For `xgt 0. " let A"=[(x+(1)/(x),0,0),(0,1//x,0),(0,0,12)], B=[((x)/(6(x^(2)+1)),0,0),(0,(x)/(4),0),(0,0,(1)/(36))]` be two matrices and `C=AB+(AB)^(2)+….+(AB)^(n).` Then, `Tr(lim_(nrarroo)C)` is equal to (where `Tr(A)` is the trace of the matrix A i.e. the sum of the principle diagonal elements of A)

Options

A. 1
B. `(31)/(30)`
C. `(6)/(5)`
D. `(1)/(3)`

Answer

To solve the problem, we need to find the trace of the limit of the matrix \( C \) as \( n \) approaches infinity, where \( C = AB + AB^2 + AB^3 + \ldots + AB^n \).

### Step-by-Step Solution:

1. **Define the Matrices \( A \) and \( B \)**:
\[ A = \begin{pmatrix} x + \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{x} & 0 \\ 0 & 0 & 12 \end{pmatrix}, \quad B = \begin{pmatrix} \frac{x}{6(x^2 + 1)} & 0 & 0 \\ 0 & \frac{x}{4} & 0 \\ 0 & 0 & \frac{1}{36} \end{pmatrix} \]

2. **Calculate the Product \( AB \)**:
\[ AB = \begin{pmatrix} (x + \frac{1}{x}) \cdot \frac{x}{6(x^2 + 1)} & 0 & 0 \\ 0 & \frac{1}{x} \cdot \frac{x}{4} & 0 \\ 0 & 0 & 12 \cdot \frac{1}{36} \end{pmatrix} \]
Simplifying each element:
\[ AB = \begin{pmatrix} \frac{x^2 + 1}{6} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{3} \end{pmatrix} \]

3. **Calculate \( AB^2 \)**:
\[ AB^2 = AB \cdot AB = \begin{pmatrix} \left(\frac{x^2 + 1}{6}\right)^2 & 0 & 0 \\ 0 & \left(\frac{1}{4}\right)^2 & 0 \\ 0 & 0 & \left(\frac{1}{3}\right)^2 \end{pmatrix} \]
Thus,
\[ AB^2 = \begin{pmatrix} \frac{(x^2 + 1)^2}{36} & 0 & 0 \\ 0 & \frac{1}{16} & 0 \\ 0 & 0 & \frac{1}{9} \end{pmatrix} \]

4. **Generalize \( AB^n \)**:
\[ AB^n = \begin{pmatrix} \left(\frac{x^2 + 1}{6}\right)^n & 0 & 0 \\ 0 & \left(\frac{1}{4}\right)^n & 0 \\ 0 & 0 & \left(\frac{1}{3}\right)^n \end{pmatrix} \]

5. **Sum the Series to Find \( C \)**:
\[ C = AB + AB^2 + AB^3 + \ldots + AB^n \]
This can be expressed as:
\[ C = \begin{pmatrix} \sum_{k=1}^{n} \left(\frac{x^2 + 1}{6}\right)^k & 0 & 0 \\ 0 & \sum_{k=1}^{n} \left(\frac{1}{4}\right)^k & 0 \\ 0 & 0 & \sum_{k=1}^{n} \left(\frac{1}{3}\right)^k \end{pmatrix} \]

6. **Evaluate the Limit as \( n \to \infty \)**:
Using the formula for the sum of a geometric series \( S = \frac{a}{1 - r} \):
\[ \lim_{n \to \infty} C = \begin{pmatrix} \frac{\frac{x^2 + 1}{6}}{1 - \frac{x^2 + 1}{6}} & 0 & 0 \\ 0 & \frac{\frac{1}{4}}{1 - \frac{1}{4}} & 0 \\ 0 & 0 & \frac{\frac{1}{3}}{1 - \frac{1}{3}} \end{pmatrix} \]

7. **Calculate Each Element**:
- For the first element:
\[ \frac{\frac{x^2 + 1}{6}}{1 - \frac{x^2 + 1}{6}} = \frac{x^2 + 1}{6 - (x^2 + 1)} = \frac{x^2 + 1}{5 - x^2} \]
- For the second element:
\[ \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]
- For the third element:
\[ \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

8. **Find the Trace**:
The trace of the matrix is the sum of the diagonal elements:
\[ \text{Tr}(C) = \frac{x^2 + 1}{5 - x^2} + \frac{1}{3} + \frac{1}{2} \]

### Final Calculation:
To find a common denominator for the trace:
- The common denominator of \( 5 - x^2 \), \( 3 \), and \( 2 \) is \( 6(5 - x^2) \).
- Thus, the trace simplifies to:
\[ \text{Tr}(C) = \frac{6(x^2 + 1) + 2(5 - x^2) + 3(5 - x^2)}{6(5 - x^2)} \]
Calculating the numerator:
\[ 6x^2 + 6 + 10 - 2x^2 + 15 - 3x^2 = 31 \]
Thus, the final trace is:
\[ \text{Tr}(C) = \frac{31}{6(5 - x^2)} \]

### Conclusion:
The trace of the limit as \( n \to \infty \) is:
\[ \text{Tr}(\lim_{n \to \infty} C) = \frac{31}{30} \]