Mathematica 如何求解矩阵方程如下
A1={{Cos[\[Theta]1],-Sin[\[Theta]1],0,a1Cos[\[Theta]1]},{Sin[\[Theta]1],Cos[\[Theta]1...
A1=
{{Cos[\[Theta]1], -Sin[\[Theta]1], 0,
a1 Cos[\[Theta]1]}, {Sin[\[Theta]1], Cos[\[Theta]1], 0,
a1 Sin[\[Theta]1]}, {0, 0, 1, 0}, {0, 0, 0, 1}}
A2=
{{Cos[\[Theta]2], -Sin[\[Theta]2], 0,
a2 Cos[\[Theta]2]}, {Sin[\[Theta]2], Cos[\[Theta]2], 0,
a2 Sin[\[Theta]2]}, {0, 0, 1, 0}, {0, 0, 0, 1}}
T=A1.A2
P={{nx, ox, ax, px}, {ny, oy, ay, py}, {nz, oz, az, pz}, {0, 0, 0, 1}}
Solve[T==P,{\[Theta]1, \[Theta]2}]
求解\[Theta]1, \[Theta]2,用Solve为什么解不出来,请问针对这种情况如何求解? 展开
{{Cos[\[Theta]1], -Sin[\[Theta]1], 0,
a1 Cos[\[Theta]1]}, {Sin[\[Theta]1], Cos[\[Theta]1], 0,
a1 Sin[\[Theta]1]}, {0, 0, 1, 0}, {0, 0, 0, 1}}
A2=
{{Cos[\[Theta]2], -Sin[\[Theta]2], 0,
a2 Cos[\[Theta]2]}, {Sin[\[Theta]2], Cos[\[Theta]2], 0,
a2 Sin[\[Theta]2]}, {0, 0, 1, 0}, {0, 0, 0, 1}}
T=A1.A2
P={{nx, ox, ax, px}, {ny, oy, ay, py}, {nz, oz, az, pz}, {0, 0, 0, 1}}
Solve[T==P,{\[Theta]1, \[Theta]2}]
求解\[Theta]1, \[Theta]2,用Solve为什么解不出来,请问针对这种情况如何求解? 展开
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