A) \[\left[ \begin{matrix} 4/3 & 4/3 \\ 4/3 & 4/3 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 3/4 & 3/4 \\ 3/4 & 3/4 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 3 & 3 \\ 3 & 3 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
Given, \[G=\left[ \begin{matrix} x & x \\ x & x \\ \end{matrix} \right]\] 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 \[\left[ \begin{matrix} 1/2 & 1/2 \\ 1/2 & 1/2 \\ \end{matrix} \right]=I'\] For inverse; \[A\,{{A}^{-1}}=I'\] Given, \[\left[ \begin{matrix} 1/3 & 1/3 \\ 1/3 & 1/3 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1/2 & 1/2 \\ 1/2 & 1/2 \\ \end{matrix} \right]\] Apply \[{{R}_{1}}\to 3/2{{R}_{1}}\] and \[{{R}_{2}}\to 3/2{{R}_{2}}\] \[\left[ \begin{matrix} 1/2 & 1/2 \\ 1/2 & 1/2 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 3/4 & 3/4 \\ 3/4 & 3/4 \\ \end{matrix} \right]\] \[I'\,{{A}^{-1}}=\left[ \begin{matrix} 3/4 & 3/4 \\ 3/4 & 3/4 \\ \end{matrix} \right]={{A}^{-1}}\] Which is the required inverse.You need to login to perform this action.
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