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January 7th, 2018, 02:46 PM   #1
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Best way to calculate a 5x5 determinant?

I have this 5×5 matrix $\displaystyle A=\begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
4 & 3 & 4 & 0 & 0\\
8 & 6 & 7 & 2 & 0\\
12 & 9 & 10 & 3 & 0\\
16 & 12 & 13 & 4 & 0
\end{bmatrix}$

and I inquire: What would be the "best" way to solve it? (faster, easier).

From what I see in most examples on the internet, they send the reader to choose the row with the smallest number of 0, choose a column of it to be discarded, and then calculate the determinant of matrix that was left from these eliminations. Then repeat for the other elements of the row. Nonetheless, in these examples there is normally a nearly null row, with all its columns being zero, except one, but this is not what we have here. With no row having more than two zeros, this process would have to be repeated many times, and the necessary work would be massive.

Nonetheless, there is a column formed by nearly only zero. So I was wondering if using some property of matrix, I can invert it, obtaining a fifth row that would be formed by nearly all zero, then calculate the determinant and if necessary modify this value for compensating its inversion.

Is that possible? If so, it would modify the determinant calculation in which way?

If it's not, which way would you recommend? I also thought about reducing it to the row echelon form, leaving superior half being all zeros, but I prefer to avoid doing so, since I find its difficulty unpredictable.

Last edited by skipjack; January 7th, 2018 at 09:52 PM.
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January 7th, 2018, 06:22 PM   #2
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What do you get if you add the third row to the fifth row?

What do you notice about the first and second columns?
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January 7th, 2018, 06:31 PM   #3
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Hi !

We can say that every matrix has its own faster and easier method to compute its determinant.
In your post, we can remark that the fifth column has four zeros, then we can calculate the determinant by expanding alone the fifth column :
$\displaystyle
\det A=5\begin{vmatrix}1&2&3&4\\4&3&4&0\\8&7&6&2\\12&9& 10&3\end{vmatrix}
$
And now we get a determinant of $\displaystyle 4\times 4$ matrix.

In this case, we don't have many zeros as previous case, it is better to use elementary operations on the rows - or the columns - of that last matrix to get a many zeros in one row - or one column-. We will get a determinant of $\displaystyle 3\times3$ matrix which can be computed directly by the rule of Sarrus.
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January 7th, 2018, 07:47 PM   #4
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The 4 × 4 determinant wouldn't be what you gave.
Thanks from Snair
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January 7th, 2018, 08:12 PM   #5
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January 8th, 2018, 02:53 AM   #6
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Hi !

Quote:
Originally Posted by Snair View Post
Hi !

We can say that every matrix has its own faster and easier method to compute its determinant.
In your post, we can remark that the fifth column has four zeros, then we can calculate the determinant by expanding alone the fifth column :
$\displaystyle
\det A=5\begin{vmatrix}1&2&3&4\\4&3&4&0\\8&7&6&2\\12&9& 10&3\end{vmatrix}
$
And now we get a determinant of $\displaystyle 4\times 4$ matrix.

In this case, we don't have many zeros as previous case, it is better to use elementary operations on the rows - or the columns - of that last matrix to get a many zeros in one row - or one column-. We will get a determinant of $\displaystyle 3\times3$ matrix which can be computed directly by the rule of Sarrus.

OK ! I made a mistake , it should be :
$\displaystyle \det A=5\begin{vmatrix}4&3&4&0\\8&7&6&2\\12&9&10&3\\16& 12&13&4\end{vmatrix}$
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January 8th, 2018, 04:13 AM   #7
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Quote:
Originally Posted by Snair View Post
Hi !

We can say that every matrix has its own faster and easier method to compute its determinant.
In your post, we can remark that the fifth column has four zeros, then we can calculate the determinant by expanding alone the fifth column :
$\displaystyle
\det A=5\begin{vmatrix}1&2&3&4\\4&3&4&0\\8&7&6&2\\12&9& 10&3\end{vmatrix}
$
And now we get a determinant of $\displaystyle 4\times 4$ matrix.

In this case, we don't have many zeros as previous case, it is better to use elementary operations on the rows - or the columns - of that last matrix to get a many zeros in one row - or one column-. We will get a determinant of $\displaystyle 3\times3$ matrix which can be computed directly by the rule of Sarrus.
Very correct. Every matrix has its own optimal method. Usually we measure efficiency by the worst case scenario, in which case something like LU decomposition or row reduction works really ok. But we can focus on other cases, like sparse matrices, which have their own special methods and for which row reduction would be very bad (for large matrices).

So yes, I definitely agree with Snair. Given any algorithm, you can find a specific situation where this particular algorithm is not so good and another is much better. But overall, the method described in the OP is ok, on average.
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January 8th, 2018, 04:43 AM   #8
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Quote:
Originally Posted by Snair View Post
OK ! I made a mistake , . . .
You corrected one mistake, but left another.
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January 8th, 2018, 05:30 AM   #9
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$\displaystyle \begin{vmatrix}
1& 2 & 3 & 4 & 5\\
4& 3 & 4 & 0 &0 \\
8& 6 & 7 & 2 &0 \\
12& 9& 10 &3 &0 \\
16& 12 & 13 & 4 & 0
\end{vmatrix} = 5\begin{vmatrix}
4 & 3 & 4 & 0\\
8 & 6 & 7 & 2\\
12 & 9 & 10 &3 \\
16 &12 &13 & 4
\end{vmatrix} = 5\begin{vmatrix}
4 & 3 &4 &0 \\
0 & 0 & -1 &2 \\
0& 0 & -2 & 3\\
0 & 0 & -3 & 4
\end{vmatrix}=0 $

After the first step you can see it by inspection.

Last edited by zylo; January 8th, 2018 at 05:47 AM. Reason: typo in line 4 of second matrix
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January 9th, 2018, 09:37 AM   #10
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Thanks to all for the suggestions. I read them quickly at first and decided to try to calculate it by my own. I started doing
$\displaystyle 5\begin{bmatrix}
4 & 3 & 4 & 0\\
8 & 6 & 7 & 2\\
12 & 9 & 10 &3 \\
16 &12 &13 & 4
\end {bmatrix}$

Then used the first row to transform most of the other rows in zeros and I achieved

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

Then decided to give another look at the answers and saw that was what zylo had done.

However, I still have a last question. Technically it would not be necessary to "erase" the -3 on the last row, before one could calculate the determinant ? I realize that one can see that the second or third row can be used to transform the -3 in 0, while the effect on the main diagonal is irrelevant, since there is a 0 on it, but in others kind of matrix this has to be done, right?

Last edited by coltson; January 9th, 2018 at 09:49 AM.
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