| * | a | b | c | d |
| a | a | b | c | d |
| b | b | a | d | c |
| c | c | d | a | b |
| d | d | c | b | a |
| * | a | b | c | d |
| a | a | b | c | d |
| b | b | a | d | c |
| c | c | d | a | b |
| d | d | c | b | a |
The table is symmetrical about the leading element. It means * is commutative on S.
Associativity:
a * (b * c) = a * d
= d
(a * b) * c = b * c
=d
Therefore,
a * (b * c) = (a * b) * c $\forall\text{ a, b, c}\in\text{S}$
So, * is Associative on S.
Finding identity element:
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at a.
⇒ x * a = a * x = x, $\forall\text{ x}\in\text{S}$
So, a is the identity element:
a * a = a
⇒ a-1 = a
b * b = a
⇒ b-1 = b
c * c = a
⇒ c-1 = c
d * d = a
⇒ d-1 = d
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$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$