Simple 3x3 matrix inverse code (C++)


Question

What's the easiest way to compute a 3x3 matrix inverse?

I'm just looking for a short code snippet that'll do the trick for non-singular matrices, possibly using Cramer's rule. It doesn't need to be highly optimized. I'd prefer simplicity over speed. I'd rather not link in additional libraries.

1
36
2/15/2012 11:12:55 PM

Accepted Answer

Why don't you try to code it yourself? Take it as a challenge. :)

For a 3×3 matrix

alt text
(source: wolfram.com)

the matrix inverse is

alt text
(source: wolfram.com)

I'm assuming you know what the determinant of a matrix |A| is.

Images (c) Wolfram|Alpha and mathworld.wolfram (06-11-09, 22.06)

32
3/21/2019 11:59:32 PM

Here's a version of batty's answer, but this computes the correct inverse. batty's version computes the transpose of the inverse.

// computes the inverse of a matrix m
double det = m(0, 0) * (m(1, 1) * m(2, 2) - m(2, 1) * m(1, 2)) -
             m(0, 1) * (m(1, 0) * m(2, 2) - m(1, 2) * m(2, 0)) +
             m(0, 2) * (m(1, 0) * m(2, 1) - m(1, 1) * m(2, 0));

double invdet = 1 / det;

Matrix33d minv; // inverse of matrix m
minv(0, 0) = (m(1, 1) * m(2, 2) - m(2, 1) * m(1, 2)) * invdet;
minv(0, 1) = (m(0, 2) * m(2, 1) - m(0, 1) * m(2, 2)) * invdet;
minv(0, 2) = (m(0, 1) * m(1, 2) - m(0, 2) * m(1, 1)) * invdet;
minv(1, 0) = (m(1, 2) * m(2, 0) - m(1, 0) * m(2, 2)) * invdet;
minv(1, 1) = (m(0, 0) * m(2, 2) - m(0, 2) * m(2, 0)) * invdet;
minv(1, 2) = (m(1, 0) * m(0, 2) - m(0, 0) * m(1, 2)) * invdet;
minv(2, 0) = (m(1, 0) * m(2, 1) - m(2, 0) * m(1, 1)) * invdet;
minv(2, 1) = (m(2, 0) * m(0, 1) - m(0, 0) * m(2, 1)) * invdet;
minv(2, 2) = (m(0, 0) * m(1, 1) - m(1, 0) * m(0, 1)) * invdet;

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