Question

$G=\left\{\left[\begin{array}{cc}x& x\\ x& x\end{array}\right],x\text{ is a nonzero real number}\right\}$ is a group with respect to matrix multiplication. In this group, the inverse of $\left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}\end{array}\right]$ is

• A. $\left[\begin{array}{cc}4/3& 4/3\\ 4/3& 4/3\end{array}\right]$
• B. $\left[\begin{array}{cc}3/4& 3/4\\ 3/4& 3/4\end{array}\right]$
• C. $\left[\begin{array}{cc}3& 3\\ 3& 3\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

Answer 1

Answer: (B)

Given, $G=\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]$ 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 $=\left[\begin{array}{cc}1/2& 1/2\\ 1/2& 1/2\end{array}\right]=I^{\prime }$
For inverse; $A{A}^{-1}=I^{\prime }$
Given, $\left[\begin{array}{cc}1/3& 1/3\\ 1/3& 1/3\end{array}\right]A^{-1}=\left[\begin{array}{cc}1/2& 1/2\\ 1/2& 1/2\end{array}\right]$
Apply $R_1\to 3/2R_1$ and $R_2\to 3/2R_2$
$\left[\begin{array}{cc}1/2& 1/2\\ 1/2& 1/2\end{array}\right]A^{-1}=\left[\begin{array}{cc}3/4& 3/4\\ <br/><br/>3/4& 3/4\end{array}\right]$
$I^{\prime }{A}^{-1}=\left[\begin{array}{cc}3/4& 3/4\\ 3/4& 3/4\end{array}\right]=A^{-1}$
Which is the required inverse.