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    <p>To follow-up on <a href="https://stackoverflow.com/a/2624561/650012">pkhaler</a>'s and <a href="https://stackoverflow.com/a/7596981/650012">Robin Hilliard</a>'s excellent responses above, here is Robin's ActionScript 3 code converted into a C# method. Hopefully this can save some typing for other C# developers, as well as C/C++ and Java developers in need of a 4x4 matrix inversion function:</p> <pre><code>public static double[,] GetInverse(double[,] a) { var s0 = a[0, 0] * a[1, 1] - a[1, 0] * a[0, 1]; var s1 = a[0, 0] * a[1, 2] - a[1, 0] * a[0, 2]; var s2 = a[0, 0] * a[1, 3] - a[1, 0] * a[0, 3]; var s3 = a[0, 1] * a[1, 2] - a[1, 1] * a[0, 2]; var s4 = a[0, 1] * a[1, 3] - a[1, 1] * a[0, 3]; var s5 = a[0, 2] * a[1, 3] - a[1, 2] * a[0, 3]; var c5 = a[2, 2] * a[3, 3] - a[3, 2] * a[2, 3]; var c4 = a[2, 1] * a[3, 3] - a[3, 1] * a[2, 3]; var c3 = a[2, 1] * a[3, 2] - a[3, 1] * a[2, 2]; var c2 = a[2, 0] * a[3, 3] - a[3, 0] * a[2, 3]; var c1 = a[2, 0] * a[3, 2] - a[3, 0] * a[2, 2]; var c0 = a[2, 0] * a[3, 1] - a[3, 0] * a[2, 1]; // Should check for 0 determinant var invdet = 1.0 / (s0 * c5 - s1 * c4 + s2 * c3 + s3 * c2 - s4 * c1 + s5 * c0); var b = new double[4, 4]; b[0, 0] = ( a[1, 1] * c5 - a[1, 2] * c4 + a[1, 3] * c3) * invdet; b[0, 1] = (-a[0, 1] * c5 + a[0, 2] * c4 - a[0, 3] * c3) * invdet; b[0, 2] = ( a[3, 1] * s5 - a[3, 2] * s4 + a[3, 3] * s3) * invdet; b[0, 3] = (-a[2, 1] * s5 + a[2, 2] * s4 - a[2, 3] * s3) * invdet; b[1, 0] = (-a[1, 0] * c5 + a[1, 2] * c2 - a[1, 3] * c1) * invdet; b[1, 1] = ( a[0, 0] * c5 - a[0, 2] * c2 + a[0, 3] * c1) * invdet; b[1, 2] = (-a[3, 0] * s5 + a[3, 2] * s2 - a[3, 3] * s1) * invdet; b[1, 3] = ( a[2, 0] * s5 - a[2, 2] * s2 + a[2, 3] * s1) * invdet; b[2, 0] = ( a[1, 0] * c4 - a[1, 1] * c2 + a[1, 3] * c0) * invdet; b[2, 1] = (-a[0, 0] * c4 + a[0, 1] * c2 - a[0, 3] * c0) * invdet; b[2, 2] = ( a[3, 0] * s4 - a[3, 1] * s2 + a[3, 3] * s0) * invdet; b[2, 3] = (-a[2, 0] * s4 + a[2, 1] * s2 - a[2, 3] * s0) * invdet; b[3, 0] = (-a[1, 0] * c3 + a[1, 1] * c1 - a[1, 2] * c0) * invdet; b[3, 1] = ( a[0, 0] * c3 - a[0, 1] * c1 + a[0, 2] * c0) * invdet; b[3, 2] = (-a[3, 0] * s3 + a[3, 1] * s1 - a[3, 2] * s0) * invdet; b[3, 3] = ( a[2, 0] * s3 - a[2, 1] * s1 + a[2, 2] * s0) * invdet; return b; } </code></pre>
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    1. POEfficient 4x4 matrix inverse (affine transform)
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    1. PTAnswer
    1. USCommunity
    1. USAnders Gustafsson
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    1. COThanks for this, it is really helpful!
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      1. USAlexander Karatarakis
    2. COAlso, confirming that the algorithm works correctly. :)
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      1. USAlexander Karatarakis
 

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