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Question

Mathematics Question on Matrices

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 [1313\[0.3em]1313]\begin{bmatrix} \frac{1}{3} &\frac{1}{3} \\\[0.3em] \frac{1}{3} & \frac{1}{3} \end{bmatrix} is

A

[4/34/3\[0.3em]4/34/3]\begin{bmatrix} 4/3 &4/3 \\\[0.3em] 4/3 & 4/3 \end{bmatrix}

B

[3/43/4\[0.3em]3/43/4]\begin{bmatrix} 3/4 &3/4 \\\[0.3em] 3/4 & 3/4 \end{bmatrix}

C

[33\[0.3em]33]\begin{bmatrix} 3 &3 \\\[0.3em] 3 & 3 \end{bmatrix}

D

[11\[0.3em]11]\begin{bmatrix} 1 &1 \\\[0.3em] 1& 1 \end{bmatrix}

Answer

[3/43/4\[0.3em]3/43/4]\begin{bmatrix} 3/4 &3/4 \\\[0.3em] 3/4 & 3/4 \end{bmatrix}

Explanation

Solution

Given, G=[xx xx]G= \begin{bmatrix}x & x \\\ x & x\end{bmatrix} is a group with respect to matrix multiplication where xR0x \in R-\\{0\\}.
Now, the identity element of above group with respect to matrix xx.
Multiplication is =[1/21/2 1/21/2]=I= \begin{bmatrix}1 / 2 & 1 / 2 \\\ 1 / 2 & 1 / 2\end{bmatrix}=I'
For inverse; AA1=IA A^{-1}=I'
Given, [1/31/3 1/31/3]A1=[1/21/2 1/21/2]\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 R13/2R1R_{1} \rightarrow 3 / 2 R_{1} and R23/2R2R_{2} \rightarrow 3 / 2 R_{2}
[1/21/2\1/21/2]A1=[3/43/4 3/43/4]\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}
IA1=[3/43/4 3/43/4]=A1I' A^{-1}= \begin{bmatrix}3 / 4 & 3 / 4 \\\ 3 / 4 & 3 / 4 \end{bmatrix}=A^{-1}
Which is the required inverse.