Anti-symmetric relation given by a matrix Relation R is given by a matrix

$$\begin{bmatrix}
1&amp; 0&amp; 0&amp; 0\\  
1&amp; 1&amp; 0&amp; 0  \\
1&amp; 0&amp; 1&amp; 0  \\
1&amp; 1&amp; 1&amp; 1
\end{bmatrix}
$$
Is it anti-symmetric?

I'm not quite understanding this.  

My notes state that for any $a,b\in A$ in the binary relation $R$ is a subset of $A\times A$, it is anti-symmetric if $(a,b)\in R$ and $(b,a)\in R$ then $a = b$.

If we assume that the elements of the set are $0,1,2,3$
Wouldn't $(2,3)\in R$ and $(3,2) \in R$ but $3\neq2$.
So would it not be anti-symmetric or am I completely missing the ball on this concept?
No, in this case, the entry corresponding to $(2, 3)$ (the $(3, 4)$ entry of the matrix) is zero, so $(2, 3) \not\in R$.

Since $(3, 2) \in R$, however, the new relation $R' := R \cup \{(2, 3)\}$ formed by adding $(2, 3)$ to $R$ is not antisymmetric.
If both (a,b)∈R ; (b,a)∈R and if a=b, then the relation R is anti-symmetric.
Here,(2,3)∉R so no need to check the condition whether a=b  or not and consider the relation is obviously Anti-symmetric.