Which expression is equivalent to (jk)l?
1 answer:
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Define
![{x} = \left[\begin{array}{ccc}x_{1}\\x_{2}\end{array}\right]](https://tex.z-dn.net/?f=%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%5C%5Cx_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20)
Then
x₁ = cos(t) x₁(0) + sin(t) x₂(0)
x₂ = -sin(t) x₁(0) + cos(t) x₂(0)
Differentiate to obtain
x₁' = -sin(t) x₁(0) + cos(t) x₂(0)
x₂' = -cos(t) x₁(0) - sin(t) x₂(0)
That is,
![\dot{x} = \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right] x(0)](https://tex.z-dn.net/?f=%5Cdot%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20x%280%29)
Note that
![\left[\begin{array}{ccc}0&1\\-1&09\end{array}\right] \left[\begin{array}{ccc}cos(t)&sin(t)\\-sin(t)&cos(t)\end{array}\right] = \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%2609%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28t%29%26sin%28t%29%5C%5C-sin%28t%29%26cos%28t%29%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20)
Therefore
![x(t) = \left[\begin{array}{ccc}0&1\\-1&0\end{array}\right] x(t)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%260%5Cend%7Barray%7D%5Cright%5D%20x%28t%29)
Then
x₁ = cos(t) x₁(0) + sin(t) x₂(0)
x₂ = -sin(t) x₁(0) + cos(t) x₂(0)
Differentiate to obtain
x₁' = -sin(t) x₁(0) + cos(t) x₂(0)
x₂' = -cos(t) x₁(0) - sin(t) x₂(0)
That is,
Note that
Therefore