1. Understanding the Conditions for an Inconsistent System

For a system of linear equations, $A\mathbf{x} = \mathbf{b}$, to be inconsistent (meaning it has no solution), specific conditions related to determinants must be met. These conditions are derived from Cramer's Rule.

Let's define the key determinants for a system of 3 linear equations in 3 variables:

• $\Delta = \det(A)$: The determinant of the coefficient matrix.
• $\Delta_x$: The determinant formed by replacing the first column (coefficients of $x$) of $A$ with the column of constant terms $\mathbf{b}$.
• $\Delta_y$: The determinant formed by replacing the second column (coefficients of $y$) of $A$ with $\mathbf{b}$.
• $\Delta_z$: The determinant formed by replacing the third column (coefficients of $z$) of $A$ with $\mathbf{b}$.

A system of linear equations is inconsistent (has no solution) if and only if:

1. The determinant of the coefficient matrix is zero: $\Delta = 0$.
2. At least one of the determinants $\Delta_x, \Delta_y, \Delta_z$ is non-zero. That is, ($\Delta_x \ne 0$ OR $\Delta_y \ne 0$ OR $\Delta_z \ne 0$).

Important Note: It's crucial to understand the other possible scenarios for a system of linear equations:

• Unique Solution: If $\Delta \ne 0$.
• Infinitely Many Solutions: If $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$.

Our objective is to find the value of $\alpha$ for which the given system of equations satisfies the conditions for an inconsistent system.

2. Setting up the Determinants from the Given System

The given system of linear equations is: $x + 2y + z = 2 \quad \text{(Eq. 1)}$ $\alpha x + 3y - z = \alpha \quad \text{(Eq. 2)}$ $- \alpha x + y + 2z = - \alpha \quad \text{(Eq. 3)}$

First, we extract the coefficient matrix $A$ and the constant terms vector $\mathbf{b}$ from this system: $A = \begin{pmatrix} 1 & 2 & 1 \\ \alpha & 3 & -1 \\ - \alpha & 1 & 2 \end{pmatrix} \quad \text{and} \quad \mathbf{b} = \begin{pmatrix} 2 \\ \alpha \\ - \alpha \end{pmatrix}$

Now we formulate the determinants required for our analysis:

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

• $\Delta_x$ (Determinant for $x$): Formed by replacing the first column of $A$ with $\mathbf{b}$. $\Delta_x = \left| \begin{matrix} 2 & 2 & 1 \\ \alpha & 3 & -1 \\ - \alpha & 1 & 2 \end{matrix} \right|$

• $\Delta_y$ (Determinant for $y$): Formed by replacing the second column of $A$ with $\mathbf{b}$. $\Delta_y = \left| \begin{matrix} 1 & 2 & 1 \\ \alpha & \alpha & -1 \\ - \alpha & - \alpha & 2 \end{matrix} \right|$

• $\Delta_z$ (Determinant for $z$): Formed by replacing the third column of $A$ with $\mathbf{b}$. $\Delta_z = \left| \begin{matrix} 1 & 2 & 2 \\ \alpha & 3 & \alpha \\ - \alpha & 1 & - \alpha \end{matrix} \right|$

3. Step 1: Calculate $\Delta$ and Find Values of $\alpha$ for $\Delta = 0$

The first condition for an inconsistent system is $\Delta = 0$. We calculate $\Delta$ by expanding the $3 \times 3$ determinant along the first row: $\Delta = 1 \cdot \left| \begin{matrix} 3 & -1 \\ 1 & 2 \end{matrix} \right| - 2 \cdot \left| \begin{matrix} \alpha & -1 \\ - \alpha & 2 \end{matrix} \right| + 1 \cdot \left| \begin{matrix} \alpha & 3 \\ - \alpha & 1 \end{matrix} \right|$ Now, we evaluate each $2 \times 2$ sub-determinant: $= 1 \cdot ((3)(2) - (-1)(1)) - 2 \cdot ((\alpha)(2) - (-1)(-\alpha)) + 1 \cdot ((\alpha)(1) - (3)(-\alpha))$ $= 1 \cdot (6 + 1) - 2 \cdot (2\alpha - \alpha) + 1 \cdot (\alpha + 3\alpha)$ $= 7 - 2(\alpha) + 4\alpha$ $\Delta = 7 + 2\alpha$

To satisfy the first condition for inconsistency, we set $\Delta = 0$: $7 + 2\alpha = 0$ $2\alpha = -7$ $\alpha = - \frac{7}{2}$ This value of $\alpha$ is a candidate for inconsistency. If $\alpha$ were any other value, $\Delta \ne 0$, and the system would have a unique solution.

4. Step 2: Calculate $\Delta_x$ and Check for Inconsistency

Now we need to check the second condition for inconsistency: that at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero when $\alpha = -7/2$. Let's start by calculating $\Delta_x$: $\Delta_x = \left| \begin{matrix} 2 & 2 & 1 \\ \alpha & 3 & -1 \\ - \alpha & 1 & 2 \end{matrix} \right|$ We expand along the first row: $\Delta_x = 2 \cdot \left| \begin{matrix} 3 & -1 \\ 1 & 2 \end{matrix} \right| - 2 \cdot \left| \begin{matrix} \alpha & -1 \\ - \alpha & 2 \end{matrix} \right| + 1 \cdot \left| \begin{matrix} \alpha & 3 \\ - \alpha & 1 \end{matrix} \right|$ $= 2 \cdot ((3)(2) - (-1)(1)) - 2 \cdot ((\alpha)(2) - (-1)(-\alpha)) + 1 \cdot ((\alpha)(1) - (3)(-\alpha))$ $= 2 \cdot (6 + 1) - 2 \cdot (2\alpha - \alpha) + 1 \cdot (\alpha + 3\alpha)$ $= 2 \cdot 7 - 2\alpha + 4\alpha$ $\Delta_x = 14 + 2\alpha$

Now, we substitute the value $\alpha = -7/2$ into the expression for $\Delta_x$: $\Delta_x = 14 + 2 \left( - \frac{7}{2} \right)$ $\Delta_x = 14 - 7$ $\Delta_x = 7$

Since $\Delta_x = 7 \ne 0$ for $\alpha = -7/2$, the second condition for inconsistency is met. This confirms that for $\alpha = -7/2$, the system is indeed inconsistent.

Tip: Once one of $\Delta_x, \Delta_y, \Delta_z$ is found to be non-zero for the value of $\alpha$ that makes $\Delta=0$, we have sufficiently proven inconsistency. Calculating the others is optional but can serve as a good verification.

5. Step 3: Calculate $\Delta_y$ and $\Delta_z$ (Verification)

For completeness, let's also calculate $\Delta_y$ and $\Delta_z$ and evaluate them at $\alpha = -7/2$.

Calculating $\Delta_y$: $\Delta_y = \left| \begin{matrix} 1 & 2 & 1 \\ \alpha & \alpha & -1 \\ - \alpha & - \alpha & 2 \end{matrix} \right|$ Expanding along the first row: $\Delta_y = 1 \cdot \left| \begin{matrix} \alpha & -1 \\ - \alpha & 2 \end{matrix} \right| - 2 \cdot \left| \begin{matrix} \alpha & -1 \\ - \alpha & 2 \end{matrix} \right| + 1 \cdot \left| \begin{matrix} \alpha & \alpha \\ - \alpha & - \alpha \end{matrix} \right|$ $= 1 \cdot (( \alpha)(2) - (-1)(-\alpha)) - 2 \cdot (( \alpha)(2) - (-1)(-\alpha)) + 1 \cdot ((\alpha)(-\alpha) - (\alpha)(-\alpha))$ $= 1 \cdot (2\alpha - \alpha) - 2 \cdot (2\alpha - \alpha) + 1 \cdot (-\alpha^2 + \alpha^2)$ $= \alpha - 2\alpha + 0$ $\Delta_y = - \alpha$ Substituting $\alpha = -7/2$: $\Delta_y = - \left( - \frac{7}{2} \right) = \frac{7}{2}$ Since $\Delta_y = 7/2 \ne 0$, this further confirms inconsistency.

Calculating $\Delta_z$: $\Delta_z = \left| \begin{matrix} 1 & 2 & 2 \\ \alpha & 3 & \alpha \\ - \alpha & 1 & - \alpha \end{matrix} \right|$ Expanding along the first row: $\Delta_z = 1 \cdot \left| \begin{matrix} 3 & \alpha \\ 1 & - \alpha \end{matrix} \right| - 2 \cdot \left| \begin{matrix} \alpha & \alpha \\ - \alpha & - \alpha \end{matrix} \right| + 2 \cdot \left| \begin{matrix} \alpha & 3 \\ - \alpha & 1 \end{matrix} \right|$ $= 1 \cdot ((3)(-\alpha) - (\alpha)(1)) - 2 \cdot ((\alpha)(-\alpha) - (\alpha)(-\alpha)) + 2 \cdot ((\alpha)(1) - (3)(-\alpha))$ $= 1 \cdot (-3\alpha - \alpha) - 2 \cdot (-\alpha^2 + \alpha^2) + 2 \cdot (\alpha + 3\alpha)$ $= -4\alpha - 2 \cdot (0) + 2 \cdot (4\alpha)$ $= -4\alpha + 8\alpha$ $\Delta_z = 4\alpha$ Substituting $\alpha = -7/2$: $\Delta_z = 4 \left( - \frac{7}{2} \right) = -14$ Since $\Delta_z = -14 \ne 0$, this also confirms inconsistency.

6. Step 4: Verify Conditions for Inconsistency

We have found that for $\alpha = -7/2$:

• $\Delta = 0$ (from $7 + 2\alpha = 0$).
• $\Delta_x = 7 \ne 0$.
• $\Delta_y = 7/2 \ne 0$.
• $\Delta_z = -14 \ne 0$.

Since $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero (in this case, all three are non-zero), the system of equations is indeed inconsistent when $\alpha = -7/2$.

7. Conclusion and Key Takeaway

The value of $\alpha$ for which the given system of linear equations is inconsistent is $\mathbf{- \frac{7}{2}}$.

This corresponds to option (D).

Key Takeaway: To determine if a system of linear equations is inconsistent, always start by calculating the determinant of the coefficient matrix ($\Delta$). If $\Delta \ne 0$, there's a unique solution. If $\Delta = 0$, then proceed to calculate $\Delta_x, \Delta_y, \Delta_z$. If at least one of these is non-zero, the system is inconsistent. If all of them are zero, the system has infinitely many solutions.

The final answer is $\boxed{{- \frac{7}{2}}}$.