A) \[{{A}^{2}}=A\]and \[{{B}^{2}}\ne B\]
B) \[{{A}^{2}}\ne A\]and \[{{B}^{2}}=B\]
C) \[{{A}^{2}}=A\]and \[{{B}^{2}}=B\]
D) \[{{A}^{2}}\ne A\]and \[{{B}^{2}}\ne B\]
Correct Answer: C
Solution :
Given\[AB=A\], \ \[B=I\] Þ \[BA=B,\] \[\therefore \] A = I. Hence, \[\Rightarrow \] and\[\left[ \begin{matrix} 1 & -\left( \frac{x+y}{1+xy} \right) \\ -\left( \frac{x+y}{1+xy} \right) & 1 \\ \end{matrix} \right]\].
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