A 4x4 homogeneous matrix for a 90 degree rotation about Y axis? According to my book

Rotations through an angle $\theta$ about the $x$, $y$, and $z$ axes are performed using the following transformation matrices.

For example,
$$    R_x(\theta) = \left[
\begin{matrix}
1& 0 & 0 \\\
0& \cos\theta &-\sin\theta\\\
0& \sin\theta &\cos\theta
\end{matrix}\right]
$$
Ry0= | cos0 0 sin0 0|
     | 0 1 0 0|
     | -sin0 0 cos0 0|
     | 0 0 0 1|


Rz0= | cos0 -sin0 0 0 |
     | sin0 cos0 0  0|
     | 0 0 1 0|
     | 0 0 0 1|

And I just need to put the angle in it no matter which rotate about axis?
After that, how can I get a $4\times 4$ matrix?
Thank you
 A: To express ordinary $\mathbb{R}^n \to \mathbb{R}^n$ linear transformation into homogeneous coordinates just add another row and column where every term is equal to $0$ but the diagonal, which should be $1$. 
For example if $A$ is your transformation matrix, then the new matrix would be
$$A_H = \left[\begin{matrix}A&  0 \\\ 0& 1 \end{matrix}\right].$$
In your example $$A = \left[\begin{matrix}1& 0 & 0 \\\ 0& \cos\theta &-\sin\theta\\\ 0& \sin\theta &\cos\theta \end{matrix}\right],$$ so $$A_H =  \left[\begin{matrix}1& 0 & 0 & 0 \\\ 0& \cos\theta &-\sin\theta & 0 \\\ 0& \sin\theta &\cos\theta & 0 \\\ 0 & 0 & 0 & 1 \end{matrix}\right].$$
Hope that helps ;-)
A: Let the rotation axis be a line through a fixed point $(x_0,y_0,z_0,1)^T$ and an infinite point (direction of the line) $(a,b,c,0)^T$, without loss of generality, we assume $a^2+b^2+c^2=1$.  Use right-handed rule for rotation, then a general homogeneous rotation with angle $\theta$ can be obtained as:
$$
\boldsymbol{R}^{3D}\left(x_0,y_0,z_0,a,b,c,\theta\right)=\mathscr{C}_1+\left(\sin\theta\cdot\mathscr{A}_2- \left(1-\cos\theta\right)\cdot\mathscr{O}_3\right)\cdot \mathscr{T}_4
$$
where:
$$\mathscr{C}_1=
\left[\begin{array}{*{20}{c}}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&2-\cos\theta
\end{array}\right]$$
$$\mathscr{A}_2  =  {\begin{array}{c}
\underbrace{
\begin{array}{c}
 {{\left[ {\begin{array}{*{20}{c}}
{\color{blue}0}&{\color{blue}-c}&{\color{blue}b}&0\\
{\color{blue}c}&{\color{blue}0}&{\color{blue}-a}&0\\
{\color{blue}-b}&{\color{blue}a}&{\color{blue}0}&0\\
0&0&0&0
\end{array}} \right]}}
\end{array}
} \\ \text{Antisymmetric matrix}
\end{array}}
$$
$$
\quad\mathscr{O}_3= \begin{array}{c}
\underbrace{  {I - {\left[ {\begin{array}{*{20}{c}}
a\\
b\\
c\\
0
\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}
a&b&c&0
\end{array}} \right]}} }\\
\text{Orthographic parallel projection} \end{array} $$
$$ \mathscr{T}_4=\begin{array}{c}
\underbrace{\left[ {\begin{array}{*{20}{c}}
1&0&0&{ - {x_0}}\\
0&1&0&{ - {y_0}}\\
0&0&1&{ - {z_0}}\\
0&0&0&1
\end{array}}  \right]}\\ \text{Translation}\end{array}$$
The L-C formulation of homogeneous 3D rotation is similar to Rodrigues' but they are not the same in essence. Details are available here: 
A submission on homogeneous rotation to arXiv.org
For your case, substitute $x_0=y_0=z_0=0$, $a=b=0$,  $c=1$ and $\theta=\dfrac{\pi}{2}$ into the L-C formulation then you can obtain the desired homogeneous rotation matrix.
