Question
If $A = \left[ {\begin{array}{*{20}{c}}{ab}&{{b^2}}\\{ - {a^2}}&{ - ab}\end{array}} \right]$ and ${A^n} = O$, then the minimum value of $n$ is

Answer

a
(a) ${A^2} = A.\,\,A = \left[ {\begin{array}{*{20}{c}}{ab}&{{b^2}}\\{ - {a^2}}&{ - ab}\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ab}&{{b^2}}\\{ - {a^2}}&{ - ab}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{{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{array}} \right] = O$

$ \Rightarrow \,\,{A^3} = A.{A^2} = 0$ and ${A^n} = 0$, for all $n \ge 2$.

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