Key Concepts and Formulas

• Cramer's Rule for Systems of Linear Equations: For a system of $n$ linear equations in $n$ variables, represented as $AX=B$, let $\Delta = \det(A)$ be the determinant of the coefficient matrix $A$. Let $\Delta_j$ be the determinant of the matrix obtained by replacing the $j$-th column of $A$ with the constant matrix $B$.
• Condition for Infinitely Many Solutions: A system of linear equations has infinitely many solutions if and only if the determinant of the coefficient matrix is zero ($\Delta = 0$) AND all the determinants formed by replacing a column of the coefficient matrix with the constant terms are also zero ($\Delta_1 = 0, \Delta_2 = 0, \dots, \Delta_n = 0$).
• Other Solution Types:
  • If $\Delta \neq 0$, the system has a unique solution.
  • If $\Delta = 0$ but at least one $\Delta_j \neq 0$, the system has no solution.

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Step-by-Step Solution

Step 1: Identify the Coefficient Matrix and Constant Matrix We are given the system of linear equations: $\begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned}$ To apply Cramer's Rule, we first write down the coefficient matrix $A$ and the constant matrix $B$. The coefficient matrix $A$ is formed by the coefficients of $x, y, z$: $A = \begin{pmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{pmatrix}$ The constant matrix $B$ is formed by the terms on the right-hand side of the equations: $B = \begin{pmatrix} 3 \\ -3 \\ 4 \end{pmatrix}$

Step 2: Set up the Determinants $\Delta, \Delta_1, \Delta_2, \Delta_3$ We need to calculate the determinant of the coefficient matrix, $\Delta$, and the determinants $\Delta_1, \Delta_2, \Delta_3$ for Cramer's Rule.

• $\Delta$ (Determinant of $A$): $\Delta = \left|\begin{array}{ccc} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{array}\right|$
• $\Delta_1$ (Replace 1st column of $A$ with $B$): $\Delta_1 = \left|\begin{array}{ccc} 3 & 1 & \beta \\ -3 & \alpha & -1 \\ 4 & 2 & 1 \end{array}\right|$
• $\Delta_2$ (Replace 2nd column of $A$ with $B$): $\Delta_2 = \left|\begin{array}{ccc} 3 & 3 & \beta \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{array}\right|$
• $\Delta_3$ (Replace 3rd column of $A$ with $B$): $\Delta_3 = \left|\begin{array}{ccc} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{array}\right|$

Step 3: Apply Conditions for Infinitely Many Solutions The problem states that the system has infinitely many solutions. According to Cramer's Rule, this implies that all four determinants must be zero: $\Delta = 0, \quad \Delta_1 = 0, \quad \Delta_2 = 0, \quad \Delta_3 = 0$ We need to find the values of $\alpha$ and $\beta$. We can use any two of these conditions involving $\alpha$ and $\beta$ to solve for them. It's often strategic to pick determinants that isolate one variable or lead to simpler equations.

Let's start with $\Delta_2 = 0$ as it contains only $\beta$ and constants among the unknown parameters.

Step 4: Solve for $\beta$ using $\Delta_2 = 0$ Set $\Delta_2 = 0$ and expand the determinant along the first row: $\left|\begin{array}{ccc} 3 & 3 & \beta \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{array}\right| = 0$ $3 \cdot \left| \begin{array}{cc} -3 & -1 \\ 4 & 1 \end{array} \right| - 3 \cdot \left| \begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array} \right| + \beta \cdot \left| \begin{array}{cc} 2 & -3 \\ 1 & 4 \end{array} \right| = 0$ Calculate the $2 \times 2$ determinants: $3((-3)(1) - (-1)(4)) - 3((2)(1) - (-1)(1)) + \beta((2)(4) - (-3)(1)) = 0$ $3(-3 + 4) - 3(2 + 1) + \beta(8 + 3) = 0$ $3(1) - 3(3) + 11\beta = 0$ $3 - 9 + 11\beta = 0$ $-6 + 11\beta = 0$ $11\beta = 6 \implies \beta = \frac{6}{11}$

Step 5: Solve for $\alpha$ using $\Delta_3 = 0$ Next, we set $\Delta_3 = 0$ and expand the determinant along the first row. This determinant contains only $\alpha$ and constants among the unknown parameters. $\left|\begin{array}{ccc} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{array}\right| = 0$ $3 \cdot \left| \begin{array}{cc} \alpha & -3 \\ 2 & 4 \end{array} \right| - 1 \cdot \left| \begin{array}{cc} 2 & -3 \\ 1 & 4 \end{array} \right| + 3 \cdot \left| \begin{array}{cc} 2 & \alpha \\ 1 & 2 \end{array} \right| = 0$ Calculate the $2 \times 2$ determinants: $3((\alpha)(4) - (-3)(2)) - 1((2)(4) - (-3)(1)) + 3((2)(2) - (\alpha)(1)) = 0$ $3(4\alpha + 6) - 1(8 + 3) + 3(4 - \alpha) = 0$ $12\alpha + 18 - 11 + 12 - 3\alpha = 0$ Combine the terms involving $\alpha$ and the constant terms: $(12\alpha - 3\alpha) + (18 - 11 + 12) = 0$ $9\alpha + 19 = 0$ $9\alpha = -19 \implies \alpha = -\frac{19}{9}$

Step 6: Calculate the Required Expression We have found $\beta = \frac{6}{11}$ and $\alpha = -\frac{19}{9}$. The problem asks for the value of $22\beta - 9\alpha$. Substitute the calculated values into the expression: $22\beta - 9\alpha = 22\left(\frac{6}{11}\right) - 9\left(-\frac{19}{9}\right)$ Simplify the expression: $= \left(\frac{22}{11}\right) \cdot 6 - \left(\frac{9}{9}\right) \cdot (-19)$ $= 2 \cdot 6 - 1 \cdot (-19)$ $= 12 - (-19)$ $= 12 + 19$ $= 31$

Verification (Optional): To ensure consistency, we can verify that $\Delta=0$ with the obtained values of $\alpha$ and $\beta$. $\Delta = \left|\begin{array}{ccc} 3 & 1 & \frac{6}{11} \\ 2 & -\frac{19}{9} & -1 \\ 1 & 2 & 1 \end{array}\right|$ Expanding along the first row: $\Delta = 3\left((-\frac{19}{9})(1) - (-1)(2)\right) - 1\left((2)(1) - (-1)(1)\right) + \frac{6}{11}\left((2)(2) - (-\frac{19}{9})(1)\right)$ $\Delta = 3\left(-\frac{19}{9} + 2\right) - 1(2 + 1) + \frac{6}{11}\left(4 + \frac{19}{9}\right)$ $\Delta = 3\left(\frac{-19+18}{9}\right) - 3 + \frac{6}{11}\left(\frac{36+19}{9}\right)$ $\Delta = 3\left(-\frac{1}{9}\right) - 3 + \frac{6}{11}\left(\frac{55}{9}\right)$ $\Delta = -\frac{1}{3} - 3 + \frac{6 \times 5}{9} = -\frac{1}{3} - 3 + \frac{30}{9} = -\frac{1}{3} - 3 + \frac{10}{3}$ $\Delta = \frac{-1 - 9 + 10}{3} = \frac{0}{3} = 0$ The condition $\Delta=0$ is satisfied, confirming our values of $\alpha$ and $\beta$.

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Common Mistakes & Tips

• Distinguish between Solution Types: Remember that $\Delta=0$ is a necessary condition for both "no solution" and "infinitely many solutions". The crucial difference is that for infinitely many solutions, all $\Delta_j$ must also be zero, whereas for no solution, at least one $\Delta_j$ must be non-zero.
• Careful with Determinant Expansion: Errors in signs or arithmetic during the expansion of $3 \times 3$ determinants are common. Double-check each step.
• Strategic Choice of Determinants: When solving for multiple variables, select conditions (e.g., $\Delta_j=0$) that simplify the process, ideally isolating one variable or leading to simpler equations.

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Summary

To determine the values of $\alpha$ and $\beta$ for which the given system of linear equations has infinitely many solutions, we utilized Cramer's Rule. The key conditions for infinitely many solutions are that the determinant of the coefficient matrix ($\Delta$) must be zero, and all auxiliary determinants ($\Delta_1, \Delta_2, \Delta_3$) must also be zero. By setting $\Delta_2 = 0$, we found $\beta = \frac{6}{11}$. By setting $\Delta_3 = 0$, we found $\alpha = -\frac{19}{9}$. Finally, substituting these values into the expression $22\beta - 9\alpha$ yielded the result $31$.

The final answer is $\boxed{\text{31}}$, which corresponds to option (A).