Inverting a 4x4 matrix

Using C code:

float invf(int i,int j,const float *m) { int o = 2+(j-i); i += 4+o; j += 4-o; #define e(a,b) m[ ((j+b)%4)*4 + ((i+a)%4) ] float inv = + e(+1,-1)*e(+0,+0)*e(-1,+1) + e(+1,+1)*e(+0,-1)*e(-1,+0) + e(-1,-1)*e(+1,+0)*e(+0,+1) - e(-1,-1)*e(+0,+0)*e(+1,+1) - e(-1,+1)*e(+0,-1)*e(+1,+0) - e(+1,-1)*e(-1,+0)*e(+0,+1); return (o%2)?inv : -inv; #undef e } bool inverseMatrix4x4(const float *m, float *out) { float inv[16]; for(int i=0;i<4;i++) for(int j=0;j<4;j++) inv[j*4+i] = invf(i,j,m); double D = 0; for(int k=0;k<4;k++) D += m[k] * inv[k*4]; if (D == 0) return false; D = 1.0 / D; for (int i = 0; i < 16; i++) out[i] = inv[i] * D; return true; }

In the first image you can see the pattern of positive and negative factors used in the calculation of inv[n].

To evaluate each cell, the pattern is shifted right up by (2+j-i), as illustrated in the second image.