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Determinant of a 4x4 matrix

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
calculate determinant of:
\(\displaystyle
\left| {\begin{array}{cc} 2 & -2 & -3 & 8 \\ 1 & -1 & 2 & -1 \\ -3 & 4 & 1 & -1 \\ -2 & 6 & -4 & 19 \end{array} } \right|\)
so I multiplication -2 row 2 and add it to row 1, multiplication 3 to row 2 and add it to row 3, multiplicate 2 to row 2 and add it to row 4 and get
\(\displaystyle
\left| {\begin{array}{cc} 0 & 0 & -5 & -10 \\ 1 & -1 & 2 & -1 \\ 0 & 1 & 7 & -4 \\ 0 & 4 & 0 & 17 \end{array} } \right|\) and develops in columne 1 and use sarrus rule but I get wrong.

Regards,
\(\displaystyle |\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
There are several methods to do this. You can just use Jacobi's Rule as is to obtain
\begin{align*}\left| \begin{array}{cccc} 2 & -2 & -3 & 8 \\ 1 & -1 & 2 & -1 \\ -3 & 4 & 1 & -1 \\ -2 & 6 & -4 & 19 \end{array} \right|
=&(+1)(2)\left| \begin{array}{ccc}-1 & 2 & -1 \\ 4 & 1 & -1 \\ 6 & -4 & 19 \end{array} \right|
+(-1)(-2)\left| \begin{array}{ccc} 1 & 2 & -1 \\ -3 & 1 & -1 \\ -2 & -4 & 19 \end{array} \right|\\
+&(+1)(-3)\left| \begin{array}{ccc} 1 & -1 & -1 \\ -3 & 4 & -1 \\ -2 & 6 & 19 \end{array} \right|
+(-1)(8)\left| \begin{array}{ccc} 1 & -1 & 2 \\ -3 & 4 & 1 \\ -2 & 6 & -4 \end{array} \right|,
\end{align*}
and keep drilling down in each matrix.
Alternatively, you can do ERO's on the matrix, keeping track of how each ERO changes the determinant.

You can certainly do the ERO's you did. I get
$$\left| \begin{array}{cccc} 2 & -2 & -3 & 8 \\ 1 & -1 & 2 & -1 \\ -3 & 4 & 1 & -1 \\ -2 & 6 & -4 & 19 \end{array} \right| \to
\left| \begin{array}{cccc} 0 & 0 & -7 & 10 \\ 1 & -1 & 2 & -1 \\ 0 & 1 & 7 & -4 \\ 0 & 4 & 0 & 17 \end{array} \right|.$$
If I compare to yours, which is
$$\left| {\begin{array}{cc} 0 & 0 & -5 & -10 \\ 1 & -1 & 2 & -1 \\ 0 & 1 & 7 & -4 \\ 0 & 4 & 0 & 17 \end{array} } \right|,$$
I see that there are some differences. Don't forget, when you actually do the determinant, that the $-1$ in the $2,1$ position has a minus sign associated with it in the Jacobi expansion.
 

Petrus

Well-known member
Feb 21, 2013
739
There are several methods to do this. You can just use Jacobi's Rule as is to obtain
\begin{align*}\left| \begin{array}{cccc} 2 & -2 & -3 & 8 \\ 1 & -1 & 2 & -1 \\ -3 & 4 & 1 & -1 \\ -2 & 6 & -4 & 19 \end{array} \right|
=&(+1)(2)\left| \begin{array}{ccc}-1 & 2 & -1 \\ 4 & 1 & -1 \\ 6 & -4 & 19 \end{array} \right|
+(-1)(-2)\left| \begin{array}{ccc} 1 & 2 & -1 \\ -3 & 1 & -1 \\ -2 & -4 & 19 \end{array} \right|\\
+&(+1)(-3)\left| \begin{array}{ccc} 1 & -1 & -1 \\ -3 & 4 & -1 \\ -2 & 6 & 19 \end{array} \right|
+(-1)(8)\left| \begin{array}{ccc} 1 & -1 & 2 \\ -3 & 4 & 1 \\ -2 & 6 & -4 \end{array} \right|,
\end{align*}
and keep drilling down in each matrix.
Alternatively, you can do ERO's on the matrix, keeping track of how each ERO changes the determinant.

You can certainly do the ERO's you did. I get
$$\left| \begin{array}{cccc} 2 & -2 & -3 & 8 \\ 1 & -1 & 2 & -1 \\ -3 & 4 & 1 & -1 \\ -2 & 6 & -4 & 19 \end{array} \right| \to
\left| \begin{array}{cccc} 0 & 0 & -7 & 10 \\ 1 & -1 & 2 & -1 \\ 0 & 1 & 7 & -4 \\ 0 & 4 & 0 & 17 \end{array} \right|.$$
If I compare to yours, which is
$$\left| {\begin{array}{cc} 0 & 0 & -5 & -10 \\ 1 & -1 & 2 & -1 \\ 0 & 1 & 7 & -4 \\ 0 & 4 & 0 & 17 \end{array} } \right|,$$
I see that there are some differences. Don't forget, when you actually do the determinant, that the $-1$ in the $2,1$ position has a minus sign associated with it in the Jacobi expansion.
Hello Ackbach,
I am really grateful for the fast responed!:) Thanks alot! I did not see I start to misscalculate ...:( I have read about a triangle way, can I also use it?

Regards
\(\displaystyle |\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Hello Ackbach,
I am really grateful for the fast responed!:) Thanks alot! I did not see I start to misscalculate ...:( I have read about a triangle way, can I also use it?

Regards
\(\displaystyle |\rangle\)
If you make your matrix upper triangular, the determinant of the resulting matrix is the product of the entries on the main diagonal. BUT, you still have to keep track of how your ERO's affect the determinant.
 

Petrus

Well-known member
Feb 21, 2013
739
BUT, you still have to keep track of how your ERO's affect the determinant.
Hello Ackbach,
What did you mean with that? If you mean with it will be an multiplicate -1 to the determinant because of well in Swedish my book called it "schedule"?
+ - + -
- + - +
+ - + -
- + - +
Regards,
\(\displaystyle |\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
No, I mean this:

The ERO that takes a multiple of one row, adds it to another row, and stores it in that row, does not change the determinant.

The ERO that switches two rows multiplies the determinant by $-1$.

The ERO that multiplies a row by a nonzero number $m$ also multiplies the determinant by $m$.
 

Petrus

Well-known member
Feb 21, 2013
739
No, I mean this:

The ERO that takes a multiple of one row, adds it to another row, and stores it in that row, does not change the determinant.

The ERO that switches two rows multiplies the determinant by $-1$.

The ERO that multiplies a row by a nonzero number $m$ also multiplies the determinant by $m$.
Thanks again Ackbach!

Regards
\(\displaystyle |\rangle\)