$\therefore|\text{A}|=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
$=\cos^2\theta+\sin^2\theta=1\neq0$
Thus, A-1 exists.
Now,
$\text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\text{adj A}$
$\Rightarrow\text{AA}^{-1}=\text{A adj A}$
$\Rightarrow\text{AA}^{-1}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix} \ \big[\because\ \text{AA}^{-1}=\text{I}\big]$
$\Rightarrow\text{k}=1$
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$2\text{x}\frac{\text{dy}}{\text{dx}}=3\text{y},\text{y}(1)=2$