## 3x3 inverse matrices; explanations

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xepher
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### 3x3 inverse matrices; explanations

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?

NathanielJ
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### Re: 3x3 inverse matrices; explanations

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.
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Yakk
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### Re: 3x3 inverse matrices; explanations

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.

A*B
I*B

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

antonfire
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### Re: 3x3 inverse matrices; explanations

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.
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