Question

如何在mathematica中编写一个可以用Givens旋转求解线性方程组的程序。

代码尝试：

ar = CoefficientArrays[
   {x + y + z == 2, 3 x - 2 y + z == 4, x - y + 5 z == 6},
   {x, y, z}];

an = Normal[ar];
b = an[[2]];
n = Length[b];
Do[Do[a[i, j] = b[[i]][[j]], {j, 1, n}], {i, 1, n}];
r1 = (a[1, 1]^2 + a[1, 2]^2)^(1/2);
c = a[1, 1]/r1;
s = a[1, 2]/r1;

Answer 1

你在找这样的东西吗？

关注您的代码。

{r1, c, s}

{Sqrt [2]，1 / Sqrt [2]，1 / Sqrt [2]}

{a, b} = LinearSolve[{{c, -s}, {s, c}}, {r1, 0}]

{1，-1}

algorithm from wikipedia

GivensRotation[a_, b_] := Which[
   b == 0, c = Sign[a]; s = 0; r = Abs[a],
   a == 0, c = 0; s = -Sign[b]; r = Abs[b],
  Abs[a] > Abs[b],
  t = b/a; u = Sign[a]*Abs[Sqrt[1 + t*t]];
  c = 1/u; s = -c*t; r = a*u,
  True,
  t = a/b;
  u = Sign[b]*Abs[Sqrt[1 + t*t]];
  s = -1/u; c = -s*t; r = b*u
  ]

GivensRotation[a, b];

{r, c, s}

{Sqrt [2]，1 / Sqrt [2]，1 / Sqrt [2]}

修改

我不熟悉用Givens旋转解决问题。以下是解决联立方程的其他方法，只是为了感兴趣。

Solve[{ x + y + z == 2, 3 x - 2 y + z == 4, x - y + 5 z == 6}, {x, y, z}]


{{x - ＆gt; 1，y - ＆gt; 0，z - ＆gt; 1}}

也

LinearSolve[{{1, 1, 1}, {3, -2, 1}, {1, -1, 5}}, {2, 4, 6}]


{1,0,1}}

或

Inverse[{{1, 1, 1}, {3, -2, 1}, {1, -1, 5}}].{2, 4, 6}


{1,0,1}}

在mathematica中写一个程序，用Givens旋转

1 个答案: