3x3 inverse matrices; explanations

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

Posts: 83
Joined: Tue Mar 09, 2010 1:42 am UTC

3x3 inverse matrices; explanations

Postby xepher » Mon Nov 29, 2010 6:55 pm UTC

I was messing around with finding a general formula for the inverse of a 3x3 matrix this morning, and I came up with an answer from using the trick with augmenting the matrix with the identity matrix, then essentially just brute forcing my way to the answer and hoping the terms worked themselves out. Also, I just want to say that the feeling you get when you figure out something that you never knew before is pretty good.

What I am interested by are the why the values of the elements are like so. I got through it the algebraic way, but is there a somewhat more intuitive answer as to why? I'm probably not sounding clear enough; I'll put in the latex and the like later, when I have more time.

Also, a related question. Why does the "augment it with an identity matrix" method work?

User avatar
Posts: 882
Joined: Sun Jan 13, 2008 9:04 pm UTC

Re: 3x3 inverse matrices; explanations

Postby NathanielJ » Mon Nov 29, 2010 7:11 pm UTC

A general formula for 3x3 inverses (and larger inverses) can be derived from the adjugate matrix, which explains why there is such symmetry in all the terms of the inverse. Basically each entry in the inverse matrix is the determinant of a matrix that is related to A in some way, all divided by the determinant of A.
Homepage: http://www.njohnston.ca
Conway's Game of Life: http://www.conwaylife.com

User avatar
Poster with most posts but no title.
Posts: 11129
Joined: Sat Jan 27, 2007 7:27 pm UTC
Location: E pur si muove

Re: 3x3 inverse matrices; explanations

Postby Yakk » Mon Nov 29, 2010 7:28 pm UTC

As an invertible matrix, you can be expressed as a compilation of primitive elements.

When you augment the matrix and apply the primitive transformations to the left hand side, the change you do on the right-hand side is in some sense the inverse operation.

Take your matrix A. Multiply it by a matrix B, and do the same with I.


If A*B = I, then I*B is B = A-1.

When you manipulate the rows of your augmented matrix, each step is actually multiplying your matrix by a primitive matrix we'll call p_i.

ie, if you scale a row by a factor of 2, can you imagine the matrix that would do that? Or add one row to another? Or swap two rows?

Each of these primitive matrices are invertible.

So when we do:
A * p_0 * p_1 * ... * p_n
where each of those is a successive primitive matrix we are applying in order corresponding to your "manipulation of the matrix to turn it into the identity"...

We also calculate this:
I * p_0 * ... * p_n
which is the right-hand side of your augmented matrix.

Let B := p_0 * ... * p_n.

Then if A*B = I, what is I*B? (hint: I already told you!)
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

User avatar
Posts: 1772
Joined: Thu Apr 05, 2007 7:31 pm UTC

Re: 3x3 inverse matrices; explanations

Postby antonfire » Tue Nov 30, 2010 1:52 am UTC

Screw algebra.

If v1, v2, ..., vn are the columns of your matrix M, then finding the first coefficient of M-1w amounts to finding the component of w in the direction of v1. This component is the ratio of how far w is removed from the plane spanned by v2, ..., vn to how far v1 is removed from that plane. This is just the ratio of the volumes of the parallelepipeds given by w, v2, ..., vn and by v1, v2, ..., vn. Similarly for other coefficients.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Return to “Mathematics”

Who is online

Users browsing this forum: mashnut and 5 guests