This problem tests your understanding of the conditions for unique solutions, no solutions, and infinitely many solutions for a system of linear equations using determinants (Cramer's Rule).

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1. Key Concepts: Cramer's Rule for Systems of Linear Equations

For a system of three linear equations in three variables $x, y, z$: $a_1x + b_1y + c_1z = d_1$ $a_2x + b_2y + c_2z = d_2$ $a_3x + b_3y + c_3z = d_3$ We define the following determinants:

• $\Delta$ (Determinant of the Coefficient Matrix): $\Delta = \left| \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right|$
• $\Delta_x$: Replace the coefficients of $x$ in $\Delta$ with the constant terms ($d_1, d_2, d_3$).
• $\Delta_y$: Replace the coefficients of $y$ in $\Delta$ with the constant terms.
• $\Delta_z$: Replace the coefficients of $z$ in $\Delta$ with the constant terms.

The nature of the solutions is determined by the values of these determinants:

1. Unique Solution: If $\Delta \ne 0$, the system has a unique solution.
2. No Solution (Inconsistent): If $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system is inconsistent.
3. Infinitely Many Solutions: If $\Delta = 0$ AND $\Delta_x = 0$ AND $\Delta_y = 0$ AND $\Delta_z = 0$, the system has infinitely many solutions.

Tip: Always calculate $\Delta$ first. Its value dictates the initial classification of the system's solvability. Only if $\Delta=0$ do you need to calculate $\Delta_x, \Delta_y, \Delta_z$.

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2. Define the Given System and its Determinants

The given system of linear equations is: $2x - y + 3z = 5$ $3x + 2y - z = 7$ $4x + 5y + \alpha z = \beta$

From this system, we set up the required determinants:

$\Delta$ (Determinant of the Coefficient Matrix): $\Delta = \left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{matrix} \right|$

$\Delta_x$ (Determinant by replacing x-coefficients with constants): $\Delta_x = \left| \begin{matrix} 5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & \alpha \end{matrix} \right|$

$\Delta_y$ (Determinant by replacing y-coefficients with constants): $\Delta_y = \left| \begin{matrix} 2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & \alpha \end{matrix} \right|$

$\Delta_z$ (Determinant by replacing z-coefficients with constants): $\Delta_z = \left| \begin{matrix} 2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta \end{matrix} \right|$

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3. Step-by-Step Calculation of Determinants

We will now calculate each determinant by expanding along the first row.

Step 3.1: Calculate $\Delta$ We calculate $\Delta$ to find the condition for a unique solution or for $\Delta=0$. $\Delta = 2\left| \begin{matrix} 2 & -1 \\ 5 & \alpha \end{matrix} \right| - (-1)\left| \begin{matrix} 3 & -1 \\ 4 & \alpha \end{matrix} \right| + 3\left| \begin{matrix} 3 & 2 \\ 4 & 5 \end{matrix} \right|$ $\Delta = 2(2\alpha - (-1)(5)) + 1(3\alpha - (-1)(4)) + 3(3 \cdot 5 - 2 \cdot 4)$ $\Delta = 2(2\alpha + 5) + (3\alpha + 4) + 3(15 - 8)$ $\Delta = 4\alpha + 10 + 3\alpha + 4 + 3(7)$ $\Delta = 7\alpha + 14 + 21$ $\mathbf{\Delta = 7\alpha + 35}$

Step 3.2: Calculate $\Delta_x$ We calculate $\Delta_x$ to check conditions when $\Delta=0$. $\Delta_x = 5\left| \begin{matrix} 2 & -1 \\ 5 & \alpha \end{matrix} \right| - (-1)\left| \begin{matrix} 7 & -1 \\ \beta & \alpha \end{matrix} \right| + 3\left| \begin{matrix} 7 & 2 \\ \beta & 5 \end{matrix} \right|$ $\Delta_x = 5(2\alpha - (-1)(5)) + 1(7\alpha - (-1)\beta) + 3(7 \cdot 5 - 2\beta)$ $\Delta_x = 5(2\alpha + 5) + (7\alpha + \beta) + 3(35 - 2\beta)$ $\Delta_x = 10\alpha + 25 + 7\alpha + \beta + 105 - 6\beta$ $\mathbf{\Delta_x = 17\alpha - 5\beta + 130}$

Step 3.3: Calculate $\Delta_y$ We calculate $\Delta_y$ to check conditions when $\Delta=0$. $\Delta_y = 2\left| \begin{matrix} 7 & -1 \\ \beta & \alpha \end{matrix} \right| - 5\left| \begin{matrix} 3 & -1 \\ 4 & \alpha \end{matrix} \right| + 3\left| \begin{matrix} 3 & 7 \\ 4 & \beta \end{matrix} \right|$ $\Delta_y = 2(7\alpha - (-1)\beta) - 5(3\alpha - (-1)(4)) + 3(3\beta - 7 \cdot 4)$ $\Delta_y = 2(7\alpha + \beta) - 5(3\alpha + 4) + 3(3\beta - 28)$ $\Delta_y = 14\alpha + 2\beta - 15\alpha - 20 + 9\beta - 84$ $\mathbf{\Delta_y = -\alpha + 11\beta - 104}$

Step 3.4: Calculate $\Delta_z$ We calculate $\Delta_z$ to check conditions when $\Delta=0$. $\Delta_z = 2\left| \begin{matrix} 2 & 7 \\ 5 & \beta \end{matrix} \right| - (-1)\left| \begin{matrix} 3 & 7 \\ 4 & \beta \end{matrix} \right| + 5\left| \begin{matrix} 3 & 2 \\ 4 & 5 \end{matrix} \right|$ $\Delta_z = 2(2\beta - 7 \cdot 5) + 1(3\beta - 7 \cdot 4) + 5(3 \cdot 5 - 2 \cdot 4)$ $\Delta_z = 2(2\beta - 35) + (3\beta - 28) + 5(15 - 8)$ $\Delta_z = 4\beta - 70 + 3\beta - 28 + 5(7)$ $\Delta_z = 7\beta - 98 + 35$ $\mathbf{\Delta_z = 7\beta - 63}$

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4. Analyze Each Option

Now we evaluate each option by substituting the given values of $\alpha$ and $\beta$ into our calculated determinants and applying Cramer's Rule conditions.

Option (A): The system has infinitely many solutions for $\alpha=-6$ and $\beta=9$.

• Substitute $\alpha=-6$: $\Delta = 7(-6) + 35 = -42 + 35 = -7$.
• Analysis: Since $\Delta = -7 \ne 0$, the system has a unique solution.
• Conclusion for (A): The statement claims infinitely many solutions, but our calculation shows a unique solution. Therefore, statement (A) is NOT correct.

Since the question asks which statement is NOT correct, (A) is our answer. Let's quickly verify the other options to ensure consistency and reinforce understanding.

Option (B): The system has a unique solution for $\alpha \ne -5$ and $\beta=8$.

• If $\alpha \ne -5$, then $\Delta = 7\alpha + 35 \ne 7(-5) + 35 = 0$.
• Analysis: Since $\Delta \ne 0$, the system indeed has a unique solution, regardless of the value of $\beta$.
• Conclusion for (B): Statement (B) is correct.

Option (C): The system is inconsistent for $\alpha=-5$ and $\beta=8$.

• Substitute $\alpha=-5$: $\Delta = 7(-5) + 35 = -35 + 35 = 0$.
• Now, we must check $\Delta_x, \Delta_y, \Delta_z$. Let's check $\Delta_x$: Substitute $\alpha=-5, \beta=8$: $\Delta_x = 17(-5) - 5(8) + 130 = -85 - 40 + 130 = -125 + 130 = 5$.
• Analysis: We have $\Delta = 0$ and $\Delta_x = 5 \ne 0$. This condition implies the system is inconsistent (has no solution).
• Conclusion for (C): Statement (C) is correct.

Option (D): The system has infinitely many solutions for $\alpha=-5$ and $\beta=9$.

• Substitute $\alpha=-5$: $\Delta = 7(-5) + 35 = -35 + 35 = 0$.
• Now, we must check $\Delta_x, \Delta_y, \Delta_z$: Substitute $\alpha=-5, \beta=9$: $\Delta_x = 17(-5) - 5(9) + 130 = -85 - 45 + 130 = -130 + 130 = 0$. $\Delta_y = -(-5) + 11(9) - 104 = 5 + 99 - 104 = 104 - 104 = 0$. $\Delta_z = 7(9) - 63 = 63 - 63 = 0$.
• Analysis: We have $\Delta = 0$, $\Delta_x = 0$, $\Delta_y = 0$, and $\Delta_z = 0$. This condition implies the system has infinitely many solutions.
• Conclusion for (D): Statement (D) is correct.

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5. Final Answer and Key Takeaway

Based on our analysis, statement (A) is the only one that is NOT correct.

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

Key Takeaway: For systems of linear equations, Cramer's Rule provides a systematic way to determine the nature of solutions. Always start by calculating the determinant of the coefficient matrix ($\Delta$). If $\Delta \ne 0$, it's a unique solution. If $\Delta = 0$, then calculate $\Delta_x, \Delta_y, \Delta_z$ to distinguish between no solution (if any of them is non-zero) and infinitely many solutions (if all of them are zero). Missing even one of these checks when $\Delta=0$ is a common mistake.