# 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 $A$ 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!

2. 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

3. 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:

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).