Discussion
If \(A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\), then \(A^{50}\) is
Answer: C
\(A^{50}\) is the matrix with first column \((1, 25, 25)\) and the identity elsewhere.
Square \(A\) first:
\[ A^{2} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = I + N, \]
where \(N\) has 1 in positions (2,1) and (3,1) and zeros elsewhere. Then \(N^{2} = 0\), so by the binomial theorem
\[ A^{50} = (I + N)^{25} = I + 25N. \]
That gives entries 25 below the diagonal in the first column. The option with 50 is \(I + 50N\), which forgets that only \(A^{2}\), not \(A\), has the form \(I + N\); 48 and 24 come from miscounting the power in the same way.
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