A) more than 2
B) 2
C) 0
D) 1
Correct Answer: A
Solution :
[a] Let \[X=\left( \begin{matrix} a & b \\ c & d \\ \end{matrix} \right)\] \[\Rightarrow {{X}^{2}}=\left( \begin{matrix} {{a}^{2}}+bc & ab+bd \\ ac+cd & bc+{{d}^{2}} \\ \end{matrix} \right)\] \[\Rightarrow {{a}^{2}}+bc=1\] and \[ab+bd=1\]\[\Rightarrow b(a+d)=1\] \[ac+cd=2\Rightarrow c(a+d)=2\Rightarrow 2b=c\] Also. \[bc+{{d}^{2}}=3\Rightarrow {{d}^{2}}-{{a}^{2}}=2\] \[\Rightarrow (d-a)(a+d)=2\Rightarrow d-a=2b\] (Using \[bc=1-{{a}^{2}}\]) \[\Rightarrow 2d=2b+1/b,\]\[2a=1/b-2b\] \[d=b+1/2b,\]\[a=1/(2b)-b\] \[c=2b\] \[\Rightarrow \left( {{b}^{2}}+\frac{1}{4{{b}^{2}}}+1 \right)+2{{b}^{2}}=3\] \[\Rightarrow 3{{b}^{2}}+\frac{1}{4{{b}^{2}}}=2\] \[\Rightarrow 3x+\frac{1}{4x}=2\] Or \[b=\pm \frac{1}{\sqrt{6}}\] or \[b=\pm \frac{1}{\sqrt{2}}\] Therefore, matrices are \[\left( \begin{matrix} 0 & 1/\sqrt{2} \\ \sqrt{2} & \sqrt{2} \\ \end{matrix} \right),\left( \begin{matrix} 0 & -1\sqrt{2} \\ -\sqrt{2} & -\sqrt{2} \\ \end{matrix} \right),\left( \begin{matrix} 2\sqrt{6} & -1/\sqrt{6} \\ 2/\sqrt{6} & 4\sqrt{6} \\ \end{matrix} \right)\]
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