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Subject: Re: 4x4 matrix inversion



The following holds for an invertible matrix:
M^-1 = adj(M) / det(M)

This is the direct formula and while not practical for large matrices I
guess it is ok for such a small matrix. To compute the inverse then, you
need a way to compute det(M) and adj(M).

Minors
------

Let M be a nxn matrix. Let Sij of M be a (n-1)x(n-1) matrix obtained
from M by removing the row i and the column j. An Sij is called a minor
of M.

The set of det(Sij) for all i and j are called the subdeterminants of
the matrix M.

Adjungate
---------

Form a matrix A such that A[i, j] = -1^(i + j) * det(Sji). This is
called the adjungate of M, adj(M).

Determinant
-----------

If B is a 1x1 matrix:
B = [b]
then
det(B) = b

Let M be a nxn matrix, with n > 1. Let [x1, x2, ... , xn] be the first
row of M. Then det(M) = sum[j = 1 to n](-1^(1 + j) * xj * det(Sij))

Example
-------

Let n = 3

M = [a b c]
[d e f]
[g h i]

Minors and subdeterminants

S11 = [e f] => det(S11) = e * det(i) - f * det(h) = e * i - f * h
[h i]
S12 = [d f] => det(S12) = d * det(i) - f * det(g) = d * i - f * g
[g i]
S13 = [d e] => det(S13) = d * det(h) - e * det(g) = d * h - e * g
[g h]

Similarly
det(S21) = bi - ch
det(S22) = ai - cg
det(S23) = ah - bg
det(S31) = bf - ce
det(S32) = af - dc
det(S33) = ae - bd

Determinant

det(M) = a * det(S11) - b * det(S12) + c * det(S13)
= a * (ei - fh) - b * (di - fg) + c * (dh - eg)

Adjungate

adj(M) = [+det(S11), -det(S21), +det(S31)]
[-det(S12), +det(S22), -det(S32)]
[+det(S13), -det(S23), +det(S33)]

Finally, we are ready to calculate M^-1:
M^-1 = adj(M) / det(M)

Deriving the n = 4 version is left as an exercise:)

--
Kalle Rutanen
http://kaba.hilvi.org

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