Question

Let \( A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix} \). If \( B = [b_{ij}]_{3 \times 3} \) and \( B = A^{99} - I \), then find \( \frac{b_{31} - b_{21}}{b_{32}} \).

Hint:

For a matrix where sub-diagonal elements are \(k\) and the element at (3,1) is \(k^2\), the power \(A^n\) follows a predictable pattern related to arithmetic series.

Answer 1

Correct Answer: 149

Step 1: Understanding the Concept:
We need to find the power of a lower triangular matrix. We can decompose \( A \) into \( I + N \), where \( N \) is a nilpotent matrix. Then we use the Binomial expansion: \( A^n = (I + N)^n = I + nN + \frac{n(n-1)}{2}N^2 + \dots \).

Step 2: Key Formula or Approach:
Let \( A = I + N \), where \( N = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} \). Calculate \( N^2 \): \[ N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} \] Note that \( N^3 = O \).

Step 3: Detailed Explanation:
1. Apply Binomial Expansion for \( A^{99} \): \[ A^{99} = (I + N)^{99} = I + 99N + \frac{99 \times 98}{2} N^2 \] 2. Find \( B = A^{99} - I \): \[ B = 99 \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + 4851 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 891 + 43659 & 297 & 0 \end{bmatrix} \] 3. Identify relevant elements: - \( b_{32} = 297 \) - \( b_{21} = 297 \) - \( b_{31} = 44550 \) 4. Calculate the ratio: \[ \frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} = 149 \]
Step 4: Final Answer:
The value is 149.