Key Concepts and Formulas

For a system of three linear equations in three variables ($x, y, z$): $\begin{aligned} a_1 x + b_1 y + c_1 z &= d_1 \\ a_2 x + b_2 y + c_2 z &= d_2 \\ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned}$ to have infinitely many solutions, we use Cramer's Rule conditions, which state that:

• The determinant of the coefficient matrix ($\Delta$) must be zero.
• The determinants formed by replacing each column of coefficients with the constant terms ($\Delta_x, \Delta_y, \Delta_z$) must also be zero.

That is, $\Delta = 0$ AND $\Delta_x = 0$ AND $\Delta_y = 0$ AND $\Delta_z = 0$. If $\Delta=0$ but any of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.

Step-by-Step Solution

Step 1: Identify the system and set up the determinants. The given system of equations is:

1. $2x + 3y - z = 5$
2. $x + \alpha y + 3z = -4$
3. $3x - y + \beta z = 7$

Based on the coefficients and constant terms, we define the following determinants:

• Main Determinant ($\Delta$): Formed by the coefficients of $x, y, z$. $\Delta = \begin{vmatrix} 2 & 3 & -1 \\ 1 & \alpha & 3 \\ 3 & -1 & \beta \end{vmatrix}$
• Determinant $\Delta_x$: Formed by replacing the first column (coefficients of $x$) with the constant terms. $\Delta_x = \begin{vmatrix} 5 & 3 & -1 \\ -4 & \alpha & 3 \\ 7 & -1 & \beta \end{vmatrix}$
• Determinant $\Delta_y$: Formed by replacing the second column (coefficients of $y$) with the constant terms. $\Delta_y = \begin{vmatrix} 2 & 5 & -1 \\ 1 & -4 & 3 \\ 3 & 7 & \beta \end{vmatrix}$
• Determinant $\Delta_z$: Formed by replacing the third column (coefficients of $z$) with the constant terms. $\Delta_z = \begin{vmatrix} 2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7 \end{vmatrix}$ For the system to have infinitely many solutions, all these determinants must be equal to zero. We will use this condition to find the values of $\alpha$ and $\beta$.

Step 2: Calculate $\Delta_y$ and solve for $\beta$. We strategically choose to calculate $\Delta_y$ first because, upon expansion, it will only contain $\beta$ as an unknown, allowing us to solve for $\beta$ directly. Expanding $\Delta_y$ along the first row: $\Delta_y = 2((-4)\beta - (3)(7)) - 5((1)\beta - (3)(3)) + (-1)((1)(7) - (-4)(3))$ $\Delta_y = 2(-4\beta - 21) - 5(\beta - 9) - 1(7 - (-12))$ $\Delta_y = -8\beta - 42 - 5\beta + 45 - 1(19)$ $\Delta_y = -13\beta + 3 - 19$ $\Delta_y = -13\beta - 16$ Since the system has infinitely many solutions, we must have $\Delta_y = 0$: $-13\beta - 16 = 0$ $13\beta = -16 \implies \beta = -\frac{16}{13}$

Step 3: Calculate $\Delta_z$ and solve for $\alpha$. Next, we calculate $\Delta_z$. Similar to $\Delta_y$, $\Delta_z$ will only contain $\alpha$ as an unknown, making it easy to solve for $\alpha$. Expanding $\Delta_z$ along the first row: $\Delta_z = 2(\alpha(7) - (-1)(-4)) - 3((1)(7) - (3)(-4)) + 5((1)(-1) - (\alpha)(3))$ $\Delta_z = 2(7\alpha - 4) - 3(7 + 12) + 5(-1 - 3\alpha)$ $\Delta_z = 14\alpha - 8 - 3(19) - 5 - 15\alpha$ $\Delta_z = 14\alpha - 8 - 57 - 5 - 15\alpha$ $\Delta_z = -\alpha - 70$ Since the system has infinitely many solutions, we must have $\Delta_z = 0$: $-\alpha - 70 = 0 \implies \alpha = -70$

Step 4: Verify with $\Delta=0$. We have found $\alpha = -70$ and $\beta = -\frac{16}{13}$. Now, we must verify that these values satisfy the condition $\Delta = 0$. This step ensures consistency. First, let's calculate $\Delta$: $\Delta = 2(\alpha\beta - (-1)(3)) - 3((1)\beta - (3)(3)) + (-1)((1)(-1) - (\alpha)(3))$ $\Delta = 2(\alpha\beta + 3) - 3(\beta - 9) - 1(-1 - 3\alpha)$ $\Delta = 2\alpha\beta + 6 - 3\beta + 27 + 1 + 3\alpha$ $\Delta = 2\alpha\beta + 3\alpha - 3\beta + 34$ Substitute $\alpha = -70$ and $\beta = -\frac{16}{13}$ into the expression for $\Delta$: $\Delta = 2(-70)\left(-\frac{16}{13}\right) + 3(-70) - 3\left(-\frac{16}{13}\right) + 34$ $\Delta = \frac{2240}{13} - 210 + \frac{48}{13} + 34$ $\Delta = \frac{2240+48}{13} - (210 - 34)$ $\Delta = \frac{2288}{13} - 176$ $\Delta = 176 - 176 = 0$ The condition $\Delta = 0$ is satisfied.

Step 5: Verify with $\Delta_x=0$. As a final consistency check, we verify that $\Delta_x = 0$ with the found values of $\alpha$ and $\beta$. First, let's calculate $\Delta_x$: $\Delta_x = 5(\alpha\beta - (-1)(3)) - 3((-4)\beta - (3)(7)) + (-1)((-4)(-1) - (\alpha)(7))$ $\Delta_x = 5(\alpha\beta + 3) - 3(-4\beta - 21) - 1(4 - 7\alpha)$ $\Delta_x = 5\alpha\beta + 15 + 12\beta + 63 - 4 + 7\alpha$ $\Delta_x = 5\alpha\beta + 7\alpha + 12\beta + 74$ Substitute $\alpha = -70$ and $\beta = -\frac{16}{13}$ into the expression for $\Delta_x$: $\Delta_x = 5(-70)\left(-\frac{16}{13}\right) + 7(-70) + 12\left(-\frac{16}{13}\right) + 74$ $\Delta_x = \frac{5600}{13} - 490 - \frac{192}{13} + 74$ $\Delta_x = \frac{5600-192}{13} - (490 - 74)$ $\Delta_x = \frac{5408}{13} - 416$ $\Delta_x = 416 - 416 = 0$ The condition $\Delta_x = 0$ is also satisfied. All conditions for infinitely many solutions are met.

Step 6: Calculate the required expression $13\alpha\beta$. We need to find the value of $13 \alpha \beta$. Substitute the determined values of $\alpha = -70$ and $\beta = -\frac{16}{13}$: $13 \alpha \beta = 13 \times (-70) \times \left(-\frac{16}{13}\right)$ The factor of $13$ in the numerator and denominator cancels out: $13 \alpha \beta = (-70) \times (-16)$ $13 \alpha \beta = 1120$

Common Mistakes & Tips

• Incomplete Conditions: A common mistake is only checking $\Delta=0$ for infinitely many solutions. Remember that $\Delta_x, \Delta_y, \Delta_z$ must also be zero. If $\Delta=0$ but any of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.
• Arithmetic Errors: Determinant calculations involve many multiplications and subtractions. Carefully double-check each step, especially signs, to avoid calculation errors.
• Strategic Calculation: When parameters are involved, prioritize calculating determinants that isolate a single parameter (like $\Delta_y$ for $\beta$ and $\Delta_z$ for $\alpha$ in this problem). This simplifies the process of finding the parameter values before verifying with more complex determinants like $\Delta$ or $\Delta_x$.

Summary

To solve a system of linear equations for parameters that lead to infinitely many solutions, we apply Cramer's Rule conditions: the determinant of the coefficient matrix ($\Delta$) and all associated determinants ($\Delta_x, \Delta_y, \Delta_z$) must be zero. By systematically calculating $\Delta_y$ and $\Delta_z$, we found the values of $\beta = -\frac{16}{13}$ and $\alpha = -70$. We then verified that these values also satisfy the conditions $\Delta=0$ and $\Delta_x=0$. Finally, we computed the required expression $13\alpha\beta$, which resulted in $1120$.

Final Answer

The final answer is $\boxed{\text{1120}}$, which corresponds to option (B).