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find the inverse $$\begin{pmatrix} 5 e^{2 t} \sin(2 t) & 5 e^{3 t} \cos(2 t)\\ -6 e^{2 t} \cos(2 t)& 6 e^{3 t} \sin(2 t) \end{pmatrix}$$

I understand the inverse of

$$\begin{pmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \\ \end{pmatrix} $$

being

$$\begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \\ \end{pmatrix} $$

But not am confused with the extra parts.

so would my set up be $$ \begin{pmatrix} \sin(t) & -\cos(t) \\ \cos(t) & \sin(t) \\ \end{pmatrix} $$

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3 Answers 3

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The straightforward thing would be to apply the formula $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$

If you want to use the inverse you already know, you can think of your matrix $A$ as the product $$A=\begin{bmatrix} 5 & 0 \\ 0 & 6 \end{bmatrix} \begin{bmatrix}\sin 2t & \cos 2t\\ -\cos 2t & \sin 2t \end{bmatrix} \begin{bmatrix} e^{2t} & 0 \\ 0 & e^{3t} \end{bmatrix} $$ and so (see https://www.lem.ma/ZN) $$A^{-1}= \begin{bmatrix} e^{2t} & 0 \\ 0 & e^{3t} \end{bmatrix}^{-1}\begin{bmatrix}\sin 2t & \cos 2t\\ -\cos 2t & \sin 2t \end{bmatrix}^{-1}\begin{bmatrix} 5 & 0 \\ 0 & 6 \end{bmatrix} ^{-1}\\ = \begin{bmatrix} e^{-2t} & 0 \\ 0 & e^{-3t} \end{bmatrix} \begin{bmatrix}\sin 2t & -\cos 2t\\ \cos 2t & \sin 2t \end{bmatrix}\begin{bmatrix} 1/5 & 0 \\ 0 & 1/6 \end{bmatrix}$$ and now you can multiply the product out.

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A $2$x$2$ matrix has an inverse defined as follows:$$\begin{bmatrix} a & b\\ c & d\end{bmatrix}^{-1}=\frac{1}{\underbrace{ad-bc}_{\text{determinant}}}\begin{bmatrix} d & -b\\ -c & a\end{bmatrix}$$ It only exists if the determinant is non-zero.

Replace $a$, $b$, $c$ and $d$ by your expressions and turn the handle...

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    $\begingroup$ I believe there's an extra $^{-1}$ in your formula. $\endgroup$ Apr 23, 2016 at 16:06
  • $\begingroup$ @YoTengoUnLCD - thanks for spotting my typo! :) $\endgroup$
    – Mufasa
    Apr 23, 2016 at 16:07
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Hint you may make everything in e ie $$\sin(x)=Ie^{ix},\cos(x) =R(e^{ix})$$ then finding inverse can be done.

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