If `{:A=[(-4,-1),(3,1)]:}`, then the determint of the matrix `(A^2016-2A^2015-A^2014)`,is

Options

A. `-175`
B. `2014`
C. 2016
D. `-25`

Answer

To solve the problem, we need to find the determinant of the expression \( A^{2016} - 2A^{2015} - A^{2014} \) where the matrix \( A \) is given as:

\[ A = \begin{pmatrix} -4 & -1 \\ 3 & 1 \end{pmatrix} \]

### Step 1: Factor out \( A^{2014} \)

We can factor out \( A^{2014} \) from the expression:

\[ A^{2016} - 2A^{2015} - A^{2014} = A^{2014}(A^2 - 2A - I) \]

### Step 2: Calculate the determinant

Using the property of determinants, we have:

\[ \text{det}(A^{2016} - 2A^{2015} - A^{2014}) = \text{det}(A^{2014}) \cdot \text{det}(A^2 - 2A - I) \]

### Step 3: Calculate \( \text{det}(A^{2014}) \)

The determinant of \( A \) is calculated as follows:

\[ \text{det}(A) = (-4)(1) - (-1)(3) = -4 + 3 = -1 \]

Thus,

\[ \text{det}(A^{2014}) = (-1)^{2014} = 1 \]

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

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

\[ A^2 = A \cdot A = \begin{pmatrix} -4 & -1 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} -4 & -1 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} (-4)(-4) + (-1)(3) & (-4)(-1) + (-1)(1) \\ (3)(-4) + (1)(3) & (3)(-1) + (1)(1) \end{pmatrix} \]

Calculating each entry:

- First row, first column: \( 16 - 3 = 13 \)
- First row, second column: \( 4 - 1 = 3 \)
- Second row, first column: \( -12 + 3 = -9 \)
- Second row, second column: \( -3 + 1 = -2 \)

Thus,

\[ A^2 = \begin{pmatrix} 13 & 3 \\ -9 & -2 \end{pmatrix} \]

Next, we calculate \( -2A \):

\[ -2A = -2 \begin{pmatrix} -4 & -1 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 8 & 2 \\ -6 & -2 \end{pmatrix} \]

Now, we calculate \( -I \):

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

Now, we combine these results to find \( A^2 - 2A - I \):

\[ A^2 - 2A - I = \begin{pmatrix} 13 & 3 \\ -9 & -2 \end{pmatrix} + \begin{pmatrix} 8 & 2 \\ -6 & -2 \end{pmatrix} + \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

Calculating each entry:

- First row, first column: \( 13 + 8 - 1 = 20 \)
- First row, second column: \( 3 + 2 + 0 = 5 \)
- Second row, first column: \( -9 - 6 + 0 = -15 \)
- Second row, second column: \( -2 - 2 - 1 = -5 \)

Thus,

\[ A^2 - 2A - I = \begin{pmatrix} 20 & 5 \\ -15 & -5 \end{pmatrix} \]

### Step 5: Calculate the determinant of \( A^2 - 2A - I \)

Now, we calculate the determinant:

\[ \text{det}(A^2 - 2A - I) = (20)(-5) - (5)(-15) = -100 + 75 = -25 \]

### Step 6: Combine results

Now we combine the results:

\[ \text{det}(A^{2016} - 2A^{2015} - A^{2014}) = \text{det}(A^{2014}) \cdot \text{det}(A^2 - 2A - I) = 1 \cdot (-25) = -25 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{-25} \]