January 7th, 2018, 03:46 PM  #1 
Newbie Joined: Jan 2018 From: brazil Posts: 4 Thanks: 0  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 10:52 PM. 
January 7th, 2018, 07:22 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,974 Thanks: 1850 
What do you get if you add the third row to the fifth row? What do you notice about the first and second columns? 
January 7th, 2018, 07:31 PM  #3 
Newbie Joined: Jan 2018 From: Tunis, Tunisia Posts: 10 Thanks: 4 Math Focus: Analysis 
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 And now we get a determinant of $\displaystyle 4\times 4$ matrix.\det A=5\begin{vmatrix}1&2&3&4\\4&3&4&0\\8&7&6&2\\12&9& 10&3\end{vmatrix} $ 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. 
January 7th, 2018, 08:47 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,974 Thanks: 1850 
The 4 × 4 determinant wouldn't be what you gave.

January 7th, 2018, 09:12 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 520 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics 
matlab

January 8th, 2018, 03:53 AM  #6  
Newbie Joined: Jan 2018 From: Tunis, Tunisia Posts: 10 Thanks: 4 Math Focus: Analysis 
Hi ! Quote:
OK ! I made a mistake , it should be : $\displaystyle \det A=5\begin{vmatrix}4&3&4&0\\8&7&6&2\\12&9&10&3\\16& 12&13&4\end{vmatrix}$  
January 8th, 2018, 05:13 AM  #7  
Senior Member Joined: Oct 2009 Posts: 629 Thanks: 192  Quote:
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.  
January 8th, 2018, 05:43 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,974 Thanks: 1850  
January 8th, 2018, 06:30 AM  #9 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
$\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 06:47 AM. Reason: typo in line 4 of second matrix 
January 9th, 2018, 10:37 AM  #10 
Newbie Joined: Jan 2018 From: brazil Posts: 4 Thanks: 0 
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 10:49 AM. 

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