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

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

Hi guys,

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

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

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

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

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

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

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

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

Thanks a bunch!

2. Originally Posted by mathsToday
Hi guys,

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

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

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

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

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

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

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

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

3. Originally Posted by mathsToday
Hi guys,

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

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

$\displaystyle + 0 \begin{vmatrix} \ -2 && 0 && 0 \\ 1 && 2 && 3\\ 1 && -2 && -1\\ \end{vmatrix} - 5 \begin{vmatrix} \ -2 && 0 && 4 \\ 1 && 2 && -1\\ 1 && -2 && 1\\ \end{vmatrix}$ Correct as far as here.
$\displaystyle = 1det(4)\begin{vmatrix} \ 2 && 3 \\ -2 && -1\\ \end{vmatrix}$
$\displaystyle - 3 det(-2) \begin{vmatrix} \ -1 && 3 \\ 1 && -1\\ \end{vmatrix} - det(4) \begin{vmatrix} \ 1 && 3 \\ 1 && -1\\ \end{vmatrix}$
$\displaystyle - 5det(-2) \begin{vmatrix} \ 2 && -1 \\ -2 && 1\\ \end{vmatrix} + det(4) \begin{vmatrix} \ 1 && 2 \\ 1 && -2\\ \end{vmatrix}$
This is where it all starts to go haywire. In the first term, $\displaystyle 1det(4)\begin{vmatrix} \ 2 && 3 \\ -2 && -1\\ \end{vmatrix}$, 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 $\displaystyle - 3 \left(-2\begin{vmatrix} \ -1 && 3 \\ 1 && -1\\ \end{vmatrix} - 4 \begin{vmatrix} \ 1 && 3 \\ 1 && -1\\ \end{vmatrix}\right)$.
Similarly, the whole of the last row of the calculation should have the factor of –5, so it should read $\displaystyle - 5\left(- \begin{vmatrix} \ 2 && -1 \\ -2 && 1\\ \end{vmatrix} + 4\begin{vmatrix} \ 1 && 2 \\ 1 && -2\\ \end{vmatrix}\right)$

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:

Originally Posted by mathsToday
$\displaystyle = 1 (-4(-2+6) - 3 \bigl(-2(1-3) - 4(-1-3)\bigr)-5\bigl(-2(2-2)+4(-2-2)\bigr)$

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