1. Key Concepts and Formulas

• System of Linear Equations: A system of $n$ linear equations in $n$ variables can be represented in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constant terms.
• Cramer's Rule Conditions for Solutions: For a system of three linear equations in three variables ($x, y, z$):
  • Let $D$ be the determinant of the coefficient matrix.
  • Let $D_x, D_y, D_z$ be the determinants obtained by replacing the first, second, and third columns of $D$ respectively with the constant terms.
  • Infinitely Many Solutions: The system has infinitely many solutions if and only if $D = 0$ AND $D_x = 0$ AND $D_y = 0$ AND $D_z = 0$.
  • No Solution: The system has no solution if $D = 0$ AND at least one of $D_x, D_y, D_z$ is non-zero.
  • Unique Solution: The system has a unique solution if $D \neq 0$.

2. Step-by-Step Solution

The given system of linear equations is:

1. $x + y + z = 5$
2. $x + 2y + 3z = 9$
3. $x + 3y + \alpha z = \beta$

Step 1: Set up the Determinants First, we write down the determinant of the coefficient matrix, $D$, and the determinants $D_x, D_y, D_z$ for the given system.

The coefficient matrix $A$ is: $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{pmatrix}$ The constant terms vector $B$ is: $B = \begin{pmatrix} 5 \\ 9 \\ \beta \end{pmatrix}$

The determinants are: $D = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{matrix} \right|$ $D_x = \left| \begin{matrix} 5 & 1 & 1 \\ 9 & 2 & 3 \\ \beta & 3 & \alpha \end{matrix} \right|$ $D_y = \left| \begin{matrix} 1 & 5 & 1 \\ 1 & 9 & 3 \\ 1 & \beta & \alpha \end{matrix} \right|$ $D_z = \left| \begin{matrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \beta \end{matrix} \right|$

Step 2: Calculate $D$ and Determine $\alpha$ For the system to have infinitely many solutions, the first condition is $D=0$. Let's calculate $D$: $D = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{matrix} \right|$ To simplify the calculation, we perform elementary row operations to create zeros in the first column. This allows for easier expansion along that column.

• Apply $R_2 \to R_2 - R_1$
• Apply $R_3 \to R_3 - R_1$

The determinant $D$ becomes: $D = \left| \begin{matrix} 1 & 1 & 1 \\ 1-1 & 2-1 & 3-1 \\ 1-1 & 3-1 & \alpha-1 \end{matrix} \right| = \left| \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & \alpha-1 \end{matrix} \right|$ Now, expand the determinant along the first column: $D = 1 \cdot \left| \begin{matrix} 1 & 2 \\ 2 & \alpha-1 \end{matrix} \right| - 0 + 0$ $D = (1)(\alpha-1) - (2)(2)$ $D = \alpha - 1 - 4$ $D = \alpha - 5$ Since the system has infinitely many solutions, $D$ must be zero: $\alpha - 5 = 0 \implies \alpha = 5$

Step 3: Calculate $D_z$ and Determine $\beta$ For infinitely many solutions, $D_x=D_y=D_z=0$. We can choose any of these to find $\beta$. Let's choose $D_z$ as it often leads to simpler calculations due to the constant terms being in the last column. $D_z = \left| \begin{matrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \beta \end{matrix} \right|$ Again, perform elementary row operations to simplify:

• Apply $R_2 \to R_2 - R_1$
• Apply $R_3 \to R_3 - R_1$

The determinant $D_z$ becomes: $D_z = \left| \begin{matrix} 1 & 1 & 5 \\ 1-1 & 2-1 & 9-5 \\ 1-1 & 3-1 & \beta-5 \end{matrix} \right| = \left| \begin{matrix} 1 & 1 & 5 \\ 0 & 1 & 4 \\ 0 & 2 & \beta-5 \end{matrix} \right|$ Now, expand the determinant along the first column: $D_z = 1 \cdot \left| \begin{matrix} 1 & 4 \\ 2 & \beta-5 \end{matrix} \right| - 0 + 0$ $D_z = (1)(\beta-5) - (4)(2)$ $D_z = \beta - 5 - 8$ $D_z = \beta - 13$ Since the system has infinitely many solutions, $D_z$ must be zero: $\beta - 13 = 0 \implies \beta = 13$

Step 4: Calculate $\beta - \alpha$ We have found the values $\alpha = 5$ and $\beta = 13$. Now, we calculate the required expression: $\beta - \alpha = 13 - 5 = 8$

3. Common Mistakes & Tips

• Incomplete Conditions: A common mistake is to only check $D=0$ for infinitely many solutions. Remember, $D=0$ alone implies either no solution or infinitely many solutions. You must also verify that $D_x=D_y=D_z=0$ for infinitely many solutions. If $D=0$ but any of $D_x, D_y, D_z$ are non-zero, the system has no solution.
• Determinant Calculation Errors: Be meticulous with arithmetic and signs when calculating determinants, especially with $3 \times 3$ matrices. Row/column operations are powerful tools to simplify determinants but must be applied correctly.
• Choosing the Easiest Determinant: After finding $D=0$ and the value of $\alpha$, you need to find $\beta$. You can choose any of $D_x=0$, $D_y=0$, or $D_z=0$. Opt for the one that seems to involve simpler numbers or fewer variables to minimize calculation errors.

4. Summary

To find the value of $\beta - \alpha$ for a system of linear equations with infinitely many solutions, we utilized Cramer's Rule. This rule states that for infinitely many solutions, the determinant of the coefficient matrix ($D$) and all related determinants ($D_x, D_y, D_z$) must simultaneously be zero. We first calculated $D$, set it to zero, and found $\alpha = 5$. Then, we calculated $D_z$ (an alternative would be $D_x$ or $D_y$), set it to zero, and determined $\beta = 13$. Finally, we computed the difference $\beta - \alpha = 13 - 5 = 8$.

5. Final Answer

The final answer is $\boxed{8}$. This corresponds to option (A).