MCQ
If $A = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]$, then ${A^4}$ is equal to
  • A
    $\left[ {\begin{array}{*{20}{c}}1&{{a^4}}\\0&1\end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}4&{4a}\\0&4\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}4&{{a^4}}\\0&4\end{array}} \right]$
  • $\left[ {\begin{array}{*{20}{c}}1&{4a}\\0&1\end{array}} \right]$

Answer

Correct option: D.
$\left[ {\begin{array}{*{20}{c}}1&{4a}\\0&1\end{array}} \right]$
d
(d) ${A^2} = A.\,A = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{2a}\\0&1\end{array}} \right]$

${A^3} = A.\,{A^2} = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&{2a}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{3a}\\0&1\end{array}} \right]$

${A^4} = A.\,{A^3} = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&{3a}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{4a}\\0&1\end{array}} \right]$.

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