if `A[{:(5,-3),(111,336):}]` and det`(-3A^(2013)+A^(2014))=alpha^(alpha)+beta^(2)(1+gamma+gamma^(2))` where `alpha,beta,gamma` are natural numbers and `(alphagtgammagtbeta)` then

Options

A. `alpha+beta=2016`
B. `alpha+beta=2023`
C. `alpha+beta+gamma=2026`
D. `beta+gamma=13`

Explanation

To solve the given problem step by step, we need to find the determinant of the expression \(-3A^{2013} + A^{2014}\) and relate it to the expression given in terms of \(\alpha\), \(\beta\), and \(\gamma\).

### Step 1: Define the matrix \(A\)

Given:
\[ A = \begin{pmatrix} 5 & -3 \\ 111 & 336 \end{pmatrix} \]

### Step 2: Calculate \(A - 3I\)

The identity matrix \(I\) for a \(2 \times 2\) matrix is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Thus, \(3I\) is:
\[ 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]
Now, we calculate \(A - 3I\):
\[ A - 3I = \begin{pmatrix} 5 & -3 \\ 111 & 336 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 5 - 3 & -3 \\ 111 & 336 - 3 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 111 & 333 \end{pmatrix} \]

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

The determinant of a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is calculated as:
\[ \det(A) = ad - bc \]
For our matrix:
\[ \det(A - 3I) = (2)(333) - (-3)(111) = 666 + 333 = 999 \]

### Step 4: Calculate \(\det(A)\)

Now, we calculate the determinant of \(A\):
\[ \det(A) = (5)(336) - (-3)(111) = 1680 + 333 = 2013 \]

### Step 5: Calculate \(\det(-3A^{2013} + A^{2014})\)

Using the properties of determinants:
\[ \det(-3A^{2013} + A^{2014}) = \det(A^{2013}(-3I + A)) = \det(A^{2013}) \cdot \det(-3I + A) \]
We know:
\[ \det(A^{2013}) = (\det(A))^{2013} = (2013)^{2013} \]
Next, we calculate \(\det(-3I + A)\):
\[ -3I + A = \begin{pmatrix} 5 - 3 & -3 \\ 111 & 336 - 3 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 111 & 333 \end{pmatrix} \]
Thus:
\[ \det(-3I + A) = 999 \]

### Step 6: Combine results

Putting it all together:
\[ \det(-3A^{2013} + A^{2014}) = (2013)^{2013} \cdot 999 \]

### Step 7: Express \(999\) in terms of \(\alpha\), \(\beta\), and \(\gamma\)

We need to express \(999\) in the form given in the problem:
\[ 999 = 9 \times 111 = 3^2 \times (1 + 10 + 10^2) \]
This gives us:
\[ \alpha = 2013, \quad \beta = 3, \quad \gamma = 10 \]

### Step 8: Verify conditions

We have:
\[ \alpha > \gamma > \beta \quad \text{(i.e., } 2013 > 10 > 3\text{)} \]

### Final Result

Thus, the values of \(\alpha\), \(\beta\), and \(\gamma\) are:
\[ \alpha = 2013, \quad \beta = 3, \quad \gamma = 10 \]