If A = [
\begin{matrix}
4 & -1 \
3 & 1
\end{matrix}
], then the determinant of the matrix (A2016 - A2015 - A2014) is:

A=\begin{bmatrix} 4 & -1\ 3 & 1 \end{bmatrix} \implies A^2=\begin{bmatrix} 4 & -1\ 3 & 1 \end{bmatrix}\begin{bmatrix} 4 & -1\ 3 & 1 \end{bmatrix}=\begin{bmatrix} 13 & -5\ 15 & -2 \end{bmatrix} and |A|=7
Let B = A^{2016} - A^{2015} - A^{2014}
Then |B| = |A^{2016} - A^{2015} - A^{2014}| = |A^{2014}(A^2 - A - I)|
Since |AB| = |A||B|, we have |B| = |A^{2014}||A^2 - A - I| = |A|^{2014}|A^2 - A - I|
|A| = 4(1) - (-1)(3) = 7
|A^2 - A - I| = \begin{vmatrix} 13-4-1 & -5+1\ 15-3 & -2-1 \end{vmatrix} = \begin{vmatrix} 8 & -4\ 12 & -3 \end{vmatrix} = 8(-3) - (-4)(12) = -24 + 48 = 24
Therefore, |B| = 7^{2014}(24)
However, this is not one of the options.
Let's reconsider the problem. The eigenvalues of A are solutions to the characteristic equation:
|A - λI| = 0
\begin{vmatrix} 4-λ & -1\ 3 & 1-λ \end{vmatrix} = (4-λ)(1-λ) - (-3) = λ^2 - 5λ + 7 = 0
Let λ1 and λ2 be the eigenvalues. Then λ1 + λ2 = 5 and λ1λ2 = 7.
Then A2016 - A2015 - A2014 = A2014(A2 - A - I)
|A2016 - A2015 - A2014| = |A|2014 |A2 - A - I| = 72014 |A2 - A - I|
The eigenvalues of A2 - A - I are λ12 - λ1 - 1 and λ22 - λ2 - 1.
|A2 - A - I| = (λ12 - λ1 - 1)(λ22 - λ2 - 1) = (λ1λ2)2 - (λ1 + λ2)λ1λ2 + λ1λ2 + (λ1 + λ2) - 1
= 72 - 5(7) + 7 + 5 - 1 = 49 - 35 + 11 = 25
Therefore |A2016 - A2015 - A2014| = 72014(25)
This is still not among the options.
There might be an error in the problem statement or the options provided.