Let `P=[[1,0,0],[4,1,0],[16,4,1]]`and `I` be the identity matrix of order `3`. If `Q = [q_()ij ]` is a matrix, such that `P^(50)-Q=I`, then `(q_(31)+q_(32))/q_(21)` equals

Options

A. 52
B. 103
C. 201
D. 205

Answer

To solve the problem, we need to find the value of \((q_{31} + q_{32}) / q_{21}\) given the matrix \(P\) and the equation \(P^{50} - Q = I\), where \(I\) is the identity matrix of order 3.

1. **Define the matrix \(P\)**:
\[ P = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} \]

2. **Calculate \(P^2\)**:
\[ P^2 = P \cdot P = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} \]
Performing the multiplication:
\[ P^2 = \begin{bmatrix} 1 & 0 & 0 \\ 4 + 4 & 1 & 0 \\ 16 + 16 + 4 & 4 + 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 36 & 8 & 1 \end{bmatrix} \]

3. **Calculate \(P^3\)**:
\[ P^3 = P^2 \cdot P = \begin{bmatrix} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 36 & 8 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} \]
Performing the multiplication:
\[ P^3 = \begin{bmatrix} 1 & 0 & 0 \\ 8 + 32 & 1 & 0 \\ 36 + 144 + 32 & 8 + 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 40 & 1 & 0 \\ 212 & 12 & 1 \end{bmatrix} \]

4. **Identify the pattern**:
From the calculations, we observe that:
- The first column remains \([1, 4k, 16k^2]\).
- The second column remains \([0, 1, 4]\).
- The third column remains \([0, 0, 1]\).

Continuing this pattern, we can conjecture that:
\[ P^n = \begin{bmatrix} 1 & 0 & 0 \\ 4n & 1 & 0 \\ 16n(n+1)/2 & 4n & 1 \end{bmatrix} \]

5. **Calculate \(P^{50}\)**:
Using the pattern:
\[ P^{50} = \begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ 4000 & 200 & 1 \end{bmatrix} \]

6. **Find \(Q\)**:
From the equation \(P^{50} - Q = I\):
\[ Q = P^{50} - I = \begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ 4000 & 200 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Thus,
\[ Q = \begin{bmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 4000 & 200 & 0 \end{bmatrix} \]

7. **Calculate \((q_{31} + q_{32}) / q_{21}\)**:
From the matrix \(Q\):
- \(q_{31} = 4000\)
- \(q_{32} = 200\)
- \(q_{21} = 200\)

Therefore:
\[ \frac{q_{31} + q_{32}}{q_{21}} = \frac{4000 + 200}{200} = \frac{4200}{200} = 21 \]

Thus, the final answer is:
\[ \boxed{21} \]