Question:

Let \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 3 \\ 1 & 0 & 0 \end{bmatrix} \), where \( a, c \in \mathbb{R} \). If \( A^n = A \) and the positive value of \( a \) belongs to the interval \( (n-1, n] \), where \( n \in \mathbb{N} \), then \( n \) is equal to:

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Matrix powers are important tools in determining periodicity and solving recurrence relations. Look for repeating patterns when calculating powers of matrices.
Updated On: Mar 22, 2025
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Correct Answer: 2

Solution and Explanation

We are given the matrix \[ A = \begin{bmatrix} 0 & 1 & 2
1 & 0 & 3
1 & 0 & 0 \end{bmatrix} \] First, compute powers of \( A \): 1. \( A^2 = A \times A = \begin{bmatrix} 0 & 1 & 2
1 & 0 & 3
1 & 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 & 2
1 & 0 & 3
1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 6
3 & 2 & 9
2 & 3 & 6 \end{bmatrix} \)
2. \( A^3 = A \times A^2 = \begin{bmatrix} 0 & 1 & 2
1 & 0 & 3
1 & 0 & 0 \end{bmatrix} \times \begin{bmatrix} 2 & 3 & 6
3 & 2 & 9
2 & 3 & 6 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 6
3 & 2 & 9
2 & 3 & 6 \end{bmatrix} \) We observe that \( A^2 = A^3 \), indicating that \( n = 3 \).
Thus, the value of \( n \) is \( \boxed{3} \).
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