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If $A = \left[ {\begin{array}{*{20}{c}}a&0&0\\0&b&0\\0&0&c\end{array}} \right]$, then ${A^n} = $
$\left[ {\begin{array}{*{20}{c}}{na}&0&0\\0&{nb}&0\\0&0&{nc}\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}a&0&0\\0&b&0\\0&0&c\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}{{a^n}}&0&0\\0&{{b^n}}&0\\0&0&{{c^n}}\end{array}} \right]$
None of these
Solution
(c) Since ${A^2} = A.A = \left[ {\begin{array}{*{20}{c}}a&0&0\\0&b&0\\0&0&c\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}a&0&0\\0&b&0\\0&0&c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a^2}}&0&0\\0&{{b^2}}&0\\0&0&{{c^2}}\end{array}} \right]$
And ${A^3} = \left[ {\begin{array}{*{20}{c}}{{a^3}}&0&0\\0&{{b^3}}&0\\0&0&{{c^3}}\end{array}} \right]$,….
==> ${A^n} = {A^{n – 1}}.A = \left[ {\begin{array}{*{20}{c}}{{a^{n – 1}}}&0&0\\0&{{b^{n – 1}}}&0\\0&0&{{c^{n – 1}}}\end{array}} \right]{\rm{ }}\left[ {\begin{array}{*{20}{c}}a&0&0\\0&b&0\\0&0&c\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}{{a^n}}&0&0\\0&{{b^n}}&0\\0&0&{{c^n}}\end{array}} \right]$.
Note: Students should remember this question as a formula.
