First of all, you will need to understand how to interpret these matrices. They are Boolean matrices where entry $M_{ij}=1$ if $(i,j)$ is in the relation and $0$ otherwise. As an example, the relation $R$ is
\begin{align*}
R=\{(0,3),(2,1),(3,2)\}.
\end{align*}
Question 1
For the inverse relation, try writing the the pairs contained in $R^{-1}$ and represent this in matrix form. How does this matrix relate to $M_R$?
It's the transposed matrix. We have $(a,b)\in R^{-1}$ if $(b,a)\in R$, which corresponds to transposing the matrix.
The intersection of the relations is defined such that $(a,b)\in R\cap S$ if both $(a,b)\in R$ and $(a,b)\in S$. How does this translate to matrices? Consider the entry $(i,j)$ in $M_{R\cap S}$: When it is $1$, what must be true about the entries $(i,j)$ in $M_R$ and $M_S$?
Both of the entries must be $1$. Thus, $M_{R\cap S}=0_{4\times 4}$ in this example.
To find the composition, it can be shown that this corresponds to the Boolean product of the matrices.
Question 2
First, you have to find the matrix $M_{R\cup S}$. In the case of the intersection the entries of both matrices $M_R$ and $M_S$ had to be $1$, since both relations should contain the pair. In the union, only one of the relations need to contain the pair. How does this translate to the matrices?
Entry $(i,j)$ in $M_{R\cup S}$ is $1$ if entry $(i,j)$ is $1$ in either $M_R$ or $M_S$.
For $n\times n$ matrices, Warshall's algorithm consist of $n$ steps. First, we fix $k=1$ and construct a matrix $W$ where $W_{ij}=M_{ij}\vee (M_{ik}\wedge M_{kj})$ (here $M$ denotes the matrix representation of the relation). Consider a small example:
\begin{align*}
M=\left[\begin{array}{c c c}
1 & 0 & 0\\
0 & 1 & 1\\
1 & 0 & 0
\end{array}\right]
\end{align*}
Fix $k=1$. We can transfer the first row and column immediately. Further, all $1$'s can be transferred. For the remaining entries, we check the corresponding entries in the $k$'th row and column -- here marked with a box. If these are both $1$, we write $1$, otherwise $0$.
\begin{align*}
W_1=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 1\\
1 & ? & ?
\end{array}\right]=\left[\begin{array}{ccc}
1 & \boxed{0} & 0\\
0 & 1 & 1\\
\boxed{1} & \mathbf{0} & ?
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & \boxed{0}\\
0 & 1 & 1\\
\boxed{1} & 0 & \mathbf{0}
\end{array}\right].
\end{align*}
Next, fix $k=2$ and repeat the process on $W_1$. Again transfer the $k$'th row and column and any entries containing $1$.
\begin{align*}
W_2=\left[\begin{array}{ccc}
1 & 0 & ?\\
0 & 1 & 1\\
1 & 0 & ?
\end{array}\right]=\left[\begin{array}{ccc}
1 & \boxed{0} & \mathbf{0}\\
0 & 1 & \boxed{1}\\
1 & 0 & ?
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & \boxed{1}\\
1 & \boxed{0} & \mathbf{0}
\end{array}\right].
\end{align*}
Finally, fix $k=3$ to find the last entries.
\begin{align*}
W_3=\left[\begin{array}{ccc}
1 & ? & 0\\
? & 1 & 1\\
1 & 0 & 0
\end{array}\right]=\left[\begin{array}{ccc}
1 & \mathbf{0} & \boxed{0}\\
? & 1 & 1\\
1 & \boxed{0} & 0
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0\\
\mathbf{1} & 1 & \boxed{1}\\
\boxed{1} & 0 & 0
\end{array}\right].
\end{align*}
This final matrix $W_3$ is the transitive closure.
The relation indicated by this matrix can be described as
$$
(x,x) \quad x = 1,2,3\\
(2,3)\\
(3,1)
$$
In order for the relation to be transitively closed, you need to add $(2,1)$. So, the matrix we want is
$$
\pmatrix{
1&0&0\\
1&1&1\\
1&0&1
}
$$
Best Answer
Using standard matrix multiplication,
$$\pmatrix{1&0&1\\1&1&0\\0&0&0}\pmatrix{0&1&0\\0&0&1\\1&0&1}=\pmatrix{1&1&1\\0&1&1\\0&0&0}.$$