Question

$G = \left\{\begin{bmatrix} x&x \\[0.3em] x & x \end{bmatrix} , x \text{ is a nonzero real number} \right\}$ is a group with respect to matrix multiplication. In this group, the inverse of $\begin{bmatrix} \frac{1}{3} &\frac{1}{3} \\[0.3em] \frac{1}{3} & \frac{1}{3} \end{bmatrix}$ is

• A. $\begin{bmatrix} 4/3 &4/3 \\[0.3em] 4/3 & 4/3 \end{bmatrix}$
• B. $\begin{bmatrix} 3/4 &3/4 \\[0.3em] 3/4 & 3/4 \end{bmatrix}$
• C. $\begin{bmatrix} 3 &3 \\[0.3em] 3 & 3 \end{bmatrix}$
• D. $\begin{bmatrix} 1 &1 \\[0.3em] 1& 1 \end{bmatrix}$

Answer 1

The Correct Option is B

Given, $G= \begin{bmatrix}x & x \\ x & x\end{bmatrix}$ 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 $= \begin{bmatrix}1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2\end{bmatrix}=I'$
For inverse; $A A^{-1}=I'$
Given, $\begin{bmatrix}1 / 3 & 1 / 3 \\ 1 / 3 & 1 / 3\end{bmatrix} A^{-1}= \begin{bmatrix}1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2\end{bmatrix}$
Apply $R_{1} \rightarrow 3 / 2 R_{1}$ and $R_{2} \rightarrow 3 / 2 R_{2}$
$\begin{bmatrix}1 / 2 & 1 / 2 \\1 / 2 & 1 / 2 \end{bmatrix} A^{-1}= \begin{bmatrix} 3 / 4 & 3 / 4 \\
3 / 4 & 3 / 4 \end{bmatrix}$
$I' A^{-1}= \begin{bmatrix}3 / 4 & 3 / 4 \\ 3 / 4 & 3 / 4 \end{bmatrix}=A^{-1}$
Which is the required inverse.