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 \( C = AB + (AB)^2 + \ldots + (AB)^n \) as \( n \) approaches infinity, where \( A \) and \( B \) are given matrices.

### Step 1: Define the matrices \( A \) and \( B \)

The matrices are given as:
\[ A = \begin{pmatrix} x + \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{x} & 0 \\ 0 & 0 & 12 \end{pmatrix}, \quad B = \begin{pmatrix} \frac{x}{6(x^2 + 1)} & 0 & 0 \\ 0 & \frac{x}{4} & 0 \\ 0 & 0 & \frac{1}{36} \end{pmatrix} \]

### Step 2: Calculate the product \( AB \)

To find \( AB \), we multiply the two matrices:
\[ AB = \begin{pmatrix} (x + \frac{1}{x}) \cdot \frac{x}{6(x^2 + 1)} & 0 & 0 \\ 0 & \frac{1}{x} \cdot \frac{x}{4} & 0 \\ 0 & 0 & 12 \cdot \frac{1}{36} \end{pmatrix} \]

Calculating each element:
- First element: \( \frac{(x + \frac{1}{x})x}{6(x^2 + 1)} = \frac{x^2 + 1}{6(x^2 + 1)} = \frac{1}{6} \)
- Second element: \( \frac{1}{x} \cdot \frac{x}{4} = \frac{1}{4} \)
- Third element: \( 12 \cdot \frac{1}{36} = \frac{1}{3} \)

Thus, we have:
\[ AB = \begin{pmatrix} \frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{3} \end{pmatrix} \]

### Step 3: Calculate \( (AB)^2 \)

Since \( AB \) is a diagonal matrix, squaring it gives:
\[ (AB)^2 = \begin{pmatrix} \left(\frac{1}{6}\right)^2 & 0 & 0 \\ 0 & \left(\frac{1}{4}\right)^2 & 0 \\ 0 & 0 & \left(\frac{1}{3}\right)^2 \end{pmatrix} = \begin{pmatrix} \frac{1}{36} & 0 & 0 \\ 0 & \frac{1}{16} & 0 \\ 0 & 0 & \frac{1}{9} \end{pmatrix} \]

### Step 4: Generalize \( (AB)^n \)

For any \( n \):
\[ (AB)^n = \begin{pmatrix} \left(\frac{1}{6}\right)^n & 0 & 0 \\ 0 & \left(\frac{1}{4}\right)^n & 0 \\ 0 & 0 & \left(\frac{1}{3}\right)^n \end{pmatrix} \]

### Step 5: Calculate \( C \)

The matrix \( C \) can be expressed as a geometric series:
\[ C = AB + (AB)^2 + \ldots + (AB)^n = \sum_{k=1}^{n} (AB)^k \]

This is a geometric series where:
\[ C = \begin{pmatrix} \frac{1/6}{1 - 1/6} & 0 & 0 \\ 0 & \frac{1/4}{1 - 1/4} & 0 \\ 0 & 0 & \frac{1/3}{1 - 1/3} \end{pmatrix} = \begin{pmatrix} \frac{1/6}{5/6} & 0 & 0 \\ 0 & \frac{1/4}{3/4} & 0 \\ 0 & 0 & \frac{1/3}{2/3} \end{pmatrix} \]

Calculating each term:
- First element: \( \frac{1}{5} \)
- Second element: \( \frac{1}{3} \)
- Third element: \( \frac{1}{2} \)

Thus,
\[ C = \begin{pmatrix} \frac{1}{5} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{2} \end{pmatrix} \]

### Step 6: Calculate the trace of \( C \)

The trace of a matrix is the sum of its diagonal elements:
\[ \text{Tr}(C) = \frac{1}{5} + \frac{1}{3} + \frac{1}{2} \]

Finding a common denominator (which is 30):
\[ \text{Tr}(C) = \frac{6}{30} + \frac{10}{30} + \frac{15}{30} = \frac{31}{30} \]

### Final Answer

Thus, the trace of \( \lim_{n \to \infty} C \) is:
\[ \text{Tr}\left(\lim_{n \to \infty} C\right) = \frac{31}{30} \]