Answer
See below
Work Step by Step
Given:
$A=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}$
Let's reduce $A$ to the row-echelon form
$\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}\approx \begin{bmatrix}
1 & -\tan \theta\\
\sin \theta& \cos \theta
\end{bmatrix} \approx \begin{bmatrix}
1 & -\tan \theta\\
0&\frac{1}{ \cos \theta}
\end{bmatrix}\approx \begin{bmatrix}
1 & -\tan \theta\\
0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$
where $1.M_1(\frac{1}{\cos \theta}) \\
2.M_2(-\sin \theta)\\
3. M_2(\cos \theta)\\
4.A_{12}(\tan \theta)$
Thus, $T(x)=Ax=M_1(\cos \theta)A_{12}(\sin \theta)M_2(\frac{1}{\cos \theta})A_{12}(-\tan \theta)x$
For $\theta \in [<0,\frac{\pi}{2}>\cup ]\rightarrow \cos \theta \gt 0$ and $\frac{1}{\cos \theta} \gt 0$, $A$ is a product of a shear parallel to the y-axis followed by a linear stretch in the y direction, followed by a shear parallel to the x-axis, followed by a linear stretch in the x-direction.
For $\theta \in [\rightarrow \cos \theta \lt 0$ and $\frac{1}{\cos \theta} \lt 0$, $A$ is a product of a shear parallel to the y-axis followed by a reflection in the x axis followed by a linear stretch in the y direction, followed by a shear parallel in the x-axis, followed by a reflection in the y-axis followed by a linear stretch in the x-direction.
