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This question already has an answer here:

Edit: I don't think this is a duplicate. That question is generally about efficient algorithms for calculating the determinant. While my question is about practical tips and tricks which can be used when doing this sort of thing by hand.


During one of the questions I was solving I came across the following determinant:

$\left|\matrix{t-1&3&0&-3\\2&t+6&0&-13\\0&3&t-1&-3\\1&4&0&t-8}\right|$

I solved it directly by expanding from the third column and got the result $(t-1)^4$.

However, along the way I made a couple of calculation errors and finding them took me a while. This got me thinking that maybe there are some tips and tricks I could use to calculate this in a faster less error prone method. Or maybe, at least know that I got the right answer in some way?

Any tips that could interest me?

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marked as duplicate by Namaste linear-algebra Jul 25 '18 at 18:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Perhaps merge with math.stackexchange.com/questions/2857224/… $\endgroup$ – Shogun Jul 25 '18 at 17:58
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    $\begingroup$ @Geronimo I don't think these questions are the same. That question is generally about efficient algorithms for calculating the determinant. While my question is about practical tips and tricks which can be used when doing this sort of thing by hand. $\endgroup$ – Jason Aug 2 '18 at 16:35
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    $\begingroup$ I see. I don't think it's a duplicate then. $\endgroup$ – Shogun Aug 2 '18 at 17:00
  • $\begingroup$ The intent is obviously to develop by the third colon. For this matrix that is the way to go. You reduce at basically no effort to a smaller matrix. Then the question remains how to compute a det for 3 by 3 matrix efficiently, which likely could be asked separately. $\endgroup$ – quid Sep 3 '18 at 12:08
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If you put the matrix in triangular form you can take the determinant by taking the product of the diagonal entries

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  • $\begingroup$ Wonderful suggestion. $\endgroup$ – Faraad Armwood Jul 25 '18 at 18:01
  • $\begingroup$ That's a good suggestion, I'll try it on my example and see if it works well for that one. $\endgroup$ – Jason Jul 25 '18 at 18:04

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