Anti-symmetric relation given by a matrix Relation R is given by a matrix
$$\begin{bmatrix}
1& 0& 0& 0\\  
1& 1& 0& 0  \\
1& 0& 1& 0  \\
1& 1& 1& 1
\end{bmatrix}
$$
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?  
 A: 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.
A: Your wording is slightly off. You say:

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 $a =
b$.

Which makes it sound like $R$ could be anti-symmetric for one pair $a,b$ and not so for another.
What you should write is

A binary relation $R$ is anti-symmetric if for every pair $a,b\in A$, if $(a,b)\in R$ and $(b,a)\in R$, then $a=b$.

This second wording makes it much more clear that anti-symmetric is a global property.
In your particular case, it is easy to see that if $(a,b)\in R$ and $a\neq b$, then $(b,a)$ is not in $R$. This means that if $(a,b)\in R$ and $(b,a)$ is in $R$, it must follow that $a\neq b$ is not true, therefore $a=b$. 
The case with $(3,2)$ is confuzing me. Yes, indeed, if $(3,2)$ and $(2,3)$ would be elements of $R$, then the relation would not be anti-symmetric. But in your case, neither $(2,3)$ nor $(3,2)$ is an element of $R$!
A: 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.
