Understanding Inconsistent Systems of Linear Equations

For a system of linear equations, such as $A\mathbf{x} = \mathbf{b}$, to be inconsistent (meaning it has no solution), we use the concept of determinants. A fundamental principle states:

• If the determinant of the coefficient matrix, $\Delta$, is non-zero ($\Delta \neq 0$), then the system has a unique solution.
• If $\Delta = 0$, the system either has infinitely many solutions or no solution (is inconsistent). To distinguish between these two cases when $\Delta = 0$, we need to perform further checks, typically by examining the determinants $\Delta_x, \Delta_y, \Delta_z$ (from Cramer's Rule) or by using Gaussian elimination (row operations).
  • If $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system is inconsistent (no solution).
  • If $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$, the system has infinitely many solutions.

Step-by-Step Solution

1. Write Down the Given System of Equations The given system of linear equations is: $\begin{aligned} x + y + z &= \alpha \quad \quad \quad \quad &(1) \\ \alpha x + 2\alpha y + 3z &= -1 \quad \quad \quad &(2) \\ x + 3\alpha y + 5z &= 4 \quad \quad \quad \quad &(3) \end{aligned}$

2. Form the Coefficient Matrix and Calculate its Determinant ($\Delta$) The coefficient matrix of this system is formed by the coefficients of $x, y,$ and $z$: $A = \begin{pmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{pmatrix}$ Now, we calculate the determinant of this matrix, denoted as $\Delta$: $\Delta = \left|\begin{array}{ccc} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{array}\right|$ We expand the determinant along the first row: $\begin{aligned} \Delta &= 1 \cdot \left|\begin{array}{cc} 2\alpha & 3 \\ 3\alpha & 5 \end{array}\right| - 1 \cdot \left|\begin{array}{cc} \alpha & 3 \\ 1 & 5 \end{array}\right| + 1 \cdot \left|\begin{array}{cc} \alpha & 2\alpha \\ 1 & 3\alpha \end{array}\right| \\ &= 1((2\alpha)(5) - (3)(3\alpha)) - 1((\alpha)(5) - (3)(1)) + 1((\alpha)(3\alpha) - (2\alpha)(1)) \\ &= 1(10\alpha - 9\alpha) - 1(5\alpha - 3) + 1(3\alpha^2 - 2\alpha) \\ &= \alpha - 5\alpha + 3 + 3\alpha^2 - 2\alpha \\ &= 3\alpha^2 - 6\alpha + 3 \end{aligned}$

3. Find Values of $\alpha$ for which $\Delta = 0$ For the system to be inconsistent, a necessary condition is that $\Delta = 0$. So, we set our calculated determinant to zero: $3\alpha^2 - 6\alpha + 3 = 0$ We can factor out a 3: $3(\alpha^2 - 2\alpha + 1) = 0$ Recognize the term in the parenthesis as a perfect square trinomial: $3(\alpha - 1)^2 = 0$ This equation implies: $(\alpha - 1)^2 = 0$ Therefore, the only value of $\alpha$ for which $\Delta = 0$ is: $\alpha = 1$ This means that for any $\alpha \neq 1$, $\Delta \neq 0$, and the system will have a unique solution. Thus, if the system is inconsistent, it must be when $\alpha = 1$.

4. Check for Inconsistency when $\alpha = 1$ Now we substitute $\alpha = 1$ back into the original system of equations to check if it leads to an inconsistency (no solution) or infinitely many solutions. The system becomes: $\begin{aligned} x + y + z &= 1 \quad \quad \quad \quad &(1') \\ x + 2y + 3z &= -1 \quad \quad \quad &(2') \\ x + 3y + 5z &= 4 \quad \quad \quad \quad &(3') \end{aligned}$ We can use elementary row operations (Gaussian elimination) or combine equations to look for a contradiction.

• Subtract (1') from (2'): $(x + 2y + 3z) - (x + y + z) = -1 - 1$ y + 2z = -2 \quad \quad \quad \quad &(4')

• Subtract (1') from (3'): $(x + 3y + 5z) - (x + y + z) = 4 - 1$ 2y + 4z = 3 \quad \quad \quad \quad &(5')

Now, let's examine equations (4') and (5'). Notice that equation (5') can be written as $2(y + 2z) = 3$. If we substitute the expression for $(y+2z)$ from (4') into this modified (5'): $2(-2) = 3$ $-4 = 3$ This is a false statement, a clear contradiction! This means that there are no values of $x, y, z$ that can satisfy all three equations simultaneously when $\alpha=1$.

Conclusion: Since $\Delta = 0$ for $\alpha = 1$, and substituting $\alpha = 1$ into the system leads to a contradiction, the system of equations is inconsistent for $\alpha = 1$.

5. Count the Number of Values of $\alpha$ We found only one value of $\alpha$, which is $\alpha = 1$, for which the system is inconsistent.

The number of values of $\alpha$ for which the system is inconsistent is 1.

Relevant Tips and Common Mistakes

• Don't Stop at $\Delta = 0$: A very common mistake is to assume that if $\Delta = 0$, the system is automatically inconsistent. Remember, $\Delta = 0$ only tells you that the system doesn't have a unique solution. It could have infinitely many solutions. You must perform further checks (like the one we did in Step 4, or using $\Delta_x, \Delta_y, \Delta_z$) to distinguish between infinitely many solutions and no solution.
• Systematic Approach: When checking for inconsistency (or infinite solutions) after $\Delta=0$, use a systematic method like Gaussian elimination or Cramer's rule. Randomly combining equations can sometimes miss crucial details or lead to errors.
• Careful Calculation: Determinant calculations can be prone to sign errors. Double-check your arithmetic!

Summary / Key Takeaway

For a system of linear equations, inconsistency (no solution) occurs when the determinant of the coefficient matrix ($\Delta$) is zero, AND upon substituting that value back into the system, a contradiction arises (e.g., $0 = k$ where $k \neq 0$). In this problem, we found that $\Delta = 0$ only for $\alpha=1$. Upon checking this value, we confirmed a contradiction, indicating inconsistency. Thus, there is only one value of $\alpha$ for which the system is inconsistent.

The final answer is $\boxed{\text{1}}$.