Since I'm late to the game, see this question
But you want the numerical values.
The transformation matrices you describe are .. strange.
here is a great resource on how to set up and use transformation matrices
Which matches my understanding of:
$$^jT_i=\begin{bmatrix}
^jR(\theta_i)&^jp_i\\
0&1
\end{bmatrix}$$
So, for example,
$$
^1T_2=\begin{bmatrix}
\cos(\theta_2')&-\sin(\theta_2')&\ell_1\\
\sin(\theta_2')&\cos(\theta_2')&0\\
0&0&1
\end{bmatrix}
$$
Just sub in the values for the angles and lengths to find the matrices.
The position of the point is:
$$x=\ell_1\cos\theta_1 + \ell_2\cos(\pi/2-\theta_2+\theta_1)+x_1$$
$$y=\ell_1\sin\theta_1 + \ell_2\sin(\pi/2-\theta_2+\theta_1)+y_1$$
Where $\ell$ is the length of each arm, and $\theta$ is the angles you describe.
You can see this by setting the arm to be fully outstretched along the x-axis, and noting that the position of the end point must be $(\ell_1+\ell_2,0)$. Also, $\theta_2'$ can be temporarily set to $\pi-\theta_2$ to make this arrangement correspond to the "laying along the x-axis" one for ease of derivation.
Or:
$$^{o}p_3=^oT_1\cdot ^1T_2\cdot ^2p_3 +p_1$$
read as "point three with respect to origin equals point of joint three with respect to joint 2, transformed from coordinate frame two to coordinate frame 1, then from coordinate frame 1 to original frame".
And then,
$$^2p_3=(\ell_2,0)$$
When solving the full system, remember to use homogeneous coordinates then, so you have to express the end point as
$$^2p_3=(\ell_2,0,1)$$