Matrix 为什么';t反替换Solve[]的结果是否给出预期结果?
我有这个矩阵Matrix 为什么';t反替换Solve[]的结果是否给出预期结果?,matrix,wolfram-mathematica,Matrix,Wolfram Mathematica,我有这个矩阵 a = {{2, -2, -4}, {-2, 5, -2}, {-4, -2, 2}} 然后我解了一个方程,其中有一个条目缺失。方程的形式如下 逆[p].a.p==q 式中,p是带有缺失项(x5)的3x3矩阵,q是给定的3x3矩阵 Solve[Inverse[( { {1/Sqrt[5], 4/(3 Sqrt[5]), -2/3}, {-2/Sqrt[5], 2/(3 Sqrt[5]), -2/6}, {0, x5, -2/3} }
a = {{2, -2, -4}, {-2, 5, -2}, {-4, -2, 2}}
然后我解了一个方程,其中有一个条目缺失。方程的形式如下
逆[p].a.p==q
式中,p是带有缺失项(x5)的3x3矩阵,q是给定的3x3矩阵
Solve[Inverse[( {
{1/Sqrt[5], 4/(3 Sqrt[5]), -2/3},
{-2/Sqrt[5], 2/(3 Sqrt[5]), -2/6},
{0, x5, -2/3}
} )].a.( {
{1/Sqrt[5], 4/(3 Sqrt[5]), -2/3},
{-2/Sqrt[5], 2/(3 Sqrt[5]), -2/6},
{0, x5, -2/3}
} ) == ( {
{6, 0, 0},
{0, 6, 0},
{0, 0, -3}
} )]
Mathematica可以很容易地解决这个问题,结果是x5->-(Sqrt[5]/3)。
但是如果我检查一下,结果会很奇怪:
In[2]:= Inverse[( {
{1/Sqrt[5], 4/(3 Sqrt[5]), -2/3},
{-2/Sqrt[5], 2/(3 Sqrt[5]), -2/6},
{0, -Sqrt[5]/3, -2/3}
} )].a.( {
{1/Sqrt[5], 4/(3 Sqrt[5]), -2/3},
{-2/Sqrt[5], 2/(3 Sqrt[5]), -2/6},
{0, -Sqrt[5]/3, -2/3}
} )
Out[2]= {{6/5 - (2 (-(2/Sqrt[5]) - 2 Sqrt[5]))/Sqrt[5],
8/5 + (2 (-(2/Sqrt[5]) - 2 Sqrt[5]))/(3 Sqrt[5]), -(4/Sqrt[5]) +
1/3 (2/Sqrt[5] + 2 Sqrt[5])}, {-((
2 (-(8/(3 Sqrt[5])) + (4 Sqrt[5])/3))/Sqrt[5]) + (
4/(3 Sqrt[5]) + (4 Sqrt[5])/3)/Sqrt[5],
10/3 + (2 (-(8/(3 Sqrt[5])) + (4 Sqrt[5])/3))/(3 Sqrt[5]) + (
4 (4/(3 Sqrt[5]) + (4 Sqrt[5])/3))/(3 Sqrt[5]), (4 Sqrt[5])/3 +
1/3 (8/(3 Sqrt[5]) - (4 Sqrt[5])/3) -
2/3 (4/(3 Sqrt[5]) + (4 Sqrt[5])/3)}, {0, 0, -3}}
预期的结果应该是
( {
{6, 0, 0},
{0, 6, 0},
{0, 0, -3}
} )
就像方程式中的一样。如果我手工计算,我会得到这个结果。我在这里遗漏了什么?只是
简化或扩展结果
以下是一个例子:
In[1]:= a = {{2, -2, -4}, {-2, 5, -2}, {-4, -2, 2}}
Out[1]= {{2, -2, -4}, {-2, 5, -2}, {-4, -2, 2}}
In[2]:= p = {{1/Sqrt[5], 4/(3 Sqrt[5]), -(2/3)}, {-(2/Sqrt[5]), 2/(
3 Sqrt[5]), -(2/6)}, {0, x5, -(2/3)}}
Out[2]= {{1/Sqrt[5], 4/(3 Sqrt[5]), -(2/3)}, {-(2/Sqrt[5]), 2/(
3 Sqrt[5]), -(1/3)}, {0, x5, -(2/3)}}
In[3]:= sol =
Solve[Inverse[p].a.p == {{6, 0, 0}, {0, 6, 0}, {0, 0, -3}}]
Out[3]= {{x5 -> -(Sqrt[5]/3)}}
In[4]:= Inverse[p].a.p /. sol[[1]]
Out[4]= <big output removed>
In[5]:= Simplify[%]
Out[5]= {{6, 0, 0}, {0, 6, 0}, {0, 0, -3}}
有人能解释为什么吗NSolve
按预期工作
In[8]:= NSolve[
Inverse[p].a.p == {{6, 0, 0}, {0, 6, 0}, {0, 0, -3}}, x5]
Out[8]= {{x5 -> -0.745356}}
这是三个方程,都在x中。(3个方程式,一个未知!)
先数值求解
Map[NSolve[eqs[[#]],x]&,Range[3]]
Out[465]= {{{x->0.}},{{x->0.}},{{x->-0.745356}}}
要使Solve接受x
,首先不要进行数值运算,将其保留为符号:
eqs = Expand@ Map[Total[lhs[[#, All]]] - q[[#]] == 0 &, Range[Length[q]]]
给
{-6 - 8/(3 (-(4/9) + (Sqrt[5] x)/3)) + (2 Sqrt[5] x)/(-(4/9) + (Sqrt[5] x)/3) ==
0,
-6 - 8/(3 (-(4/9) + (Sqrt[5] x)/3)) + (2 Sqrt[5] x)/(-(4/9) + (Sqrt[5] x)/3) == 0,
3 + 4/(3 (-(4/9) + (Sqrt[5] x)/3)) - (8 Sqrt[5])/(9 (-(4/9) + (Sqrt[5] x)/3))
+ (2 x)/(3 (-(4/9) + (Sqrt[5] x)/3)) - (
Sqrt[5] x)/(-(4/9) + (Sqrt[5] x)/3) + (
2 Sqrt[5] x^2)/(-(4/9) + (Sqrt[5] x)/3) == 0}
现在使用Solve,其中包含显式x
,现在就可以了
Map[Solve[eqs[[#]], x] &, Range[3]]
{{{}}, {{}}, {{x -> -(Sqrt[5]/3)}}}
--纳赛尔是的,我也注意到了这一点。当我说解[…,x]时,它不起作用!,奇怪,我现在正在看它。我不明白的是,为什么我们在RHS中使用{{6,0,0},{0,6,0},{0,0,-3}
?有3个方程,所以RHS应该是{6,6,-3}
。当我这样做时,我得到了x的额外解决方案。请看我的答复。
eqs = Expand@ Map[Total[lhs[[#, All]]] - q[[#]] == 0 &, Range[Length[q]]]
{-6 - 8/(3 (-(4/9) + (Sqrt[5] x)/3)) + (2 Sqrt[5] x)/(-(4/9) + (Sqrt[5] x)/3) ==
0,
-6 - 8/(3 (-(4/9) + (Sqrt[5] x)/3)) + (2 Sqrt[5] x)/(-(4/9) + (Sqrt[5] x)/3) == 0,
3 + 4/(3 (-(4/9) + (Sqrt[5] x)/3)) - (8 Sqrt[5])/(9 (-(4/9) + (Sqrt[5] x)/3))
+ (2 x)/(3 (-(4/9) + (Sqrt[5] x)/3)) - (
Sqrt[5] x)/(-(4/9) + (Sqrt[5] x)/3) + (
2 Sqrt[5] x^2)/(-(4/9) + (Sqrt[5] x)/3) == 0}
Map[Solve[eqs[[#]], x] &, Range[3]]
{{{}}, {{}}, {{x -> -(Sqrt[5]/3)}}}