高斯消元法求逆矩阵 matlab,求救！用高斯消元法求二进制矩阵的逆！
float Inverse(CLAYMATRIX& mOut, const CLAYMATRIX& rhs)
{
CLAYMATRIX m(rhs);
DWORD is[4];
DWORD js[4];
float fDet = 1.0f;
int f = 1;
for (int k = 0; k < 4; k ++)
{
// 第一步，全选主元
float fMax = 0.0f;
for (DWORD i = k; i < 4; i ++)
{
for (DWORD j = k; j < 4; j ++)
{
const float f = Abs(m(i, j));
if (f > fMax)
{
fMax = f;
is[k] = i;
js[k] = j;
}
}
}
if (Abs(fMax) < 0.0001f)
return 0;
if (is[k] != k)
{
f = -f;
swap(m(k, 0), m(is[k], 0));
swap(m(k, 1), m(is[k], 1));
swap(m(k, 2), m(is[k], 2));
swap(m(k, 3), m(is[k], 3));
}
if (js[k] != k)
{
f = -f;
swap(m(0, k), m(0, js[k]));
swap(m(1, k), m(1, js[k]));
swap(m(2, k), m(2, js[k]));
swap(m(3, k), m(3, js[k]));
}
// 计算行列值
fDet *= m(k, k);
// 计算逆矩阵
// 第二步
m(k, k) = 1.0f / m(k, k);
// 第三步
for (DWORD j = 0; j < 4; j ++)
{
if (j != k)
m(k, j) *= m(k, k);
}
// 第四步
for (DWORD i = 0; i < 4; i ++)
{
if (i != k)
{
for (j = 0; j < 4; j ++)
{
if (j != k)
m(i, j) = m(i, j) - m(i, k) * m(k, j);
}
}
}
// 第五步
for (i = 0; i < 4; i ++)
{
if (i != k)
m(i, k) *= -m(k, k);
}
}
for (k = 3; k >= 0; k --)
{
if (js[k] != k)
{
swap(m(k, 0), m(js[k], 0));
swap(m(k, 1), m(js[k], 1));
swap(m(k, 2), m(js[k], 2));
swap(m(k, 3), m(js[k], 3));
}
if (is[k] != k)
{
swap(m(0, k), m(0, is[k]));
swap(m(1, k), m(1, is[k]));
swap(m(2, k), m(2, is[k]));
swap(m(3, k), m(3, is[k]));
}
}
mOut = m;
return fDet * f;
}