Determinants (do you notice any thing wrong??)

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Apr 2nd 2008, 05:17 AM
mathsToday

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} \-1 && 3 && 0 && 5 \\ -2 && 0 && 4 && 0\\ 1 && 2 && -1 && 3\\ 1 && -2 && 1 && -1\\ \end{vmatrix} = 1 \begin{vmatrix} \ 0 && 4 && 0 \\ 2 && -1 && 3\\ -2 && 1 && -1\\ \end{vmatrix} - 3 \begin{vmatrix} \ -2 && 4 && 0 \\ 1 && -1 && 3\\ 1 && 1 && -1\\ \end{vmatrix} $

$ + 0 \begin{vmatrix} \ -2 && 0 && 0 \\ 1 && 2 && 3\\ 1 && -2 && -1\\ \end{vmatrix} - 5 \begin{vmatrix} \ -2 && 0 && 4 \\ 1 && 2 && -1\\ 1 && -2 && 1\\ \end{vmatrix} $

$ = 1det(4)\begin{vmatrix} \ 2 && 3 \\ -2 && -1\\ \end{vmatrix} $
$ - 3 det(-2) \begin{vmatrix} \ -1 && 3 \\ 1 && -1\\ \end{vmatrix} - det(4) \begin{vmatrix} \ 1 && 3 \\ 1 && -1\\ \end{vmatrix} $
$ - 5det(-2) \begin{vmatrix} \ 2 && -1 \\ -2 && 1\\ \end{vmatrix} + det(4) \begin{vmatrix} \ 1 && 2 \\ 1 && -2\\ \end{vmatrix} $

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

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

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

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

Thanks a bunch!
Apr 2nd 2008, 07:45 AM
topsquark

Quote:

Originally Posted by mathsToday

Hi guys,

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

$A = \begin{vmatrix} \-1 && 3 && 0 && 5 \\ -2 && 0 && 4 && 0\\ 1 && 2 && -1 && 3\\ 1 && -2 && 1 && -1\\ \end{vmatrix} = 1 \begin{vmatrix} \ 0 && 4 && 0 \\ 2 && -1 && 3\\ -2 && 1 && -1\\ \end{vmatrix} - 3 \begin{vmatrix} \ -2 && 4 && 0 \\ 1 && -1 && 3\\ 1 && 1 && -1\\ \end{vmatrix} $

$ + 0 \begin{vmatrix} \ -2 && 0 && 0 \\ 1 && 2 && 3\\ 1 && -2 && -1\\ \end{vmatrix} - 5 \begin{vmatrix} \ -2 && 0 && 4 \\ 1 && 2 && -1\\ 1 && -2 && 1\\ \end{vmatrix} $

$ = 1det(4)\begin{vmatrix} \ 2 && 3 \\ -2 && -1\\ \end{vmatrix} $
$ - 3 det(-2) \begin{vmatrix} \ -1 && 3 \\ 1 && -1\\ \end{vmatrix} - det(4) \begin{vmatrix} \ 1 && 3 \\ 1 && -1\\ \end{vmatrix} $
$ - 5det(-2) \begin{vmatrix} \ 2 && -1 \\ -2 && 1\\ \end{vmatrix} + det(4) \begin{vmatrix} \ 1 && 2 \\ 1 && -2\\ \end{vmatrix} $

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

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

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

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
Apr 2nd 2008, 07:50 AM
Opalg

Quote:

Originally Posted by mathsToday

Hi guys,

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

$A = \begin{vmatrix} \-1 && 3 && 0 && 5 \\ -2 && 0 && 4 && 0\\ 1 && 2 && -1 && 3\\ 1 && -2 && 1 && -1\\ \end{vmatrix} = 1 \begin{vmatrix} \ 0 && 4 && 0 \\ 2 && -1 && 3\\ -2 && 1 && -1\\ \end{vmatrix} - 3 \begin{vmatrix} \ -2 && 4 && 0 \\ 1 && -1 && 3\\ 1 && 1 && -1\\ \end{vmatrix} $

$ + 0 \begin{vmatrix} \ -2 && 0 && 0 \\ 1 && 2 && 3\\ 1 && -2 && -1\\ \end{vmatrix} - 5 \begin{vmatrix} \ -2 && 0 && 4 \\ 1 && 2 && -1\\ 1 && -2 && 1\\ \end{vmatrix} $ Correct as far as here.
$ = 1det(4)\begin{vmatrix} \ 2 && 3 \\ -2 && -1\\ \end{vmatrix} $
$ - 3 det(-2) \begin{vmatrix} \ -1 && 3 \\ 1 && -1\\ \end{vmatrix} - det(4) \begin{vmatrix} \ 1 && 3 \\ 1 && -1\\ \end{vmatrix} $
$ - 5det(-2) \begin{vmatrix} \ 2 && -1 \\ -2 && 1\\ \end{vmatrix} + det(4) \begin{vmatrix} \ 1 && 2 \\ 1 && -2\\ \end{vmatrix} $

This is where it all starts to go haywire. In the first term, $ 1det(4)\begin{vmatrix} \ 2 && 3 \\ -2 && -1\\ \end{vmatrix} $ , 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 $ - 3 \left(-2\begin{vmatrix} \ -1 && 3 \\ 1 && -1\\ \end{vmatrix} - 4 \begin{vmatrix} \ 1 && 3 \\ 1 && -1\\ \end{vmatrix}\right) $ .
Similarly, the whole of the last row of the calculation should have the factor of –5, so it should read $ - 5\left(- \begin{vmatrix} \ 2 && -1 \\ -2 && 1\\ \end{vmatrix} + 4\begin{vmatrix} \ 1 && 2 \\ 1 && -2\\ \end{vmatrix}\right) $

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

$= 1 (-4(-2+6) - 3 \bigl(-2(1-3) - 4(-1-3)\bigr)-5\bigl(-2(2-2)+4(-2-2)\bigr)$

$= 1 (4(4) - 3 (-2(-2) - 4(-4))-5(-2(0) + 4(-4))$
$= 1 (16) - 3 (4 + 16) +5(-16)$ 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).