Discussion
Let \(A = \begin{pmatrix} 5 & -3 \\ 6 & -4 \end{pmatrix}\). Then the trace of \(A^{1000}\) equals
Answer: B
The trace of \(A^{1000}\) is \(2^{1000} + 1\).
The eigenvalues of A satisfy \(\lambda^{2} - (\text{tr}A)\lambda + \det A = 0\), with \(\text{tr}A = 5 - 4 = 1\) and \(\det A = -20 + 18 = -2\):
\[ \lambda^{2} - \lambda - 2 = 0 \Rightarrow \lambda = 2,\ -1 \]
The eigenvalues of \(A^{1000}\) are \(2^{1000}\) and \((-1)^{1000} = 1\), and the trace is their sum, \(2^{1000} + 1\).
\(2^{1000} - 1\) takes the second eigenvalue's power as -1 (an odd exponent would), and \(2^{1000}\) leaves it out; 1 counts only the second eigenvalue.
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