Let's say we have a manipulator with the following DH parameter table:
and transformation matrices:
$$A_{01}=
\begin{bmatrix}
\cos(q_1) & \sin(q_1) & 0 & 95\cos(q_1)\\
\sin(q_1) & -\cos(q_1) & 0 & 95\sin(q_1)\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}$$
$$A_{12}=
\begin{bmatrix}
\cos(q_2) & 0 & \sin(q_2) & 89\cos(q_2)\\
\sin(q_2) & 0 & -\cos(q_2) & 89\sin(q_2)\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}$$
$$A_{23}=
\begin{bmatrix}
\cos(q_3) & -\sin(q_3) & 0 & 147\cos(q_3)\\
\sin(q_2) & \cos(q_2) & 0 & 147\sin(q_2)\\
0 & 0 & 1 & -28\\
0 & 0 & 0 & 1
\end{bmatrix}$$
When multiplied, they form the absolute transformation matrix of the end-effector relative to the reference frame:
$$A_{03}=A_{01}\cdot A_{12}\cdot A_{23}=$$
$$\begin{bmatrix}
\cos(q_3)\cos(q_1-q_2) & -\sin(q_3)\cos(q_1-q_2) & -\sin(q_1-q_2) & 95\cos(q_1)+28\sin(q_1-q_2)+89\cos(q_1-q_2)+147\cos(q_3)\cos(q_1-q_2)\\
\cos(q_3)\sin(q_1-q_2) & -\sin(q_3)\sin(q_1-q_2) & \cos(q_1-q_2) & 95\sin(q_1)-28\cos(q_1-q_2)+89\sin(q_0-q_1)+147\cos(q_3)\sin(q_1-q_2)\\
-\sin(q_3) & -\cos(q_3) & 0 & -147\sin(q_3)\\
0 & 0 & 0 & 1
\end{bmatrix}$$
We can easily read the position of the end-effector relative to the reference frame just from the fourth column of the $A_{03}$ matrix:
$\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} = $
$\begin{bmatrix}
95\cos(q_1)+28\sin(q_1-q_2)+89\cos(q_1-q_2)+147\cos(q_3)\cos(q_1-q_2)\\
95\sin(q_1)-28\cos(q_1-q_2)+89\sin(q_0-q_1)+147\cos(q_3)\sin(q_1-q_2)\\
-147\sin(q_3)
\end{bmatrix}$
How can we read the orientation of the end-effector relative to the reference frame?
