A) \[\alpha ={{a}^{2}}+{{b}^{2}},\beta =ab\]
B) \[\alpha ={{a}^{2}}+{{b}^{2}},\beta =2ab\]
C) \[\alpha ={{a}^{2}}+{{b}^{2}},\beta ={{a}^{2}}-{{b}^{2}}\]
D) \[\alpha =2ab,\,\,\beta ={{a}^{2}}+{{b}^{2}}\]
Correct Answer: B
Solution :
[b] \[{{A}^{2}}=\left[ \begin{matrix} \alpha & \beta \\ \beta & \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} a & b \\ b & a \\ \end{matrix} \right]\left[ \begin{matrix} a & b \\ b & a \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & 2ab \\ 2ab & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right]\] \[\therefore \alpha ={{a}^{2}}+{{b}^{2}},\beta =2ab\]
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