Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Need help to compute $\det A$ where $$A=\left(\begin{matrix}36&60&72&37\\43&71&78&34\\44&69&73&32\\30&50&65&38\end{matrix} \right)$$

How would one use elementary operations to calculate the determinant easily?

I know that $\det A=1$

share|improve this question
    
+1 $~~~~~~~~~~~~~$ –  Babak S. Feb 1 '13 at 14:51
    
If you care about the number of operations, here's a trick with 30 multiplications: bitbucket.org/eigen/eigen/src/… –  user161143 Jul 1 at 20:49

3 Answers 3

I suggest Gaussian Elimination till upper triangle form or further but keep track of the effect of each elementary. see here for elementary's effect on det

share|improve this answer
    
or use softwares such as matlab or free sources like wolframalpha if dimension is too high –  mez Jan 24 '13 at 19:41
    
Trying to do something like this by hand is a real waste of time. Any computer algebra system will give you the answer instantly. –  Robert Israel Jan 24 '13 at 19:44
    
@RobertIsrael It might not be a waste of time if you are in exam.:) –  mez Jan 24 '13 at 19:45
    
Nobody in their right mind would ask for a $4 \times 4$ determinant with all distinct $2$-digit entries on an exam. –  Robert Israel Jan 24 '13 at 19:55
    
@RobertIsrael You cannot nullify this person's curiosity for a method that is of use by saying it is a waste of time. 4*4 determinants have appeared on my exam though not all 2-digit, but complicated enough. You are assuming too much, who are you to judge what is a waste of time for him anyway? –  mez Jan 24 '13 at 19:59

Here's one way to do it without fractions. You could start by subtracting row $2$ from row $3$ to get $$ \left[ \begin {array}{cccc} 36&60&72&37\\ 43&71&78&34\\ 1&-2&-5&-2\\ 30&50&65&38 \end {array} \right]$$ Then subtract $36$, $43$, and $30$ times row $3$ from rows $1$, $2$ and $4$ respectively to get $$ \left[ \begin {array}{cccc} 0&132&252&109\\ 0&157&293&120\\ 1&-2&-5&-2\\ 0&110&215& 98\end {array} \right]$$ Expanding by minors in the first column, we just need one $3 \times 3$ determinant, which is $$132 \times 293 \times 98 + 252 \times 120 \times 110 + 109 \times 157 \times 215 - 132 \times 120 \times 215 - 252 \times 157 \times 98 - 109 \times 293 \times 110 = 1$$ I hope you're allowed to use a calculator for that...

share|improve this answer

For a 4x4 determinant I would probably use the method of minors: the 3x3 subdeterminants have a convenient(ish) mnemonic as a sum of products of diagonals and broken diagonals, with all the diagonals in one direction positive and all the diagonals in the other direction negative; this lets you compute the determinant of e.g. the bottom-right 3x3 as 71*73*38 + 78*32*50 + 34*69*65 - 34*73*50 - 71*32*65 - 78*69*38. That's probably slightly less than a 5-minute calculation with pencil and paper and a 1-minute calculation with a calculator, which means you could find the overall determinant in maybe 5 minutes with calculator, 15-20 with pencil and paper. Not blazingly fast, of course, but for me I suspect it'd be marginally faster than Gaussian Elimination, and the all-integer nature of it is (for me, at least) a minor plus. Alternately, the subdeterminants can be computed by taking minors again; this cuts down slightly on the number of multiplications per subdeterminant(from 12 to 9) and gives a total of 40 multiplications to compute the 4x4 determinant.

share|improve this answer
    
just out of curiosity, what you mentioned is what I thought to be the standard way, do you know any other way? –  mez Jan 24 '13 at 20:16
    
@mezhang Well, there's the direct approach in terms of summing over all the permutations, but you're right that this is more or less the canonical brute-force method. I'll add another method that actually works pretty well here... –  Steven Stadnicki Jan 24 '13 at 20:28
    
Could you explain or show me a link to the summing over all permutations method? I don't know it. –  mez Jan 24 '13 at 20:33
    
@mezhang It's one of the definitions of the determinant; see en.wikipedia.org/wiki/Leibniz_formula_for_determinants , for instance. –  Steven Stadnicki Jan 24 '13 at 21:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.