If matrix A is given by `A=[[6,11] , [2,4]]` then determinant of `A^(2005)-6A^(2004)` is

Options

A. `2^(2006)`
B. `(-11)2^(2005)`
C. `-2^(2005).7`
D. `(-9) 2^(2004)`

Explanation

To solve the problem, we need to find the determinant of the expression \( A^{2005} - 6A^{2004} \), where the matrix \( A \) is given by:

\[ A = \begin{bmatrix} 6 & 11 \\ 2 & 4 \end{bmatrix} \]

### Step 1: Calculate the determinant of matrix \( A \)

The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is calculated using the formula:

\[ \text{det}(A) = ad - bc \]

For our matrix \( A \):

\[ \text{det}(A) = (6)(4) - (2)(11) = 24 - 22 = 2 \]

### Step 2: Express \( A^{2005} - 6A^{2004} \)

We can factor out \( A^{2004} \) from the expression:

\[ A^{2005} - 6A^{2004} = A^{2004}(A - 6I) \]

where \( I \) is the identity matrix.

### Step 3: Calculate \( A - 6I \)

The identity matrix \( I \) for a 2x2 matrix is:

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

Thus,

\[ 6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \]

Now, we can compute \( A - 6I \):

\[ A - 6I = \begin{bmatrix} 6 & 11 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 0 & 11 \\ 2 & -2 \end{bmatrix} \]

### Step 4: Calculate the determinant of \( A - 6I \)

Using the determinant formula for the matrix \( A - 6I \):

\[ \text{det}(A - 6I) = (0)(-2) - (2)(11) = 0 - 22 = -22 \]

### Step 5: Calculate the determinant of \( A^{2005} - 6A^{2004} \)

Using the property of determinants:

\[ \text{det}(A^{2005} - 6A^{2004}) = \text{det}(A^{2004}) \cdot \text{det}(A - 6I) \]

Since \( \text{det}(A^{n}) = (\text{det}(A))^n \):

\[ \text{det}(A^{2004}) = (\text{det}(A))^{2004} = 2^{2004} \]

Now, substituting back:

\[ \text{det}(A^{2005} - 6A^{2004}) = 2^{2004} \cdot (-22) \]

### Final Result

Thus, the determinant of \( A^{2005} - 6A^{2004} \) is:

\[ \text{det}(A^{2005} - 6A^{2004}) = -22 \cdot 2^{2004} \]