Let \(A\) be a \(10 \times 10\) matrix such that \(A^{5}\) is a null matrix, and let \(I\) be the \(10 \times 10\) identity matrix. The determinant of \(A + I\) is \(\underline{\hspace{1cm}}\).
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Correct Answer: 1
Solution and Explanation
Nilpotent matrices have all eigenvalues equal to zero.
Thus eigenvalues of \(A + I\) are: \[ 1, 1, 1, \dots, 1 (10\ \text{times}) \] Hence, \[ \det(A+I) = 1^{10} = 1. \]
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