1. Key Concept: Sum of Powers Determinant and Vandermonde Determinants

This problem involves evaluating a determinant whose elements are sums of powers. A powerful technique for such determinants is to express them as the product of two matrices, typically a matrix and its transpose, where one of these matrices is a Vandermonde matrix.

A Vandermonde determinant for distinct elements $x_1, x_2, \dots, x_n$ is given by: $V(x_1, x_2, \dots, x_n) = \left| \begin{matrix} 1 & x_1 & x_1^2 & \dots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \dots & x_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & x_n^2 & \dots & x_n^{n-1} \end{matrix} \right| = \prod_{1 \le i < j \le n} (x_j - x_i)$ For a $3 \times 3$ matrix with elements $x_1, x_2, x_3$, this simplifies to $V(x_1, x_2, x_3) = (x_2-x_1)(x_3-x_1)(x_3-x_2)$.

A key identity for determinants whose elements are sums of powers is: If $D$ is an $n \times n$ determinant with elements $D_{ij} = \sum_{k=1}^m x_k^{i+j-2}$, then $D = \det(A A^T)$ where $A$ is an $n \times m$ matrix with $A_{ik} = x_k^{i-1}$. If $m=n$, then $D = (\det A)^2$, where $A$ is a Vandermonde-like matrix.

2. Step 1: Expressing Elements in Terms of Powers

We are given $f(n) = \alpha^n + \beta^n$. The determinant elements are of the form $1 + f(k)$. Let's list the elements of the given determinant: D = \left| {\matrix{ 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr } } \right| Substitute the definition of $f(n)$:

• $1 + f(1) = 1 + \alpha^1 + \beta^1$
• $1 + f(2) = 1 + \alpha^2 + \beta^2$
• $1 + f(3) = 1 + \alpha^3 + \beta^3$
• $1 + f(4) = 1 + \alpha^4 + \beta^4$ The first element is $a_{11} = 3$. We can express this as $1 + \alpha^0 + \beta^0 = 1^0 + \alpha^0 + \beta^0$. Notice that each element $a_{ij}$ in the determinant can be written as $1^{i+j-2} + \alpha^{i+j-2} + \beta^{i+j-2}$. Let $x_1 = 1$, $x_2 = \alpha$, $x_3 = \beta$. Then $a_{ij} = x_1^{i+j-2} + x_2^{i+j-2} + x_3^{i+j-2}$.

3. Step 2: Recognizing the Determinant as a Matrix Product

The general form $D_{ij} = \sum_{k=1}^n x_k^{i+j-2}$ for an $n \times n$ matrix is known to be the determinant of the product of a matrix and its transpose. Specifically, let $A$ be a matrix defined by its elements $A_{ik} = x_k^{i-1}$. For our $3 \times 3$ determinant and variables $x_1=1, x_2=\alpha, x_3=\beta$, the matrix $A$ is: $A = \begin{pmatrix} x_1^0 & x_2^0 & x_3^0 \\ x_1^1 & x_2^1 & x_3^1 \\ x_1^2 & x_2^2 & x_3^2 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{pmatrix}$ The transpose of $A$ is: $A^T = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2 \end{pmatrix}$ Now, let's compute the product $A A^T$: $A A^T = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2 \end{pmatrix}$ The element at position $(i,j)$ of $A A^T$ is $\sum_{k=1}^3 A_{ik} (A^T)_{kj} = \sum_{k=1}^3 x_k^{i-1} x_k^{j-1} = \sum_{k=1}^3 x_k^{i+j-2}$. Let's verify a few elements:

• $(A A^T)_{11} = 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3$
• $(A A^T)_{12} = 1 \cdot 1 + 1 \cdot \alpha + 1 \cdot \beta = 1+\alpha+\beta$
• $(A A^T)_{22} = 1 \cdot 1 + \alpha \cdot \alpha + \beta \cdot \beta = 1+\alpha^2+\beta^2$ This confirms that the given determinant $D$ is equal to $\det(A A^T)$. From the property of determinants, $\det(A A^T) = \det(A) \det(A^T)$. Since $\det(A^T) = \det(A)$, we have $D = (\det(A))^2$.

4. Step 3: Evaluating the Determinant of the Vandermonde Matrix

Now we need to calculate $\det(A)$: $\det(A) = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{matrix} \right|$ This is a Vandermonde determinant with elements $1, \alpha, \beta$ (in the order of columns). Using the formula $\prod_{1 \le i < j \le n} (x_j - x_i)$, with $x_1=1, x_2=\alpha, x_3=\beta$, we get: $\det(A) = (\alpha - 1)(\beta - 1)(\beta - \alpha)$ To explicitly show the calculation: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $\det(A) = \left| \begin{matrix} 1 & 1 & 1 \\ 0 & \alpha-1 & \beta-1 \\ 0 & \alpha^2-1 & \beta^2-1 \end{matrix} \right|$ Expand along the first column: $\det(A) = 1 \cdot \left| \begin{matrix} \alpha-1 & \beta-1 \\ (\alpha-1)(\alpha+1) & (\beta-1)(\beta+1) \end{matrix} \right|$ Factor out $(\alpha-1)$ from the first column and $(\beta-1)$ from the second column of the $2 \times 2$ matrix: $\det(A) = (\alpha-1)(\beta-1) \left| \begin{matrix} 1 & 1 \\ \alpha+1 & \beta+1 \end{matrix} \right|$ $\det(A) = (\alpha-1)(\beta-1) [ (\beta+1) - (\alpha+1) ]$ $\det(A) = (\alpha-1)(\beta-1) (\beta - \alpha)$

5. Step 4: Calculating the Determinant D

We found that $D = (\det(A))^2$. $D = [ (\alpha-1)(\beta-1)(\beta-\alpha) ]^2$ We can rewrite the terms to match the form given in the problem:

• $(\alpha-1)^2 = (-(1-\alpha))^2 = (1-\alpha)^2$
• $(\beta-1)^2 = (-(1-\beta))^2 = (1-\beta)^2$
• $(\beta-\alpha)^2 = (-(\alpha-\beta))^2 = (\alpha-\beta)^2$ Substituting these back into the expression for $D$: $D = (1-\alpha)^2 (1-\beta)^2 (\alpha-\beta)^2$

6. Step 5: Comparing with the Given Expression

The problem states that $D = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}$. Comparing our derived expression for $D$ with the given expression: $(1-\alpha)^2 (1-\beta)^2 (\alpha-\beta)^2 = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}$ By direct comparison, we find that $K=1$.

Tips and Common Mistakes:

• Recognizing the pattern: The most crucial step is to identify the determinant as a sum of powers, which points to the Vandermonde determinant and the $AA^T$ factorization. Look for elements $a_{ij}$ that are sums of $x_k^{i+j-2}$ or similar power forms.
• Vandermonde determinant formula: Remember the formula for a Vandermonde determinant. The order of subtraction matters for the sign, but since we are squaring the determinant in this problem, the sign ultimately doesn't affect the final result for $K$.
• Careful with indexing: Ensure you correctly identify the $x_k$ values (here $1, \alpha, \beta$) and the powers (here $0, 1, 2, \dots$).
• Verification: If time permits, try a simple numerical example (e.g., $\alpha=2, \beta=3$) to cross-check your determinant calculation. This can quickly reveal errors. For $\alpha=2, \beta=3$, $D = (1-2)^2(1-3)^2(2-3)^2 = (-1)^2(-2)^2(-1)^2 = 1 \cdot 4 \cdot 1 = 4$. Direct calculation of the determinant in the problem for these values also yields $4$, confirming $K=1$.

7. Summary/Key Takeaway

This problem demonstrates a standard technique for evaluating determinants whose entries are sums of powers. By recognizing the structure $a_{ij} = \sum_{k=1}^m x_k^{i+j-2}$, the determinant can be factored as $\det(A A^T)$, where $A$ is a matrix whose columns are powers of $x_k$. In this specific case, the determinant simplified to the square of a Vandermonde determinant involving $1, \alpha,$ and $\beta$, leading to $K=1$.

The final answer is $\boxed{\text{C}}$.