\(G=\left\{\left[\begin{array}{ll}x & x \\ x & x\end{array}\right], x\right.\) 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

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.