This problem tests your understanding of determinants, powers of matrices, and adjoints of matrices. We will systematically apply relevant properties to simplify the given expression.

Key Concepts and Formulas

1. Determinant of a $2 \times 2$ matrix: For a matrix M = \left[ {\matrix{ a & b \cr c & d \cr } } \right], its determinant is $\det(M) = ad - bc$.
2. Determinant of a power of a matrix: For any square matrix $M$ and a positive integer $n$, $\det(M^n) = (\det M)^n$.
3. Adjoint of a $2 \times 2$ matrix: For a matrix M = \left[ {\matrix{ a & b \cr c & d \cr } } \right], its adjoint is \text{Adj}(M) = \left[ {\matrix{ d & { - b} \cr { - c} & a \cr } } \right].
4. Adjoint of a scalar multiple of a matrix: For an $n \times n$ matrix $M$ and a scalar $k$, $\text{Adj}(kM) = k^{n-1} \text{Adj } M$. For a $2 \times 2$ matrix ($n=2$), this simplifies to $\text{Adj}(kM) = k \text{Adj } M$.
5. Power of an upper triangular matrix: If M = \left[ {\matrix{ a & b \cr 0 & d \cr } } \right] is an upper triangular matrix, then for $a \ne d$, its $n$-th power is given by M^n = \left[ {\matrix{ a^n & b \frac{a^n-d^n}{a-d} \cr 0 & d^n \cr } } \right]. If $a=d$, then M^n = \left[ {\matrix{ a^n & n a^{n-1} b \cr 0 & a^n \cr } } \right].
6. Determinant of an upper triangular matrix: The determinant of an upper triangular matrix is the product of its diagonal elements.

---

Step-by-Step Solution

Given the matrix A = \left[ {\matrix{ 2 & 3 \cr 0 & { - 1} \cr } } \right]. We need to find the value of $\det(A^4) + \det(A^{10} - (\text{Adj}(2A))^{10})$.

1. Calculate the determinant of A ($\det A$) We start by finding the determinant of the given matrix $A$. $\det A = (2)(-1) - (3)(0) = -2 - 0 = -2$ Reasoning: This is a direct application of the determinant formula for a $2 \times 2$ matrix.

2. Evaluate the first term: $\det(A^4)$ Using the property $\det(M^n) = (\det M)^n$: $\det(A^4) = (\det A)^4 = (-2)^4 = 16$ Reasoning: This property allows us to calculate the determinant of a matrix power without explicitly computing the matrix power itself, which simplifies calculations significantly.

3. Simplify $\text{Adj}(2A)$ First, we find $\text{Adj}(A)$. For A = \left[ {\matrix{ 2 & 3 \cr 0 & { - 1} \cr } } \right], \text{Adj } A = \left[ {\matrix{ -1 & -3 \cr -0 & 2 \cr } } \right] = \left[ {\matrix{ -1 & -3 \cr 0 & 2 \cr } } \right] Next, we use the property $\text{Adj}(kM) = k^{n-1} \text{Adj } M$. For our $2 \times 2$ matrix, $n=2$, so $k^{n-1} = k^{2-1} = k$. Thus, $\text{Adj}(2A) = 2 \text{Adj } A$. \text{Adj}(2A) = 2 \left[ {\matrix{ -1 & -3 \cr 0 & 2 \cr } } \right] = \left[ {\matrix{ -2 & -6 \cr 0 & 4 \cr } } \right] Reasoning: This step simplifies the adjoint of a scalar multiple, making subsequent power calculations easier. The specific formula for $2 \times 2$ adjoint is applied first.

4. Evaluate the components of the second term: $A^{10}$ and $(\text{Adj}(2A))^{10}$ Both $A$ and $\text{Adj}(2A)$ are upper triangular matrices. We can use the formula for the $n$-th power of an upper triangular matrix M = \left[ {\matrix{ a & b \cr 0 & d \cr } } \right]: M^n = \left[ {\matrix{ a^n & b \frac{a^n-d^n}{a-d} \cr 0 & d^n \cr } } \right] (since $a \ne d$ for both matrices).

• For $A^{10}$: Here $a=2, b=3, d=-1$. A^{10} = \left[ {\matrix{ 2^{10} & 3 \frac{2^{10}-(-1)^{10}}{2-(-1)} \cr 0 & (-1)^{10} \cr } } \right] = \left[ {\matrix{ 2^{10} & 3 \frac{2^{10}-1}{3} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 2^{10} & 2^{10}-1 \cr 0 & 1 \cr } } \right]

• For $(\text{Adj}(2A))^{10}$: Let B = \text{Adj}(2A) = \left[ {\matrix{ -2 & -6 \cr 0 & 4 \cr } } \right]. Here $a'=-2, b'=-6, d'=4$. B^{10} = \left[ {\matrix{ (-2)^{10} & -6 \frac{(-2)^{10}-4^{10}}{(-2)-4} \cr 0 & 4^{10} \cr } } \right] = \left[ {\matrix{ 2^{10} & -6 \frac{2^{10}-4^{10}}{-6} \cr 0 & 4^{10} \cr } } \right] = \left[ {\matrix{ 2^{10} & 2^{10}-4^{10} \cr 0 & 4^{10} \cr } } \right] Reasoning: Computing high powers of matrices directly is tedious. Using the specific formula for upper triangular matrices simplifies this considerably.

5. Calculate the matrix $A^{10} - (\text{Adj}(2A))^{10}$ Now we subtract the two matrices found in the previous step: A^{10} - (\text{Adj}(2A))^{10} = \left[ {\matrix{ 2^{10} & 2^{10}-1 \cr 0 & 1 \cr } } \right] - \left[ {\matrix{ 2^{10} & 2^{10}-4^{10} \cr 0 & 4^{10} \cr } } \right] = \left[ {\matrix{ 2^{10}-2^{10} & (2^{10}-1) - (2^{10}-4^{10}) \cr 0-0 & 1-4^{10} \cr } } \right] = \left[ {\matrix{ 0 & 2^{10}-1-2^{10}+4^{10} \cr 0 & 1-4^{10} \cr } } \right] = \left[ {\matrix{ 0 & 4^{10}-1 \cr 0 & 1-4^{10} \cr } } \right] Reasoning: Matrix subtraction is performed element-wise. Since both $A^{10}$ and $(\text{Adj}(2A))^{10}$ are upper triangular, their difference is also an upper triangular matrix.

6. Evaluate the second term: $\det(A^{10} - (\text{Adj}(2A))^{10})$ The matrix we obtained is \left[ {\matrix{ 0 & 4^{10}-1 \cr 0 & 1-4^{10} \cr } } \right]. For an upper triangular matrix, the determinant is the product of its diagonal elements. $\det(A^{10} - (\text{Adj}(2A))^{10}) = (0) \cdot (1-4^{10}) - (4^{10}-1) \cdot (0) = 0$ Reasoning: As the matrix is upper triangular and has a zero on its main diagonal (specifically, the $(1,1)$ entry), its determinant is 0.

7. Calculate the final value Finally, we sum the values of the two terms: $\det(A^4) + \det(A^{10} - (\text{Adj}(2A))^{10}) = 16 + 0 = 16$

---

Tips and Common Mistakes

• Don't confuse $\det(A+B)$ with $\det A + \det B$: The determinant is not linear over matrix addition/subtraction. You must compute the resulting matrix first before finding its determinant.
• Properties for $n \times n$ vs. $2 \times 2$ matrices: Be mindful of how general matrix properties specialize for $2 \times 2$ matrices (e.g., $\text{Adj}(kM) = k^{n-1} \text{Adj } M$ becomes $\text{Adj}(kM) = k \text{Adj } M$ for $n=2$).
• Powers of triangular matrices: Using the specific formula for powers of upper/lower triangular matrices can save a lot of calculation time compared to repeated matrix multiplication.
• Eigenvalues for commutative matrices: If two matrices $X$ and $Y$ commute ($XY=YX$), then the eigenvalues of $X \pm Y$ are $\lambda_i \pm \mu_i$ (where $\lambda_i$ are eigenvalues of $X$ and $\mu_i$ are eigenvalues of $Y$). Here, $A^{10}$ and $(\text{Adj}(2A))^{10}$ commute (as $\text{Adj}(2A)$ can be expressed as a scalar multiple of $A^{-1}$), so the eigenvalues of their difference are the differences of their respective eigenvalues. This confirms the $(1,1)$ diagonal element being zero ($2^{10} - 2^{10} = 0$).

---

Summary / Key Takeaway

This problem highlights the importance of mastering matrix properties related to determinants, adjoints, and powers. By systematically applying these properties, especially those for upper triangular matrices, complex expressions can be broken down into manageable calculations. The key insight for the second term was recognizing that the resulting matrix was upper triangular with a zero on its main diagonal, immediately implying a determinant of zero.

The final answer is $\boxed{16}$.