Discussion
\(A^{35}\) if \(A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\) is :
Answer: B
\(A^{35} = \begin{pmatrix} 1 & 35 \\ 0 & 1 \end{pmatrix}\).
Write \(A = I + N\) with \(N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\). Since \(N^{2} = 0\) and \(I\) commutes with \(N\), the binomial theorem gives
\[ A^{n} = I + nN = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \]
For \(n = 35\), the top-right entry is 35. The 70 in option A is the entry of \(A^{70}\), and options C and D change the diagonal, which stays 1 because \(A\) is upper triangular with 1s on the diagonal.
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