MCQ
$A = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$ and $I = \left[ {\begin{array}{*{20}{c}}
1&0\\0&1\end{array}} \right]$ , then which of the following holds for all $n \geq 2, n \in N$ ?
  • A
    ${A^n} = {2^{n - 1}}A + \left( {n - 1} \right)I$
  • B
    ${A^n} = nA + \left( {n - 1} \right)I$
  • C
    ${A^n} = {2^{n - 1}}A - \left( {n - 1} \right)I$
  • ${A^n} = nA - \left( {n - 1} \right)I$

Answer

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

it satisfies only option $(D)$

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