A) \[\alpha ={{a}^{2}}+{{b}^{2}},\beta =ab\]
B) \[\alpha ={{a}^{2}}+{{b}^{2}},\beta ={{a}^{2}}-{{b}^{2}}\]
C) \[\alpha ={{a}^{2}}+{{b}^{2}},\beta =2ab\]
D) \[\alpha ={{a}^{2}}-{{b}^{2}},\beta ={{a}^{2}}+{{b}^{2}}\]
Correct Answer: C
Solution :
[c] \[\because A=\left[ \begin{align} & a\,\,\,b \\ & b\,\,\,a \\ \end{align} \right]\] \[{{A}^{2}}=\left[ \begin{align} & a\,\,\,b \\ & b\,\,\,a \\ \end{align} \right]\left[ \begin{align} & a\,\,\,b \\ & b\,\,\,a \\ \end{align} \right]=\left[ \begin{align} & {{a}^{2}}+{{b}^{2}}\,\,\,\,\,\,2ab \\ & 2ab\,\,\,\,\,\,\,\,\,\,\,\,{{a}^{2}}+{{b}^{2}} \\ \end{align} \right]\] \[\because {{A}^{2}}=\left[ \begin{align} & \alpha \,\,\,\beta \\ & \beta \,\,\,\alpha \\ \end{align} \right]\] Thus, \[\alpha ={{a}^{2}}+{{b}^{2}}\And \beta =2ab.\] Hence, option [c] is correct.
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