1. Key Concept: Analyzing Systems of Linear Equations using Determinants (Cramer's Rule)

For a system of $n$ linear equations in $n$ variables, represented in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, we can use determinants to determine the nature of solutions.

Let $\Delta = \det(A)$ be the determinant of the coefficient matrix. Let $\Delta_j$ be the determinant of the matrix obtained by replacing the $j$-th column of $A$ with the constant matrix $B$.

The nature of solutions is determined as follows:

1. Unique Solution: If $\Delta \ne 0$, the system has a unique solution.
2. No Solution (Inconsistent System): If $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z, \dots$ is non-zero, the system has no solutions. This means the equations are contradictory.
3. Infinite Solutions (Consistent and Dependent System): If $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = \dots = 0$, the system has infinitely many solutions. This means the equations are dependent.

Our goal is to find the value of $\alpha$ for which the system has no solutions. According to Cramer's Rule, this occurs when $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.

2. Step 1: Express the System in Matrix Form and Identify the Coefficient and Constant Matrices

The given system of linear equations is:

$$\begin{align*} \alpha x + y + z &= \alpha - 1 \\ x + \alpha y + z &= \alpha - 1 \\ x + y + \alpha z &= \alpha - 1 \end{align*}$$

We can write this in the matrix form $AX = B$, where: The coefficient matrix $A$ is: $A = \begin{pmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{pmatrix}$ The variable matrix $X$ is: $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ The constant matrix $B$ is: $B = \begin{pmatrix} \alpha - 1 \\ \alpha - 1 \\ \alpha - 1 \end{pmatrix}$

3. Step 2: Calculate the Determinant of the Coefficient Matrix, $\Delta$

To find the values of $\alpha$ for which the system might have no solution or infinite solutions, we first need to find when $\Delta = \det(A) = 0$.

Let's calculate the determinant of $A$: $\Delta = \det \begin{pmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{pmatrix}$ We can calculate this by expanding along the first row: $\Delta = \alpha(\alpha \cdot \alpha - 1 \cdot 1) - 1(1 \cdot \alpha - 1 \cdot 1) + 1(1 \cdot 1 - \alpha \cdot 1)$ $\Delta = \alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha)$ Now, we factor out common terms. Notice that $(\alpha^2 - 1) = (\alpha-1)(\alpha+1)$ and $(1-\alpha) = -(\alpha-1)$: $\Delta = \alpha(\alpha-1)(\alpha+1) - (\alpha-1) - (\alpha-1)$ Factor out $(\alpha-1)$: $\Delta = (\alpha-1) [\alpha(\alpha+1) - 1 - 1]$ $\Delta = (\alpha-1) [\alpha^2 + \alpha - 2]$ Now, factor the quadratic term $\alpha^2 + \alpha - 2$. We are looking for two numbers that multiply to $-2$ and add to $1$. These numbers are $2$ and $-1$. So, $\alpha^2 + \alpha - 2 = (\alpha+2)(\alpha-1)$. Substitute this back into the expression for $\Delta$: $\Delta = (\alpha-1) (\alpha+2)(\alpha-1)$ $\Delta = (\alpha-1)^2 (\alpha+2)$

Now, we set $\Delta = 0$ to find the critical values of $\alpha$: $(\alpha-1)^2 (\alpha+2) = 0$ This equation holds if $\alpha-1 = 0$ or $\alpha+2 = 0$. Thus, the possible values for $\alpha$ are $\alpha = 1$ or $\alpha = -2$. These are the values for which the system might have no solution or infinite solutions. We need to analyze each case further.

4. Step 3: Analyze Each Case for $\Delta = 0$

We will examine the two values of $\alpha$ found in Step 2.

Case A: When $\alpha = 1$

Substitute $\alpha = 1$ into the original system of equations:

$$\begin{align*} 1x + y + z &= 1 - 1 \implies x + y + z = 0 \\ x + 1y + z &= 1 - 1 \implies x + y + z = 0 \\ x + y + 1z &= 1 - 1 \implies x + y + z = 0 \end{align*}$$

All three equations become identical: $x+y+z=0$. This means the system is essentially a single equation in three variables. Such a system has infinitely many solutions (e.g., if $x=1, y=-1$, then $z=0$; if $x=2, y=0$, then $z=-2$, and so on).

To formally check using Cramer's Rule: For $\alpha=1$, we have $\Delta = (1-1)^2(1+2) = 0$. The constant matrix $B$ becomes: $B = \begin{pmatrix} 1-1 \\ 1-1 \\ 1-1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ Now, let's calculate $\Delta_x$: $\Delta_x = \det \begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ Since the first column consists entirely of zeros, $\Delta_x = 0$. Similarly, $\Delta_y = 0$ and $\Delta_z = 0$ because their respective columns would also be all zeros. Since $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$, the system has infinitely many solutions when $\alpha = 1$. Therefore, $\alpha = 1$ is not the answer for "no solutions".

Case B: When $\alpha = -2$

Substitute $\alpha = -2$ into the original system of equations:

$$\begin{align*} -2x + y + z &= -2 - 1 \implies -2x + y + z = -3 \\ x - 2y + z &= -2 - 1 \implies x - 2y + z = -3 \\ x + y - 2z &= -2 - 1 \implies x + y - 2z = -3 \end{align*}$$

For $\alpha = -2$, we know $\Delta = (-2-1)^2(-2+2) = (-3)^2(0) = 0$. Now we need to check if any of $\Delta_x, \Delta_y, \Delta_z$ is non-zero. The constant matrix $B$ becomes: $B = \begin{pmatrix} -2-1 \\ -2-1 \\ -2-1 \end{pmatrix} = \begin{pmatrix} -3 \\ -3 \\ -3 \end{pmatrix}$ Let's calculate $\Delta_x$: $\Delta_x = \det \begin{pmatrix} -3 & 1 & 1 \\ -3 & -2 & 1 \\ -3 & 1 & -2 \end{pmatrix}$ To simplify the calculation, we can take out a common factor of $-3$ from the first column: $\Delta_x = -3 \det \begin{pmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix}$ Now, calculate the determinant of the remaining matrix: $\det \begin{pmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix} = 1((-2)(-2) - 1(1)) - 1(1(-2) - 1(1)) + 1(1(1) - (-2)(1))$ $= 1(4 - 1) - 1(-2 - 1) + 1(1 + 2)$ $= 1(3) - 1(-3) + 1(3)$ $= 3 + 3 + 3 = 9$ So, $\Delta_x = -3 \times 9 = -27$.

Since $\Delta = 0$ AND $\Delta_x = -27 \ne 0$, according to Cramer's Rule, the system has no solutions when $\alpha = -2$.

5. Conclusion

Based on our analysis:

• If $\alpha = 1$, the system has infinitely many solutions.
• If $\alpha = -2$, the system has no solutions.

Therefore, the system of equations has no solutions if $\alpha = -2$.

6. Tips and Common Mistakes to Avoid

• Determinant Calculation: Be extremely careful with signs and arithmetic when calculating determinants. A small error here can lead to incorrect values for $\alpha$.
• Confusing Conditions: The most common mistake is to stop after finding $\Delta=0$ and not checking $\Delta_x, \Delta_y, \Delta_z$. Remember, $\Delta=0$ only tells you there's not a unique solution; it could be no solution or infinite solutions. You must check the numerators for Cramer's rule.
• Symmetry: In this problem, the system is symmetric in $x, y, z$ and the constant vector $B$ also has identical components. This implies that $\Delta_x = \Delta_y = \Delta_z$. Therefore, calculating just one of them (e.g., $\Delta_x$) is sufficient to determine if they are all zero or all non-zero. This can save time.
• Direct Substitution (for verification): For simple cases like $\alpha=1$, substituting the value back into the original equations can sometimes quickly reveal the nature of solutions (e.g., all equations becoming identical, or contradictory equations like $0=5$).

7. Summary and Key Takeaway

To determine when a system of linear equations has no solutions, we use Cramer's Rule. First, calculate the determinant of the coefficient matrix, $\Delta$. If $\Delta \neq 0$, there's a unique solution. If $\Delta = 0$, we then calculate the determinants $\Delta_j$ for each variable. If $\Delta = 0$ and at least one $\Delta_j \neq 0$, the system has no solutions. If $\Delta = 0$ and all $\Delta_j = 0$, the system has infinitely many solutions. In this problem, we found that $\Delta = 0$ for $\alpha=1$ and $\alpha=-2$. Further analysis showed that $\alpha=1$ leads to infinite solutions, while $\alpha=-2$ leads to $\Delta_x \neq 0$, indicating no solutions.

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