solve a x = b , where b is a column vector . this is more efficient than computing the inverse in one-shot cases . concode_field_sep Vec3 ex concode_elem_sep Vec3 ey concode_elem_sep Vec3 ez concode_elem_sep long serialVersionUID concode_elem_sep Mat33 IDENTITY concode_field_sep Vec2 solve22 concode_elem_sep Vec3 solve33 concode_elem_sep void set concode_elem_sep void set concode_elem_sep Vec3 mul concode_elem_sep void mul22ToOut concode_elem_sep void getSymInverse33 concode_elem_sep void getInverse22 concode_elem_sep Vec2 mul22 concode_elem_sep void setScaleTransform concode_elem_sep int hashCode concode_elem_sep boolean equals concode_elem_sep void mulToOut concode_elem_sep void solve33ToOut concode_elem_sep void setZero concode_elem_sep void mulToOutUnsafe concode_elem_sep void setIdentity concode_elem_sep void mul22ToOutUnsafe
void function ( Vec2 arg0 , Vec2 arg1 ) { final float loc0 = ex . x , loc1 = ey . x , loc2 = ex . y , loc3 = ey . y ; float loc4 = loc0 * loc3 - loc1 * loc2 ; if ( loc4 != 0.0f ) { loc4 = 1.0f / loc4 ; } arg1 . x = loc4 * ( loc3 * arg0 . x - loc1 * arg0 . y ) ; arg1 . y = loc4 * ( loc0 * arg0 . y - loc2 * arg0 . x ) ; }