Let A = `[(7,5),(4,8)] ,B= [(2,3),(3,5)]` and `C= [(5,-3),(-3,2)]` then `sum_(k=0)^(oo) (1)/(3^(k)) tr { A(BC)^(k)}`= _____

Answer

To solve the problem, we need to calculate the expression:

\[ \sum_{k=0}^{\infty} \frac{1}{3^k} \text{tr}(A(BC)^k) \]

where \( A = \begin{pmatrix} 7 & 5 \\ 4 & 8 \end{pmatrix} \), \( B = \begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix} \), and \( C = \begin{pmatrix} 5 & -3 \\ -3 & 2 \end{pmatrix} \).

### Step 1: Calculate \( BC \)

First, we need to compute the product \( BC \):

\[ BC = B \cdot C = \begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix} \cdot \begin{pmatrix} 5 & -3 \\ -3 & 2 \end{pmatrix} \]

Using the row-column multiplication rule:

\[ BC = \begin{pmatrix} (2 \cdot 5 + 3 \cdot -3) & (2 \cdot -3 + 3 \cdot 2) \\ (3 \cdot 5 + 5 \cdot -3) & (3 \cdot -3 + 5 \cdot 2) \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 2 \cdot 5 + 3 \cdot -3 = 10 - 9 = 1 \)
- First row, second column: \( 2 \cdot -3 + 3 \cdot 2 = -6 + 6 = 0 \)
- Second row, first column: \( 3 \cdot 5 + 5 \cdot -3 = 15 - 15 = 0 \)
- Second row, second column: \( 3 \cdot -3 + 5 \cdot 2 = -9 + 10 = 1 \)

Thus,

\[ BC = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

### Step 2: Substitute \( BC \) into the expression

Now that we have \( BC = I \), we can substitute this back into our original summation:

\[ \sum_{k=0}^{\infty} \frac{1}{3^k} \text{tr}(A(I)^k) = \sum_{k=0}^{\infty} \frac{1}{3^k} \text{tr}(A) \]

### Step 3: Calculate \( \text{tr}(A) \)

The trace of matrix \( A \) is the sum of its diagonal elements:

\[ \text{tr}(A) = 7 + 8 = 15 \]

### Step 4: Substitute \( \text{tr}(A) \) into the summation

Now we substitute \( \text{tr}(A) \) into the summation:

\[ \sum_{k=0}^{\infty} \frac{1}{3^k} \cdot 15 = 15 \sum_{k=0}^{\infty} \frac{1}{3^k} \]

### Step 5: Calculate the geometric series

The series \( \sum_{k=0}^{\infty} \frac{1}{3^k} \) is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \):

\[ \sum_{k=0}^{\infty} \frac{1}{3^k} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]

### Step 6: Final calculation

Now we can compute the final result:

\[ 15 \cdot \frac{3}{2} = \frac{45}{2} \]

Thus, the final answer is:

\[ \frac{45}{2} \]