Consider a binary operation $*$ on the set $\{1,2,3,4,5\}$ given by the following multiplication table. Is $^*$ commutative?
(Hint: use the following table)

$^*$	$1$	$2$	$3$	$4$	$5$
$1$	$1$	$1$	$1$	$1$	$1$
$2$	$1$	$2$	$2$	$2$	$2$
$3$	$1$	$2$	$3$	$3$	$3$
$4$	$1$	$2$	$3$	$4$	$4$
$5$	$1$	$2$	$3$	$4$	$5$

(A) binary operation $*$ on a set $S$ is commutative if $a \,^*\, b = b \,^*\, a$ for all $a, b \in S$.
To check if the operation is commutative,we check if the table is symmetric with respect to the main diagonal.
Looking at the table:
- $1 \,^*\, 2 = 1$ and $2 \,^*\, 1 = 1$. Thus,$1 \,^*\, 2 = 2 \,^*\, 1$.
- $1 \,^*\, 3 = 1$ and $3 \,^*\, 1 = 1$. Thus,$1 \,^*\, 3 = 3 \,^*\, 1$.
- $2 \,^*\, 3 = 2$ and $3 \,^*\, 2 = 2$. Thus,$2 \,^*\, 3 = 3 \,^*\, 2$.
- $2 \,^*\, 4 = 2$ and $4 \,^*\, 2 = 2$. Thus,$2 \,^*\, 4 = 4 \,^*\, 2$.
- $3 \,^*\, 4 = 3$ and $4 \,^*\, 3 = 3$. Thus,$3 \,^*\, 4 = 4 \,^*\, 3$.
- $1 \,^*\, 5 = 1$ and $5 \,^*\, 1 = 1$. Thus,$1 \,^*\, 5 = 5 \,^*\, 1$.
- $2 \,^*\, 5 = 2$ and $5 \,^*\, 2 = 2$. Thus,$2 \,^*\, 5 = 5 \,^*\, 2$.
- $3 \,^*\, 5 = 3$ and $5 \,^*\, 3 = 3$. Thus,$3 \,^*\, 5 = 5 \,^*\, 3$.
- $4 \,^*\, 5 = 4$ and $5 \,^*\, 4 = 4$. Thus,$4 \,^*\, 5 = 5 \,^*\, 4$.
Since $a \,^*\, b = b \,^*\, a$ for all $a, b \in \{1,2,3,4,5\}$,the operation $^*$ is commutative.