Q87 / 180 · KCET 2011 · Mathematics
$G=\left\{\left[\begin{array}{ll}x & x \\ x & x\end{array}\right], x\right.$ is a non-zero real number $\}$ is a group with respect to matrix multiplication.
In this group, the inverse of $\left[\begin{array}{ll}\frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3}\end{array}\right]$ is
- A$\left[\begin{array}{ll}4 / 3 & 4 / 3 \\ 4 / 3 & 4 / 3\end{array}\right]$
- Correct answerB$\left[\begin{array}{ll}3 / 4 & 3 / 4 \\ 3 / 4 & 3 / 4\end{array}\right]$
- C$\left[\begin{array}{ll}3 & 3 \\ 3 & 3\end{array}\right]$
- D$\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
Solution
Given, $G=\left[\begin{array}{ll}x & x \\ x & x\end{array}\right]$ is a group with respect to matrix multiplication where $x \in R-\{0\}$. Now, the identity element of above group with respect to matrix $x$.
Multiplication is $=\left[\begin{array}{ll}1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2\end{array}\right]=I^{\prime}$
For inverse; $\quad A A^{-1}=I^{\prime}$
Given, $\quad\left[\begin{array}{ll}1 / 3 & 1 / 3 \\ 1 / 3 & 1 / 3\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2\end{array}\right]$
Apply $R_{1} \rightarrow 3 / 2 R_{1}$ and $R_{2} \rightarrow 3 / 2 R_{2}$
$$
\begin{gathered}
{\left[\begin{array}{ll}
1 / 2 & 1 / 2 \\
1 / 2 & 1 / 2
\end{array}\right] A^{-1}=\left[\begin{array}{ll}
3 / 4 & 3 / 4 \\
3 / 4 & 3 / 4
\end{array}\right]} \\
I^{\prime} A^{1}=\left[\begin{array}{ll}
3 / 4 & 3 / 4 \\
3 / 4 & 3 / 4
\end{array}\right]=A^{1}
\end{gathered}
$$
Which is the required inverse.
Hence, the correct answer is option: 'B'.