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    <p>You should be able to exploit the fact that the matrix is affine to speed things up over a full inverse. Namely, if your matrix looks like this</p> <pre><code>A = [ M b ] [ 0 1 ] </code></pre> <p>where A is 4x4, M is 3x3, b is 3x1, and the bottom row is (0,0,0,1), then </p> <pre><code>inv(A) = [ inv(M) -inv(M) * b ] [ 0 1 ] </code></pre> <p>Depending on your situation, it may be faster to compute the result of inv(A) * x instead of actually forming inv(A). In that case, things simplify to</p> <pre><code>inv(A) * [x] = [ inv(M) * (x - b) ] [1] = [ 1 ] </code></pre> <p>where x is a 3x1 vector (usually a 3D point).</p> <p>Lastly, if M represents a rotation (i.e. its columns are orthonormal), then you can use the fact that inv(M) = transpose(M). Then computing the inverse of A is just a matter of subtracting the translation component, and multiplying by the transpose of the 3x3 part.</p> <p>Note that whether or not the matrix is orthonormal is something that you should know from the analysis of the problem. Checking it during runtime would be fairly expensive; although you might want to do it in debug builds to check that your assumptions hold.</p> <p>Hope all of that is clear...</p>
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    1. POEfficient 4x4 matrix inverse (affine transform)
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    1. POEfficient 4x4 matrix inverse (affine transform)
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    1. COSo the first formula you got from "blockwise inversion" (http://en.wikipedia.org/wiki/Invertible_matrix)? Or is there another mental trick? I'm not quite sure about the inv(A) * x = inv(M) * (x - b). First, they're different sizes - do you remove a row from A on the left or add a row on the right? Second, I'm not sure how that equation comes about. Third, I'm not sure what you're solving for in that equation. Oliver keeps mentioning not computing inverses symbolically, but I don't know what that means - I need the inverse to do inverse transform. If you have time I'd like to know.
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    2. COI edited the inv(A) * x formula to make the dimensions clearer. The first formula was from http://en.wikipedia.org/wiki/Affine_transformation. Forgetting about affine transforms for a minute, in general, when you're solving A*x = b, you want the solution inv(A)*b. But often times you don't need to actual form inv(A), just compute what the product *would* be. Back to affine transforms, in 3D applications, you might not actually need the inverse of the matrix, you just want the inverse transform *acting on* (multiplying) a vector. If that's the case, then using the formula might be faster.
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    3. COEven if you do need to store the matrix inverse, you can use the fact that it's affine to reduce the work computing the inverse, since you only need to invert a 3x3 matrix instead of 4x4. And if you know that it's a rotation, computing the transpose is *much* faster than computing the inverse, and in this case, they're equivalent.
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