3 && 0 && 5 \\<br />
-2 && 0 && 4 && 0\\<br />
1 && 2 && -1 && 3\\<br />
1 && -2 && 1 && -1\\<br />
\end{vmatrix} = 1 \begin{vmatrix}<br />
\ 0 && 4 && 0 \\<br />
2 && -1 && 3\\<br />
-2 && 1 && -1\\<br />
\end{vmatrix} - 3 \begin{vmatrix}<br />
\ -2 && 4 && 0 \\<br />
1 && -1 && 3\\<br />
1 && 1 && -1\\<br />
\end{vmatrix} <br />

    <br />
+ 0 \begin{vmatrix}<br />
\ -2 && 0 && 0 \\<br />
1 && 2 && 3\\<br />
1 && -2 && -1\\<br />
\end{vmatrix} - 5 \begin{vmatrix}<br />
\ -2 && 0 && 4 \\<br />
1 && 2 && -1\\<br />
1 && -2 && 1\\<br />
\end{vmatrix}<br />

    <br />
= 1det(4)\begin{vmatrix}<br />
\ 2 && 3 \\<br />
-2 && -1\\<br />
\end{vmatrix}<br />
    <br />
 - 3 det(-2) \begin{vmatrix}<br />
\ -1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix} - det(4) \begin{vmatrix}<br />
\ 1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix}<br />
    <br />
 - 5det(-2) \begin{vmatrix}<br />
\ 2 && -1 \\<br />
-2 && 1\\<br />
\end{vmatrix} + det(4) \begin{vmatrix}<br />
\ 1 && 2 \\<br />
1 && -2\\<br />
\end{vmatrix}<br />

    =  1 (4(-2+6) - 3 (-2(1-3) - 4(-1-3))-5(-2(2-2)+4(-2-2))

    =  1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))

    =  1 (16) - 3 (4 + 16) +5(-16)

    If anyone noticed any wrong be greatly appreciate if they could correct it saying so.

    Thanks a bunch!
    My calculator says you are wrong. (I get 4.) I would suggest you expand across the second row... you only have to do two 3 x 3 determinants that way.

    -Dan
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by mathsToday View Post
    Hi guys,

    hoping somebody can check through my calculations for the determinant of A a 4 x 4 matrix.

    A = \begin{vmatrix}<br />
\-1 && 3 && 0 && 5 \\<br />
-2 && 0 && 4 && 0\\<br />
1 && 2 && -1 && 3\\<br />
1 && -2 && 1 && -1\\<br />
\end{vmatrix} = 1 \begin{vmatrix}<br />
\ 0 && 4 && 0 \\<br />
2 && -1 && 3\\<br />
-2 && 1 && -1\\<br />
\end{vmatrix} - 3 \begin{vmatrix}<br />
\ -2 && 4 && 0 \\<br />
1 && -1 && 3\\<br />
1 && 1 && -1\\<br />
\end{vmatrix} <br />

    <br />
+ 0 \begin{vmatrix}<br />
\ -2 && 0 && 0 \\<br />
1 && 2 && 3\\<br />
1 && -2 && -1\\<br />
\end{vmatrix} - 5 \begin{vmatrix}<br />
\ -2 && 0 && 4 \\<br />
1 && 2 && -1\\<br />
1 && -2 && 1\\<br />
\end{vmatrix}<br />
Correct as far as here.
    <br />
= 1det(4)\begin{vmatrix}<br />
\ 2 && 3 \\<br />
-2 && -1\\<br />
\end{vmatrix}<br />
    <br />
 - 3 det(-2) \begin{vmatrix}<br />
\ -1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix} - det(4) \begin{vmatrix}<br />
\ 1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix}<br />
    <br />
 - 5det(-2) \begin{vmatrix}<br />
\ 2 && -1 \\<br />
-2 && 1\\<br />
\end{vmatrix} + det(4) \begin{vmatrix}<br />
\ 1 && 2 \\<br />
1 && -2\\<br />
\end{vmatrix}<br />
    This is where it all starts to go haywire. In the first term, <br />
1det(4)\begin{vmatrix}<br />
\ 2 && 3 \\<br />
-2 && -1\\<br />
\end{vmatrix}<br />
, the 4 should be 4 (because it's the second element in its row, and the +/ signs should alternate as you go along the row). The next two terms should both be multiplied by 3, so that row of the calculation should read <br />
 - 3 \left(-2\begin{vmatrix}<br />
\ -1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix} - 4 \begin{vmatrix}<br />
\ 1 && 3 \\<br />
1 && -1\\<br />
\end{vmatrix}\right)<br />
.
    Similarly, the whole of the last row of the calculation should have the factor of 5, so it should read <br />
 - 5\left(- \begin{vmatrix}<br />
\ 2 && -1 \\<br />
-2 && 1\\<br />
\end{vmatrix} + 4\begin{vmatrix}<br />
\ 1 && 2 \\<br />
1 && -2\\<br />
\end{vmatrix}\right)<br />

    Reading further on, I see that two of those errors get corrected in the next line of the calculation. The parentheses are inserted as they should be, and it's only the sign of the 4 that's still wrong:

    Quote Originally Posted by mathsToday View Post
    =  1 (-4(-2+6) - 3 \bigl(-2(1-3) - 4(-1-3)\bigr)-5\bigl(-2(2-2)+4(-2-2)\bigr)

    =  1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))
    =  1 (16) - 3 (4 + 16) +5(-16) Another wrong sign: +5 should be 5.
    So the correct answer should be 16 60 + 80 = 4.

    Final comment: this is a terrible method for evaluating a determinant. It's far easier and quicker to use elementary row and column operations (even easier to use a calculator with a determinant facility, but maybe that would be cheating).
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