<s>[INST] How to rotate a whole rectangle by an arbitrary angle around the origin using a transformation matrix? Suppose, I have a 2D rectangle ABCD like the following:
$A(0,0)$, $B(140,0)$, $C(140,100)$, $D(0,100)$.
I want to rotate the whole rectangle by $\theta = 50°$.

I want to rotate it around the Z-axis by an arbitrary angle using a rotation transformation matrix.
How to do that?
I know that, $$ A = 
        \begin{bmatrix}
        \
        0 & 0 & 1 \\
        \end{bmatrix};
B = 
        \begin{bmatrix}
        \
        140 & 0 & 1 \\
        \end{bmatrix};
C = 
        \begin{bmatrix}
        \
        140 & 100 & 1 \\
        \end{bmatrix};
D = 
        \begin{bmatrix}
        \
        0 & 100 & 1 \\
        \end{bmatrix}.
$$
And, I know that the rotation matrix is, $$R = 
        \begin{bmatrix}
        \
        cos \theta & -sin \theta & 0 \\
        sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}$$
Now, what is the calculation?
I have tried the following ways,
$$Rotation = 
        \begin{bmatrix}
        \
        0 & 0 & 1 \\
         140 & 0 & 1 \\
        140 & 100 & 1 \\
        0 & 100 & 1 \\
        \end{bmatrix}.\begin{bmatrix}
        \
        cos \theta & -sin \theta & 0 \\
        sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}$$
And,
$$
A' =  \begin{bmatrix}
        \
        0 & 0 & 1 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & -sin \theta & 0 \\
        sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}\\
B' =  \begin{bmatrix}
        \
        140 & 0 & 1 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & -sin \theta & 0 \\
        sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}\\
C' =  \begin{bmatrix}
        \
        140 & 100 & 1 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & -sin \theta & 0 \\
        sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}\\
D' =  \begin{bmatrix}
        \
        0 & 100 & 1 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & -sin \theta & 0 \\
        sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}
$$
For example,
$$B' =  \begin{bmatrix}
        \
        140 & 0 & 1 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos 50^\circ & -sin 50^\circ & 0 \\
        sin 50^\circ & cos 50^\circ & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix} = \begin{bmatrix}
        \
        89.99 & -107.24 & 1 \\
        \end{bmatrix}\\
C' =  \begin{bmatrix}
        \
        140 & 100 & 1 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos 50^\circ & -sin 50^\circ & 0 \\
        sin 50^\circ & cos 50^\circ & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix} = \begin{bmatrix}
        \
        166.59 & -42.96 & 1 \\
        \end{bmatrix}$$
What is the right way to work with?
 [/INST] I have solved my problem.
$$Rotation = 
        \begin{bmatrix}
        \
        0 & 0 & 0 \\
         140 & 0 & 0 \\
        140 & 100 & 0 \\
        0 & 100 & 0 \\
        \end{bmatrix}.\begin{bmatrix}
        \
        cos \theta & -sin \theta & 0 \\
        sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}$$
Alternatively,
$$
A' =  \begin{bmatrix}
        \
        0 & 0 & 0 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & sin \theta & 0 \\
        -sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix} 
= \begin{bmatrix}0 & 0 & 0\end{bmatrix}\\
B' =  \begin{bmatrix}
        \
        140 & 0 & 0 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & sin \theta & 0 \\
        -sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}
= \begin{bmatrix}98.9949 & 98.9949 & 0\end{bmatrix}\\
C' =  \begin{bmatrix}
        \
        140 & 100 & 0 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & sin \theta & 0 \\
        -sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}
= \begin{bmatrix}28.2843 & 169.706 & 0\end{bmatrix}\\
D' =  \begin{bmatrix}
        \
        0 & 100 & 0 \\
        \end{bmatrix} . \begin{bmatrix}
        \
        cos \theta & sin \theta & 0 \\
        -sin \theta & cos \theta & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}
= \begin{bmatrix}-70.7107 & 70.7107 & 0\end{bmatrix}\\
$$
</s>