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Inverse of a 4x4 matrix?

I'm intrigued to know how this can be found. Would anyone care to enlighten me or provide a link to a method?

Cheers :rolleyes:

And by hand. I know you can do it with a calculator.
Reply 1
Avatar for nota bene
nota bene
15
Row reduction.

set up the matrix you want to invert (A)|(I) and work to get (I)|(A^-1)

(There are other methods)
Reply 2
Avatar for DFranklin
DFranklin
18
http://en.wikipedia.org/wiki/Gauss%E2%80%93Jordan_elimination (the same as Nota's suggestion, but with some explanation)
Reply 3
Avatar for bigmo7
bigmo7
17
Ah man 3 by 3 was enough for me! (Hey that rhymes!)
tbh you can use the same method that use used for 3x3 extended to 4x4 but that takes forevarrr.
Reply 5
Avatar for Jamie Innes
Jamie Innes
OP
1
Totally Tom
tbh you can use the same method that use used for 3x3 extended to 4x4 but that takes forevarrr.


Lol that was what I was thinking.

Does Gauss-Jordan elimination still work?
Reply 6
Avatar for Phugoid
Phugoid
The determinant of a square matrix, A A is always given by:

A−1=adj(A)det⁡(A) A^{-1} = \frac{\text{adj}(A)}{\det(A)}
Jamie Innes
Lol that was what I was thinking.

Does Gauss-Jordan elimination still work?

?

err yes.
Same as with a 3x3.
bigmo7
Ah man 3 by 3 was enough for me! (Hey that rhymes!)

Seconded. Although you do it the same way.
Phugoid
The determinant of a square matrix, A A is always given by:

A−1=adj(A)det⁡(A) A^{-1} = \frac{\text{adj}(A)}{\det(A)}
You mean "inverse of a square matrix".
Reply 11
Avatar for Phugoid
Phugoid
DFranklin
You mean "inverse of a square matrix".


Woopsies. Typo indeed.
Reply 12
Avatar for Jamie Innes
Jamie Innes
OP
1
Phugoid
The determinant of a square matrix, A A is always given by:

A−1=adj(A)det⁡(A) A^{-1} = \frac{\text{adj}(A)}{\det(A)}


Sorry, what's the adj(A)?
Reply 13
Avatar for Zhen Lin
Zhen Lin
12
The adjugate matrix. Finding it is about as difficult as finding the inverse matrix though, so I don't think that's very helpful...
Zhen Lin
The adjugate matrix. Finding it is about as difficult as finding the inverse matrix though, so I don't think that's very helpful...
Agreed. In fact, I've never seen someone advocate using the adjugate matrix unless they were going to then use the recursive formulation of the determinant to calculate each term. For large n, that's about as bad a plan as you can get: if memory serves you need roughly (n+1)! calculations to find the inverse that way.
Yeah, rref is the best way to do it already
Reply 16
Avatar for lnx
lnx
http://www.intmath.com/Matrices-determinants/5_Inverse-matrix.php have a look at this it shows how you can find the inverse of 4x4 5x5 and 6x6, finding 4x4 is basically the same as 3x3.
lnx
http://www.intmath.com/Matrices-determinants/5_Inverse-matrix.php have a look at this it shows how you can find the inverse of 4x4 5x5 and 6x6, finding 4x4 is basically the same as 3x3.
I can't see the method they finally use for the 4x4 matrix, but if it involves determinants or adjoint/adjugate matrices, it's almost certainly grossly inefficient.
Reply 18
Avatar for lnx
lnx
DFranklin
I can't see the method they finally use for the 4x4 matrix, but if it involves determinants or adjoint/adjugate matrices, it's almost certainly grossly inefficient.


Why?
See post #15. Basically, to find each element of the adjugate, you need to find a 3x3 determinant. So that's 16 3x3 determinants to calculate. Each 3x3 determinant requires 12 multiplies and 5 adds, so you need 192 multiplies and 80 additions.
If you want a 5x5 it gets worse. To find each element of the adjugate you need to find a 4x4 determinant. That's 25 4x4 determinants. Each 4x4 determinant requires finding 4 3x3 determinants, so takes at least 48 multiplies and 20 adds. So you need well over a thousand multiplies to find the adjugate.
6x6, you need something like 9000 multiplies.

In contrast, Gauss-Jordan elimination will solve a 6x6 matrix in a couple of hundred multiplies. 4x4 matrix will be comfortably under a hundred ops.

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