A) 2
B) 3
C) 4
D) 5
Correct Answer: A
Solution :
\[{{A}^{2}}=A.\,\,A=\left[ \begin{matrix} ab & {{b}^{2}} \\ -{{a}^{2}} & -ab \\ \end{matrix} \right]\,\left[ \begin{matrix} ab & {{b}^{2}} \\ -{{a}^{2}} & -ab \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{a}^{2}}{{b}^{2}}-{{a}^{2}}{{b}^{2}} & a{{b}^{3}}-a{{b}^{3}} \\ -{{a}^{3}}b+{{a}^{3}}b & -{{a}^{2}}{{b}^{2}}+{{a}^{2}}{{b}^{2}} \\ \end{matrix} \right]=O\] \[\Rightarrow \,\,{{A}^{3}}=A.{{A}^{2}}=0\]and\[{{A}^{n}}=0\], for all \[n\ge 2\].
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