1. Key Concept: Condition for a Unique Solution

For a system of linear equations represented in the matrix form $AX = B$, where $A$ is the coefficient matrix, $X$ is the column matrix of variables, and $B$ is the column matrix of constants, a unique solution exists if and only if the determinant of the coefficient matrix, denoted as $\det(A)$ or $\Delta$, is non-zero. That is, $\Delta \ne 0$.

If $\Delta = 0$, the system either has no solution (inconsistent) or infinitely many solutions (consistent and dependent). In such cases, there is no unique solution.

2. Formulating the Coefficient Matrix

The given system of linear equations is:

1. $(1 + \alpha) x + \beta y + z = 2$
2. $\alpha x + (1 + \beta) y + z = 3$
3. $\alpha x + \beta y + 2z = 2$

To determine the existence of a unique solution, we first extract the coefficient matrix $A$ from this system: $A = \begin{pmatrix} 1+\alpha & \beta & 1 \\ \alpha & 1+\beta & 1 \\ \alpha & \beta & 2 \end{pmatrix}$

3. Calculating the Determinant of the Coefficient Matrix

Now, we need to calculate the determinant of matrix $A$, denoted as $\Delta = \det(A)$. We can use row operations to simplify the matrix before calculating the determinant, which often reduces the chances of calculation errors.

• Step 3.1: Apply Row Operations Perform the operation $R_1 \to R_1 - R_2$ (subtract Row 2 from Row 1). This operation does not change the value of the determinant. $\Delta = \begin{vmatrix} (1+\alpha) - \alpha & \beta - (1+\beta) & 1 - 1 \\ \alpha & 1+\beta & 1 \\ \alpha & \beta & 2 \end{vmatrix} = \begin{vmatrix} 1 & -1 & 0 \\ \alpha & 1+\beta & 1 \\ \alpha & \beta & 2 \end{vmatrix}$ Next, perform the operation $R_2 \to R_2 - R_3$ (subtract Row 3 from Row 2). This also does not change the value of the determinant. $\Delta = \begin{vmatrix} 1 & -1 & 0 \\ \alpha - \alpha & (1+\beta) - \beta & 1 - 2 \\ \alpha & \beta & 2 \end{vmatrix} = \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \alpha & \beta & 2 \end{vmatrix}$

• Step 3.2: Expand the Determinant Now, we expand the determinant along the first row (or any row/column with zeros for easier calculation). $\Delta = 1 \cdot \begin{vmatrix} 1 & -1 \\ \beta & 2 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 0 & -1 \\ \alpha & 2 \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & 1 \\ \alpha & \beta \end{vmatrix}$ $\Delta = 1 \cdot ((1)(2) - (-1)(\beta)) + 1 \cdot ((0)(2) - (-1)(\alpha)) + 0$ $\Delta = (2 + \beta) + (\alpha)$ $\Delta = \alpha + \beta + 2$

4. Applying the Unique Solution Condition

For the system to have a unique solution, the determinant $\Delta$ must be non-zero. Therefore, we must have: $\alpha + \beta + 2 \neq 0$

5. Checking the Options

We now test each given ordered pair $(\alpha, \beta)$ to see which one satisfies the condition $\alpha + \beta + 2 \neq 0$.

• (A) ($-3$, $1$) Substitute $\alpha = -3$ and $\beta = 1$ into the condition: $-3 + 1 + 2 = 0$ Since $\Delta = 0$, this pair does not result in a unique solution.

• (B) ($1$, $-3$) Substitute $\alpha = 1$ and $\beta = -3$ into the condition: $1 + (-3) + 2 = 0$ Since $\Delta = 0$, this pair does not result in a unique solution.

• (C) ($-4$, $2$) Substitute $\alpha = -4$ and $\beta = 2$ into the condition: $-4 + 2 + 2 = 0$ Since $\Delta = 0$, this pair does not result in a unique solution.

• (D) ($2$, $4$) Substitute $\alpha = 2$ and $\beta = 4$ into the condition: $2 + 4 + 2 = 8$ Since $\Delta = 8 \neq 0$, this pair does result in a unique solution.

6. Conclusion and Final Answer

Based on our calculations, the ordered pair $(\alpha, \beta) = (2, 4)$ is the only option for which the determinant of the coefficient matrix is non-zero, thus ensuring a unique solution for the system of linear equations.

The final answer is $\boxed{\text{(D)}}$.

7. Tips and Common Mistakes

• Determinant Calculation: Be extremely careful with signs and arithmetic when calculating determinants, especially for $3 \times 3$ matrices. Using row/column operations to introduce zeros can simplify the calculation significantly.
• Condition for Unique Solution: Always remember that for a system of $n$ linear equations in $n$ variables, a unique solution exists if and only if $\det(A) \neq 0$.
• Checking Options: After deriving the condition, systematically substitute the values from each option to find the correct one.
• Discrepancy Check: If your mathematically derived answer does not match the provided options or the answer key, it's crucial to re-check your determinant calculation and the problem statement itself for any potential typos or misinterpretations. In competitive exams, such discrepancies can sometimes occur.