None of these.
Solution:
Let e be the identity element in Q+ with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$
a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$
e = 3, $\forall\text{ a}\in\text{Q}^+$
Thus, 3 is the identity element in Q+ with respect to *.
Let $\text{a}\in\text{Q}^+$ and $\text{b}\in\text{Q}^+$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
$\therefore\ \frac{\text{ab}}3=3\text{ and }\frac{\text{ba}}3=3$
$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$
Thus, $\frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+$.
Given: $\text{a}*\text{b}=\frac{\text{ab}}3$
$4*6=\frac{4\times6}3=8$
Now,
$\text{a}^{-1}=\frac{9}{\text{a}}$
$(4*6)^{-1}=8^{-1}$
$=\frac{9}8$
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$\frac{5}{3}$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{9}$
Then which of the following options is/are correct?
$(1)$ For $x =1$, there exists a unit vector $\alpha \hat{ i }+\beta \hat{ j }+\gamma \hat{ k }$ for which $R \left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
$(2)$ There exists a real number $x$ such that $P Q=Q P$
$(3)$ $\operatorname{det} R=\operatorname{det}\left[\begin{array}{lll}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{array}\right]+8$, for all $x \in R$
$(4)$ For $x=0$, if $R\left[\begin{array}{l}1 \\ a \\ b\end{array}\right]=6\left[\begin{array}{l}1 \\ a \\ b\end{array}\right]$, then $a+b=5$