If $$A = \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$$, then the determinant of the matrix $$(A^{2016} - 2A^{2015} - A^{2014})$$ is:

We are given the matrix $$A=\begin{bmatrix}-4 & -1 \\ 3 & 1\end{bmatrix}$$ and we have to evaluate the determinant of the matrix expression $$A^{2016}-2A^{2015}-A^{2014}.$$

First observe that each term in the expression contains a very high power of $$A$$. It is convenient to factor out the lowest power present, namely $$A^{2014}$$. We write

$$A^{2016}-2A^{2015}-A^{2014}=A^{2014}\left(A^{2}-2A-I\right).$$

Now we recall the fundamental property of determinants:

$$\det(AB)=\det A\;\det B \quad$$ for any two square matrices $$A,B.$$

Using this, the determinant of the whole product equals the product of the determinants:

$$\det\!\bigl(A^{2016}-2A^{2015}-A^{2014}\bigr)=\det\!\bigl(A^{2014}\bigr)\;\det\!\bigl(A^{2}-2A-I\bigr).$$

We evaluate the two determinants separately.

Determinant of $$A^{2014}$$
For any square matrix, the rule $$\det\!\bigl(A^k\bigr)=(\det A)^k$$ holds. Hence

$$\det\!\bigl(A^{2014}\bigr)=(\det A)^{2014}.$$

So we need $$\det A$$. For a 2 × 2 matrix $$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$$ the determinant is $$ad-bc$$. Therefore

$$\det A=(-4)(1)-(-1)(3)=-4+3=-1.$$

Raising this to the 2014-th power gives

$$\det\!\bigl(A^{2014}\bigr)=(-1)^{2014}=1.$$

Determinant of $$A^{2}-2A-I$$

To handle this, we use the eigenvalue approach. Let the eigenvalues of $$A$$ be $$\lambda_1$$ and $$\lambda_2$$. Then the eigenvalues of any polynomial $$f(A)$$ are simply $$f(\lambda_1)$$ and $$f(\lambda_2)$$. Thus

$$\det\!\bigl(A^{2}-2A-I\bigr)=\bigl(\lambda_1^{2}-2\lambda_1-1\bigr)\bigl(\lambda_2^{2}-2\lambda_2-1\bigr).$$

To compute these numbers, we first find the characteristic polynomial of $$A$$:

$$\det(A-\lambda I)= \begin{vmatrix}-4-\lambda & -1 \\[4pt] 3 & 1-\lambda\end{vmatrix} =(-4-\lambda)(1-\lambda)-(-1)(3) =(-4-\lambda)(1-\lambda)+3.$$

Expanding,

$$( -4-\lambda)(1-\lambda)=-4(1-\lambda)-\lambda(1-\lambda)=-4+4\lambda-\lambda+\lambda^{2}=-4+3\lambda+\lambda^{2}.$$

Adding the $$3$$ already present gives

$$\lambda^{2}+3\lambda-1=0.$$

Hence the eigenvalues satisfy the quadratic equation

$$\lambda^{2}+3\lambda-1=0 \quad\Longrightarrow\quad \lambda^{2}=-3\lambda+1.$$

Now we evaluate the polynomial $$\lambda^{2}-2\lambda-1$$ at any eigenvalue $$\lambda$$. Using the relation just obtained, we substitute $$\lambda^{2}=-3\lambda+1$$:

$$\lambda^{2}-2\lambda-1=(-3\lambda+1)-2\lambda-1=-5\lambda.$$

Therefore, for $$\lambda_1$$ and $$\lambda_2$$ we have

$$\lambda_1^{2}-2\lambda_1-1=-5\lambda_1,\qquad \lambda_2^{2}-2\lambda_2-1=-5\lambda_2.$$

The required determinant now becomes

$$\det\!\bigl(A^{2}-2A-I\bigr)=(-5\lambda_1)(-5\lambda_2)=25\lambda_1\lambda_2.$$

The product $$\lambda_1\lambda_2$$ equals the constant term of the characteristic polynomial (taken with sign), i.e. it equals $$-1$$ because

$$\lambda_1\lambda_2=\det A=-1.$$

Hence

$$\det\!\bigl(A^{2}-2A-I\bigr)=25(-1)=-25.$$

Putting everything together

Finally,

$$\det\!\bigl(A^{2016}-2A^{2015}-A^{2014}\bigr)= \det\!\bigl(A^{2014}\bigr)\;\det\!\bigl(A^{2}-2A-I\bigr)= 1\times(-25)=-25.$$

Hence, the correct answer is Option D.