If \(A=\left(\begin{array}{ll}3 & -2 \\ 2 & -1\end{array}\right)\), then A20 equals

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  1. \(\left(\begin{array}{cc} 41 & 40 \\ -40 & -39 \end{array}\right)\)
  2. \(\left(\begin{array}{ll} 41 & -40 \\ 40 & -39 \end{array}\right)\)
  3. \(\left(\begin{array}{cc} 41 & -40 \\ -40 & -39 \end{array}\right)\)
  4. \(\left(\begin{array}{cc} 41 & 40 \\ 40 & -39 \end{array}\right)\)

Answer (Detailed Solution Below)

Option 2 : \(\left(\begin{array}{ll} 41 & -40 \\ 40 & -39 \end{array}\right)\)
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Detailed Solution

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Calculation: 

Given 

\(A=\left(\begin{array}{ll}3 & -2 \\ 2 & -1\end{array}\right)\)

\(A^2=\left(\begin{array}{ll}5 & -4 \\ 4 & -3\end{array}\right)\)

\(A^4=\left(\begin{array}{ll}9 & -8\\ 8 & -7\end{array}\right)\)

\(A^8=\left(\begin{array}{ll}17 & -16\\ 16 & 15\end{array}\right)\)

\(A^{16}=\left(\begin{array}{ll}33 & -32\\ 32 & 31\end{array}\right)\)

and 

20 = \(\left(\begin{array}{ll} 41 & -40 \\ 40 & -39 \end{array}\right)\)

Hence option (2) is correct 

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