Given a non-empty set $X$,let $^*: P(X) \times P(X) \rightarrow P(X)$ be defined as $A \,^*\, B = (A - B) \cup (B - A)$,$\forall A, B \in P(X)$. Show that the empty set $\Phi$ is the identity for the operation $^*$ and all the elements $A$ of $P(X)$ are invertible with $A^{-1} = A$.

(A) It is given that $^*: P(X) \times P(X) \rightarrow P(X)$ is defined as $A \,^*\, B = (A - B) \cup (B - A)$ for all $A, B \in P(X)$.
$1$. Identity Element:
For any $A \in P(X)$,we have:
$A \,^*\, \Phi = (A - \Phi) \cup (\Phi - A) = A \cup \Phi = A$
$\Phi \,^*\, A = (\Phi - A) \cup (A - \Phi) = \Phi \cup A = A$
Since $A \,^*\, \Phi = A = \Phi \,^*\, A$ for all $A \in P(X)$,the empty set $\Phi$ is the identity element for the operation $^*$.
$2$. Invertible Elements:
An element $A \in P(X)$ is invertible if there exists $B \in P(X)$ such that $A \,^*\, B = \Phi = B \,^*\, A$,where $\Phi$ is the identity element.
Consider $A \,^*\, A$:
$A \,^*\, A = (A - A) \cup (A - A) = \Phi \cup \Phi = \Phi$
Since $A \,^*\, A = \Phi$,every element $A \in P(X)$ is its own inverse.
Thus,all elements $A$ of $P(X)$ are invertible with $A^{-1} = A$.