Let `A =[[2, -1],[1, 1]]`. If the sum of the diagonal elements of `A^(13)` is `3^n`, then `n` is equal to ___________.

Answer

To solve the problem, we need to find the value of \( n \) such that the sum of the diagonal elements (trace) of the matrix \( A^{13} \) is equal to \( 3^n \). The matrix \( A \) is given as:

\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \]

### Step 1: Calculate \( A^2 \)

First, we calculate \( A^2 \):

\[ A^2 = A \cdot A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \]

Calculating the elements:

- First row, first column: \( 2 \cdot 2 + (-1) \cdot 1 = 4 - 1 = 3 \)
- First row, second column: \( 2 \cdot (-1) + (-1) \cdot 1 = -2 - 1 = -3 \)
- Second row, first column: \( 1 \cdot 2 + 1 \cdot 1 = 2 + 1 = 3 \)
- Second row, second column: \( 1 \cdot (-1) + 1 \cdot 1 = -1 + 1 = 0 \)

Thus, we have:

\[ A^2 = \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix} \]

### Step 2: Calculate \( A^4 \)

Next, we calculate \( A^4 \) by squaring \( A^2 \):

\[ A^4 = A^2 \cdot A^2 = \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix} \cdot \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix} \]

Calculating the elements:

- First row, first column: \( 3 \cdot 3 + (-3) \cdot 3 = 9 - 9 = 0 \)
- First row, second column: \( 3 \cdot (-3) + (-3) \cdot 0 = -9 + 0 = -9 \)
- Second row, first column: \( 3 \cdot 3 + 0 \cdot 3 = 9 + 0 = 9 \)
- Second row, second column: \( 3 \cdot (-3) + 0 \cdot 0 = -9 + 0 = -9 \)

Thus, we have:

\[ A^4 = \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix} \]

### Step 3: Calculate \( A^8 \)

Next, we calculate \( A^8 \):

\[ A^8 = A^4 \cdot A^4 = \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix} \cdot \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix} \]

Calculating the elements:

- First row, first column: \( 0 \cdot 0 + (-9) \cdot 9 = 0 - 81 = -81 \)
- First row, second column: \( 0 \cdot (-9) + (-9) \cdot (-9) = 0 + 81 = 81 \)
- Second row, first column: \( 9 \cdot 0 + (-9) \cdot 9 = 0 - 81 = -81 \)
- Second row, second column: \( 9 \cdot (-9) + (-9) \cdot (-9) = -81 + 81 = 0 \)

Thus, we have:

\[ A^8 = \begin{bmatrix} -81 & 81 \\ -81 & 0 \end{bmatrix} \]

### Step 4: Calculate \( A^{12} \)

Next, we calculate \( A^{12} \):

\[ A^{12} = A^8 \cdot A^4 = \begin{bmatrix} -81 & 81 \\ -81 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix} \]

Calculating the elements:

- First row, first column: \( -81 \cdot 0 + 81 \cdot 9 = 0 + 729 = 729 \)
- First row, second column: \( -81 \cdot (-9) + 81 \cdot (-9) = 729 - 729 = 0 \)
- Second row, first column: \( -81 \cdot 0 + 0 \cdot 9 = 0 + 0 = 0 \)
- Second row, second column: \( -81 \cdot (-9) + 0 \cdot (-9) = 729 + 0 = 729 \)

Thus, we have:

\[ A^{12} = \begin{bmatrix} 729 & 0 \\ 0 & 729 \end{bmatrix} = 729 I \]

### Step 5: Calculate \( A^{13} \)

Now, we calculate \( A^{13} \):

\[ A^{13} = A^{12} \cdot A = 729 I \cdot A = 729 \cdot \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \]

Thus, we have:

\[ A^{13} = \begin{bmatrix} 1458 & -729 \\ 729 & 729 \end{bmatrix} \]

### Step 6: Calculate the Trace of \( A^{13} \)

The trace of a matrix is the sum of its diagonal elements:

\[ \text{Trace}(A^{13}) = 1458 + 729 = 2187 \]

### Step 7: Express Trace in Terms of \( 3^n \)

Now, we express \( 2187 \) as a power of \( 3 \):

\[ 2187 = 3^7 \]

### Conclusion

Thus, we find that \( n = 7 \).

Therefore, the answer is:

\[ \boxed{7} \]