Key Concepts and Formulas

1. Properties of Determinants: For any two square matrices $A$ and $B$ of the same order, the determinant of their product is the product of their determinants: $\det(AB) = \det(A) \det(B)$ A direct consequence for any positive integer $n$ is: $\det(A^n) = (\det(A))^n$ Also, for a scalar $k$ and an $n \times n$ matrix $A$, $\det(kA) = k^n \det(A)$. For a $2 \times 2$ matrix, $\det(kA) = k^2 \det(A)$.

2. Matrix Algebra Operations: Basic operations such as matrix multiplication, scalar multiplication, and matrix addition/subtraction. The identity matrix $I$ plays a crucial role as the multiplicative identity in matrix algebra, where $A \cdot I = I \cdot A = A$. When factoring out a matrix from an expression like $A^2 - 2A - 1$, the scalar '1' must be replaced by the identity matrix $I$ to maintain matrix dimension compatibility, resulting in $A^2 - 2A - I$.

3. Cayley-Hamilton Theorem (Advanced Tip): Every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A$, its characteristic equation is $\det(A - \lambda I) = 0$, which can be written as $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$. By the theorem, $A^2 - \text{tr}(A)A + \det(A)I = 0$. This theorem is powerful for simplifying matrix polynomials.

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Step-by-Step Solution

Given: The matrix $A = \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$. We need to find the determinant of the matrix expression $(A^{2016} - 2A^{2015} - A^{2014})$.

Step 1: Simplify the matrix expression using factorization. The given expression involves very high powers of $A$. Direct calculation of $A^{2014}$, $A^{2015}$, or $A^{2016}$ is computationally infeasible and unnecessary. We observe that $A^{2014}$ is a common factor in all terms. Let $M = A^{2016} - 2A^{2015} - A^{2014}$. We can factor out $A^{2014}$ from the expression. Remember that when factoring out from a constant term (like $-1$ in this context), we must replace it with the identity matrix $I$ of the same order as $A$. $M = A^{2014}(A^2 - 2A - I)$ Now, we need to find $\det(M)$. Using the determinant property $\det(AB) = \det(A)\det(B)$: $\det(M) = \det(A^{2014}) \det(A^2 - 2A - I)$ And using $\det(A^n) = (\det(A))^n$: $\det(M) = (\det(A))^{2014} \det(A^2 - 2A - I)$ Our strategy is now to calculate $\det(A)$ and $\det(A^2 - 2A - I)$ separately and then combine them.

Step 2: Calculate the determinant of matrix A, $\det(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$. $\det(A) = \det \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix} = (-4)(1) - (-1)(3)$ $= -4 - (-3) = -4 + 3 = -1$ So, $\det(A) = -1$.

Step 3: Calculate the matrix expression $A^2 - 2A - I$. We will calculate $A^2$, $2A$, and $I$ separately and then combine them.

First, calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$ Perform matrix multiplication (row by column): $A^2 = \begin{bmatrix} (-4)(-4) + (-1)(3) & (-4)(-1) + (-1)(1) \\ (3)(-4) + (1)(3) & (3)(-1) + (1)(1) \end{bmatrix}$ $A^2 = \begin{bmatrix} 16 - 3 & 4 - 1 \\ -12 + 3 & -3 + 1 \end{bmatrix} = \begin{bmatrix} 13 & 3 \\ -9 & -2 \end{bmatrix}$

Next, calculate $2A$: $2A = 2 \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 2 \cdot (-4) & 2 \cdot (-1) \\ 2 \cdot 3 & 2 \cdot 1 \end{bmatrix} = \begin{bmatrix} -8 & -2 \\ 6 & 2 \end{bmatrix}$

The identity matrix $I$ for a $2 \times 2$ matrix is: $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Now, substitute these matrices into the expression $A^2 - 2A - I$: $A^2 - 2A - I = \begin{bmatrix} 13 & 3 \\ -9 & -2 \end{bmatrix} - \begin{bmatrix} -8 & -2 \\ 6 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ Perform the subtraction element by element: $A^2 - 2A - I = \begin{bmatrix} 13 - (-8) - 1 & 3 - (-2) - 0 \\ -9 - 6 - 0 & -2 - 2 - 1 \end{bmatrix}$ $A^2 - 2A - I = \begin{bmatrix} 13 + 8 - 1 & 3 + 2 - 0 \\ -15 & -5 \end{bmatrix} = \begin{bmatrix} 20 & 5 \\ -15 & -5 \end{bmatrix}$

Step 4: Calculate the determinant of the matrix $(A^2 - 2A - I)$. Let $B = A^2 - 2A - I = \begin{bmatrix} 20 & 5 \\ -15 & -5 \end{bmatrix}$. $\det(B) = (20)(-5) - (5)(-15)$ $= -100 - (-75)$ $= -100 + 75 = -25$ So, $\det(A^2 - 2A - I) = -25$.

Step 5: Combine the results to find $\det(A^{2016} - 2A^{2015} - A^{2014})$. From Step 1, we established: $\det(A^{2016} - 2A^{2015} - A^{2014}) = (\det(A))^{2014} \det(A^2 - 2A - I)$ Substitute the values we calculated: $\det(A) = -1$ (from Step 2) and $\det(A^2 - 2A - I) = -25$ (from Step 4). $\det(A^{2016} - 2A^{2015} - A^{2014}) = (-1)^{2014} \cdot (-25)$ Since 2014 is an even number, $(-1)^{2014} = 1$. $= 1 \cdot (-25) = -25$ Thus, the determinant of the matrix $(A^{2016} - 2A^{2015} - A^{2014})$ is $-25$.

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Common Mistakes & Tips

• Careful Matrix Arithmetic: Matrix multiplication, scalar multiplication, and subtraction must be performed accurately, element by element. Even a minor calculation error can lead to an incorrect final answer.
• Understanding the Identity Matrix: Remember that in matrix expressions, a scalar constant (like the $-1$ in $A^2 - 2A - 1$) must be represented as a scalar multiple of the identity matrix ($-1 \cdot I = -I$) to maintain dimensional consistency for matrix addition/subtraction.
• Leverage Determinant Properties: Always look for opportunities to simplify expressions involving determinants by using properties like $\det(AB) = \det(A)\det(B)$ and $\det(A^n) = (\det(A))^n$. This avoids the arduous task of calculating high powers of matrices directly.
• Cayley-Hamilton Theorem for Efficiency: For $2 \times 2$ matrices, the Cayley-Hamilton theorem often provides a more elegant and quicker path to simplify matrix polynomials. For matrix $A = \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$:
  1. Trace of $A$, $\text{tr}(A) = -4 + 1 = -3$.
  2. Determinant of $A$, $\det(A) = -1$ (as calculated in Step 2).
  3. The characteristic equation is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$, so $\lambda^2 - (-3)\lambda + (-1) = 0 \implies \lambda^2 + 3\lambda - 1 = 0$.
  4. By Cayley-Hamilton Theorem, $A^2 + 3A - I = 0$, which implies $A^2 = -3A + I$.
  5. Substitute this into the expression $(A^2 - 2A - I)$: $(-3A + I) - 2A - I = -5A$.
  6. So, the determinant required is $\det(A^{2014}(-5A)) = (\det(A))^{2014} \det(-5A)$.
  7. $(\det(A))^{2014} = (-1)^{2014} = 1$.
  8. For a $2 \times 2$ matrix, $\det(kA) = k^2 \det(A)$. Thus, $\det(-5A) = (-5)^2 \det(A) = 25 \cdot (-1) = -25$.
  9. The final result is $1 \cdot (-25) = -25$. This alternative method confirms the result and is generally faster.

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Summary

The problem asked for the determinant of a matrix expression involving high powers of $A$. The key to solving this efficiently was to first factor out $A^{2014}$ from the expression, simplifying it to $A^{2014}(A^2 - 2A - I)$. This allowed us to use the determinant properties $\det(AB) = \det(A)\det(B)$ and $\det(A^n) = (\det(A))^n$. We then calculated $\det(A) = -1$ and the matrix $A^2 - 2A - I$, followed by its determinant, which was $-25$. Combining these results, we found the final determinant to be $(-1)^{2014} \cdot (-25) = 1 \cdot (-25) = -25$.

The final answer is $\boxed{-25}$, which corresponds to option (D).