Key Concept: Conditions for Infinitely Many Solutions

For a system of linear equations in three variables, say: $\begin{aligned} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{aligned}$ to have infinitely many solutions, the following conditions must be met according to Cramer's Rule:

1. The determinant of the coefficient matrix ($\Delta$) must be zero. \Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0
2. All the determinants obtained by replacing a column of the coefficient matrix with the constant terms ($\Delta_x, \Delta_y, \Delta_z$) must also be zero. \Delta_x = \left| {\begin{array}{*{20}{c}} {{d_1}}&{{b_1}}&{{c_1}}\\ {{d_2}}&{{b_2}}&{{c_2}}\\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0 \Delta_y = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{d_1}}&{{c_1}}\\ {{a_2}}&{{d_2}}&{{c_2}}\\ {{a_3}}&{{d_3}}&{{c_3}} \end{array}} \right| = 0 \Delta_z = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{d_1}}\\ {{a_2}}&{{b_2}}&{{d_2}}\\ {{a_3}}&{{b_3}}&{{d_3}} \end{array}} \right| = 0 If $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.

Given System of Equations: We are given the following system of linear equations:

1. $x+y+z=6$
2. $2x+5y+\alpha z=\beta$
3. $x+2y+3z=14$

We are told that this system has infinitely many solutions. Our goal is to find the value of $\alpha+\beta$.

---

Step 1: Calculate $\Delta$ and find $\alpha$

First, we form the coefficient matrix and calculate its determinant $\Delta$. \Delta = \left| {\begin{array}{*{20}{c}} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{array}} \right| Since the system has infinitely many solutions, we must have $\Delta = 0$.

To simplify the calculation of the determinant, we can perform column operations to create zeros in a row or column. Let's aim to create zeros in the first row. Apply the operations: $C_1 \to C_1 - C_3$ and $C_2 \to C_2 - C_3$.

• The first column becomes:
  • $1 - 1 = 0$
  • $2 - \alpha$
  • $1 - 3 = -2$
• The second column becomes:
  • $1 - 1 = 0$
  • $5 - \alpha$
  • $2 - 3 = -1$

The determinant now becomes: \Delta = \left| {\begin{array}{*{20}{c}} 0 & 0 & 1 \\ {2 - \alpha} & {5 - \alpha} & \alpha \\ {-2} & {-1} & 3 \end{array}} \right| Now, we expand the determinant along the first row. Only the third element contributes, as the first two are zero. \Delta = 1 \cdot \left| {\begin{array}{*{20}{c}} {2 - \alpha} & {5 - \alpha} \\ {-2} & {-1} \end{array}} \right| $\Delta = 1 \cdot [ (2 - \alpha)(-1) - (5 - \alpha)(-2) ]$ $\Delta = [ -(2 - \alpha) + 2(5 - \alpha) ]$ $\Delta = [ -2 + \alpha + 10 - 2\alpha ]$ $\Delta = 8 - \alpha$ Since $\Delta$ must be $0$ for infinitely many solutions: $8 - \alpha = 0$ $\alpha = 8$

---

Step 2: Calculate $\Delta_x$ and find $\beta$

Now that we have found $\alpha=8$, we substitute this value back into the system of equations. The system becomes:

1. $x+y+z=6$
2. $2x+5y+8z=\beta$
3. $x+2y+3z=14$

For infinitely many solutions, we also need $\Delta_x = 0$ (and $\Delta_y = 0$, $\Delta_z = 0$). We can choose any one of these to find $\beta$. Let's use $\Delta_x$.

$\Delta_x$ is formed by replacing the first column of $\Delta$ with the constant terms $(6, \beta, 14)$: \Delta_x = \left| {\begin{array}{*{20}{c}} 6 & 1 & 1 \\ \beta & 5 & 8 \\ 14 & 2 & 3 \end{array}} \right| Since $\Delta_x$ must be $0$: \left| {\begin{array}{*{20}{c}} 6 & 1 & 1 \\ \beta & 5 & 8 \\ 14 & 2 & 3 \end{array}} \right| = 0 Again, we can simplify the determinant calculation using column operations. Let's aim to create zeros in the first row. Apply the operations: $C_1 \to C_1 - 6C_3$ and $C_2 \to C_2 - C_3$.

• The first column becomes:
  • $6 - 6(1) = 0$
  • $\beta - 6(8) = \beta - 48$
  • $14 - 6(3) = 14 - 18 = -4$
• The second column becomes:
  • $1 - 1 = 0$
  • $5 - 8 = -3$
  • $2 - 3 = -1$

The determinant now becomes: \Delta_x = \left| {\begin{array}{*{20}{c}} 0 & 0 & 1 \\ {\beta - 48} & {-3} & 8 \\ {-4} & {-1} & 3 \end{array}} \right| Expand the determinant along the first row: \Delta_x = 1 \cdot \left| {\begin{array}{*{20}{c}} {\beta - 48} & {-3} \\ {-4} & {-1} \end{array}} \right| $\Delta_x = 1 \cdot [ (\beta - 48)(-1) - (-3)(-4) ]$ $\Delta_x = [ -(\beta - 48) - 12 ]$ $\Delta_x = [ -\beta + 48 - 12 ]$ $\Delta_x = 36 - \beta$ Since $\Delta_x$ must be $0$: $36 - \beta = 0$ $\beta = 36$

---

Step 3: Calculate $\alpha + \beta$

We have found $\alpha = 8$ and $\beta = 36$. Therefore, $\alpha + \beta = 8 + 36 = 44$.

---

Tips and Common Mistakes:

• Crucial Condition: Remember that for infinitely many solutions, all determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$) must be zero. If $\Delta=0$ but any of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.
• Determinant Calculation: Be very careful with signs when expanding determinants, especially when using cofactor expansion. Row/column operations are powerful tools to simplify determinants by creating zeros, but ensure they are applied correctly.
• System Consistency: After finding $\alpha$ and $\beta$, you could (optionally) substitute them back into the original system and try to solve it using Gaussian elimination or by finding a relationship between the equations to confirm infinite solutions. For example, check if one equation is a linear combination of the others.

---

Summary:

We systematically applied the conditions for a system of linear equations to have infinitely many solutions. By setting the determinant of the coefficient matrix ($\Delta$) to zero, we found $\alpha=8$. Then, by setting the determinant $\Delta_x$ (formed by replacing the x-coefficient column with constants) to zero, we found $\beta=36$. Finally, adding these values gave us $\alpha+\beta=44$.

The final answer is $\boxed{\text{44}}$.