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Nicholas Kong

Why is Guass method better for finding determinants of a 3x3 and 4x4 matrix?

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Why is Guass method better for finding determinants of a 3x3 and 4x4 matrix? 

 

The book says it saves time... 

 

But I still find myself spending the same amount of time using the Guass method

on the matrix and then applying the long formula below:

 

aei + bfg + cdh - hfa - idb - gec

 

for calculating the determinant of a 3x3 matrix. 

 

Assuming of course the 3x3 matrix looked like this: Imagine a left and right large brackets around the alphabets below!

 

a b c

 

d e f

 

g h i

 

 

The book also put a negative sign in front of the determinant mentions when the book used the Guass method on a 3x3 matrix and on a 4x4 matrix. Why is that?

 

I never had to put a negative sign in front of my determinant before when I was NOT using Guass method.

Edited by warnexus

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The formula you have written is the fastest method to compute the determinant of a 3x3 matrix. But it is not generalizable (it only works in that form in the 3x3 case). It is however possible to compute closed form formulas for other dimensions (for example 4x4 matrices). The Gauss method has a better asymptotic complexity than other general methods, but it is not the fastest for small matrices.

EDIT: The change of sign is probably motivated by the use of row switching operations which change the sign of the determinant. Your textbook surely contains a more in depth explanation of why you need to do this (if it does not contain such explanation then change your textbook). Edited by apatriarca

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But after Gaussian elimination, the elements in d,g,h are 0, and so all terms in aei + bfg + cdh - hfa - idb - gec vanish, except for aei. But indeed, for 3x3 matrices it's probably still faster to compute it using your formula, but the number of terms in that closed form is n!, with each term consisting of n factors (where n is the number of dimensions), so the complexity grows much faster than with gaussian elimination, where after bringing your matrix to reduced form, always just a single term remains. 

 

The reason they put a negative sign before the determinant was probably because they did an odd number of row exchanges. Each time you exchange rows, the sign of the determinant changes.

Edited by quasar3d

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But after Gaussian elimination, the elements in d,g,h are 0, and so all terms in aei + bfg + cdh - hfa - idb - gec vanish, except for aei. But indeed, for 3x3 matrices it's probably still faster to compute it using your formula, but the number of terms in that closed form is n!, with each term consisting of n factors (where n is the number of dimensions), so the complexity grows much faster than with gaussian elimination, where after bringing your matrix to reduced form, always just a single term remains. 

 

The reason they put a negative sign before the determinant was probably because they did an odd number of row exchanges. Each time you exchange rows, the sign of the determinant changes.

Nice observation with the odd number of row exchange! I agree with that too!

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