如图所示，我用FindInstance求出方程组的解。问题：
1，我如何在下面的编程中引用这组解；
2，这组解怎么保存？我按吧主@xzcyr的方法保存成txt文档，文档是这样的：
{Subscript[α, 1] -> 0.049425554250005145, Subscript[α, 2] -> 0.10970009461838608, Subscript[α, 3] -> 0.031236631923210925}
这不是能够直接引用的格式呀？
3，这是一个线性方程组，可以矩阵左除；这里我只是用它来演示的，想说明我现在遇到的困惑，后面我还要用FindInstance求一个非线性的方程组。
4，我把程序粘贴到最后。
5，谢谢吧里的兄弟姐妹们。
b = FindInstance[(-2.076791871778445)*Subscript[\[Alpha],
1] + (-6.743300806281740)*Subscript[\[Alpha],
2] + (-10.485550381996774)*Subscript[\[Alpha],
3] == (-1.169920603608618) && (-2.058926832568549)*
Subscript[\[Alpha], 1] + (-0.407795804898393)*
Subscript[\[Alpha], 2] + (-1.324103988358989)*
Subscript[\[Alpha],
3] == (-0.
18785938715
4668) && (-1.972835100897579)*
Subscript[\[Alpha], 1] + (-3.067677386687139)*
Subscript[\[Alpha], 2] + (-0.607590725158021)*
Subscript[\[Alpha],
3] == (-0.453012055725708), {Subscript[\[Alpha], 1],
Subscript[\[Alpha], 2], Subscript[\[Alpha], 3]}]

更正：是
b[[1, 1, 2]]
b[[1, 2, 2]]
b[[1, 3, 2]]
不好意思，差点误导大家。

首先这种情况可以用{a1,a2,a3}/.FindInstance那一串
手机打字不方便所以就用a代替了，意会即可
其次这种情况一般不用FindInstance，你可以在帮助里搜一下LinearSolve

b = FindInstance[(-2.076791871778445)*
Subscript[\[Alpha], 1] + (-6.743300806281740)*
Subscript[\[Alpha], 2] + (-10.485550381996774)*
Subscript[\[Alpha],
3] == (-1.169920603608618) && (-2.058926832568549)*
Subscript[\[Alpha], 1] + (-0.407795804898393)*
Subscript[\[Alpha], 2] + (-1.324103988358989)*
Subscript[\[Alpha],
3] == (-0.
18785938715
4668) && (-1.972835100897579)*
Subscript[\[Alpha], 1] + (-3.067677386687139)*
Subscript[\[Alpha], 2] + (-0.607590725158021)*
Subscript[\[Alpha],
3] == (-0.453012055725708), {Subscript[\[Alpha], 1],
Subscript[\[Alpha], 2], Subscript[\[Alpha], 3]}]
Export["test.txt", %]
{Subscript[\[Alpha], 1] x + Subscript[\[Alpha], 2] y +
Subscript[\[Alpha], 3] z == 20,
Subscript[\[Alpha], 2] x - Subscript[\[Alpha], 1] y == 1,
Subscript[\[Alpha], 1] x - Subscript[\[Alpha], 3] z == -10} /.
ToExpression@Import["test.txt"]
Solve[%, {x, y, z}]
这样你看行不