If $$A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$$, then the value of $$\det(A^4) + \det\left(A^{10} - (\text{Adj}(2A))^{10}\right)$$ is equal to ________.

We have $$A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$$. The determinant is $$\det(A) = 2(-1) - 3(0) = -2$$.

Since $$\det(A^n) = (\det A)^n$$ for any square matrix, we get $$\det(A^4) = (-2)^4 = 16$$.

Now we find $$\det(A^{10} - (\text{Adj}(2A))^{10})$$. For any $$n \times n$$ matrix, the identity $$\text{Adj}(kM) = k^{n-1}\text{Adj}(M)$$ holds. With $$n = 2$$ and $$k = 2$$: $$\text{Adj}(2A) = 2^1 \cdot \text{Adj}(A) = 2\,\text{Adj}(A)$$.

For any invertible matrix, the adjugate satisfies $$A \cdot \text{Adj}(A) = \det(A) \cdot I$$, so $$\text{Adj}(A) = \det(A) \cdot A^{-1}$$. Here $$\text{Adj}(A) = (-2)A^{-1}$$.

Therefore $$\text{Adj}(2A) = 2 \cdot (-2)A^{-1} = -4A^{-1}$$, and $$(\text{Adj}(2A))^{10} = (-4)^{10}(A^{-1})^{10} = 4^{10} \cdot A^{-10}$$.

Since $$A$$ is upper triangular, all its powers are also upper triangular, with diagonal entries being the corresponding powers of the eigenvalues. The eigenvalues of $$A$$ are the diagonal entries: $$\lambda_1 = 2$$ and $$\lambda_2 = -1$$.

The diagonal entries of $$A^{10}$$ are $$2^{10} = 1024$$ and $$(-1)^{10} = 1$$. The diagonal entries of $$A^{-10}$$ are $$2^{-10} = \frac{1}{1024}$$ and $$(-1)^{-10} = 1$$.

So the diagonal entries of $$A^{10} - 4^{10} A^{-10}$$ are: first entry $$= 1024 - 4^{10} \cdot \frac{1}{1024} = 1024 - \frac{2^{20}}{2^{10}} = 1024 - 1024 = 0$$, and second entry $$= 1 - 4^{10} \cdot 1 = 1 - 4^{10}$$.

Since the matrix $$A^{10} - 4^{10}A^{-10}$$ is upper triangular (as both $$A^{10}$$ and $$A^{-10}$$ are upper triangular), its determinant equals the product of its diagonal entries: $$0 \times (1 - 4^{10}) = 0$$.

Therefore $$\det(A^4) + \det(A^{10} - (\text{Adj}(2A))^{10}) = 16 + 0 = 16$$.

The answer is 16.