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$$

\mathbf{x}' = \begin{pmatrix}

a\cos 2t & a\sin 2t\\

a\sin 2t & -a\cos 2t

\end{pmatrix}\mathbf{x},

$$

verify that under the change of variables:

$$

\mathbf{x} = \begin{pmatrix}

\cos 2t & \sin 2t\\

\sin 2t & -\cos 2t

\end{pmatrix}\mathbf{u},

$$

the equations for $u$ become a system with constant coefficients:

$$

\mathbf{u}' = \begin{pmatrix}

a & 1\\

-1 & -a

\end{pmatrix}\mathbf{u}.

$$

Here is what I have

$$

\left[\begin{pmatrix}

\cos 2t & \sin 2t\\

\sin 2t & -\cos 2t

\end{pmatrix}\mathbf{u}\right]' = \begin{pmatrix} a & 0\\ 0 & a\end{pmatrix}\mathbf{u}

$$

What do I do about that left side? I don't see how I will get

$$

\mathbf{u}' = \begin{pmatrix}

a & 1\\

-1 & -a

\end{pmatrix}\mathbf{u}

$$