1. Key Concepts and Formulas

• System of Linear Equations: A system of $n$ linear equations in $n$ variables, typically represented as $AX=B$, has infinitely many solutions if and only if:
  • The determinant of the coefficient matrix $A$ is zero ($\det(A) = 0$). This implies that the rows (and columns) of $A$ are linearly dependent.
  • The rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A|B]$, and this rank is less than the number of variables ($n$).
• Linear Dependence and Consistency: For a system to have infinitely many solutions, it means that at least one equation is a linear combination of the other equations, and this linear relationship must also hold true for the constant terms. If the determinant is zero but the constant terms do not follow the same linear relationship, the system is inconsistent (no solution).

2. Step-by-Step Solution

Let the given system of equations be: (1) $7x + 11y + \alpha z = 13$ (2) $5x + 4y + 7z = \beta$ (3) $175x + 194y + 57z = 361$

Step 1: Identify the Linear Relationship Between Equations For the system to have infinitely many solutions, the third equation must be a linear combination of the first two equations. This means there exist constants $k_1$ and $k_2$ such that: $\text{Equation (3)} = k_1 \times \text{Equation (1)} + k_2 \times \text{Equation (2)}$ We find $k_1$ and $k_2$ by comparing the coefficients of $x$ and $y$ from the three equations.

Comparing coefficients of $x$: $175 = k_1 \cdot 7 + k_2 \cdot 5 \quad \text{(Equation A)}$ Comparing coefficients of $y$: $194 = k_1 \cdot 11 + k_2 \cdot 4 \quad \text{(Equation B)}$

Now, we solve this system of two linear equations for $k_1$ and $k_2$: Multiply Equation A by 4: $4 \times (7k_1 + 5k_2) = 4 \times 175 \implies 28k_1 + 20k_2 = 700 \quad \text{(Equation C)}$ Multiply Equation B by 5: $5 \times (11k_1 + 4k_2) = 5 \times 194 \implies 55k_1 + 20k_2 = 970 \quad \text{(Equation D)}$ Subtract Equation C from Equation D: $(55k_1 + 20k_2) - (28k_1 + 20k_2) = 970 - 700$ $27k_1 = 270$ $k_1 = \frac{270}{27} \implies k_1 = 10$ Substitute $k_1 = 10$ back into Equation A: $7(10) + 5k_2 = 175$ $70 + 5k_2 = 175$ $5k_2 = 175 - 70$ $5k_2 = 105$ $k_2 = \frac{105}{5} \implies k_2 = 21$ Thus, we have found the linear relationship: Equation (3) is $10 \times \text{Equation (1)} + 21 \times \text{Equation (2)}$. This relationship must hold for all coefficients and the constant terms for infinitely many solutions to exist.

**Step 2: Calculate the value of $\alpha$} Using the derived linear relationship $10 \times \text{Equation (1)} + 21 \times \text{Equation (2)} = \text{Equation (3)}$, we compare the coefficients of $z$: $10 \cdot \alpha + 21 \cdot 7 = 57$ $10\alpha + 147 = 57$ $10\alpha = 57 - 147$ $10\alpha = -70$ $\alpha = -7$

**Step 3: Calculate the value of $\beta$} Next, we use the same linear relationship for the constant terms: $10 \times 13 + 21 \times \beta = 361$ $130 + 21\beta = 361$ $21\beta = 361 - 130$ $21\beta = 231$ $\beta = \frac{231}{21}$ $\beta = 11$

Step 4: Final Calculation We need to find the value of $\alpha + \beta + 2$. Substitute the values we found for $\alpha$ and $\beta$: $\alpha + \beta + 2 = (-7) + (11) + 2$ $\alpha + \beta + 2 = 4 + 2$ $\alpha + \beta + 2 = 6$

3. Common Mistakes & Tips

• Linear Dependence Check: Always verify the linear combination for all coefficients (x, y, z) and the constant term. If the relationship holds for the variable coefficients but not for the constant terms, the system is inconsistent (no solution), not infinitely many solutions.
• Arithmetic Precision: Be meticulous with calculations, especially when solving systems of equations. A small error can lead to incorrect values for $k_1, k_2, \alpha,$ or $\beta$.
• Determinant Method vs. Linear Combination: While Cramer's rule conditions ($\det(A)=0$ and $\Delta_x=\Delta_y=\Delta_z=0$) are fundamental, identifying linear dependence through inspection or row operations is often more efficient for 3x3 systems, especially when coefficients are related.

4. Summary

For a system of linear equations to have infinitely many solutions, one equation must be a linear combination of the others, and this linear dependence must extend to the constant terms. By comparing the coefficients of $x$ and $y$, we found that the third equation was $10 \times \text{Equation (1)} + 21 \times \text{Equation (2)}$. Applying this relationship to the $z$-coefficients and constant terms, we determined $\alpha = -7$ and $\beta = 11$. Finally, calculating the expression $\alpha + \beta + 2$ yields $(-7) + (11) + 2 = 6$.

The final answer is $\boxed{6}$, which corresponds to option (A).