Key Concept: Conditions for a System of Linear Equations to Have No Solution

For a system of linear equations to have no solution (i.e., be inconsistent), the equations must lead to a contradiction, such as $0 = k$ where $k \neq 0$. This typically occurs when:

1. The determinant of the coefficient matrix ($\det(A)$) is zero, AND
2. At least one of the determinants $D_x, D_y, D_z$ (formed by replacing a column of $A$ with the constant vector $B$) is non-zero. Alternatively, using rank analysis: Rank($A$) $\neq$ Rank($[A|B]$), where $[A|B]$ is the augmented matrix. A direct algebraic approach (elimination or substitution) is often the most robust way to determine inconsistency, especially in cases where parameters are involved.

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1. Setting Up the System

The given system of linear equations is:

1. $x + y + z = 1$
2. $x + ay + z = 1$
3. $ax + by + z = 0$

Our goal is to find the values of 'b' for which this system has no solution.

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2. Step-by-Step Analysis using Elimination

We will use algebraic elimination to simplify the system and identify conditions for its consistency.

Step 2.1: Eliminate a variable from the first two equations.

• Subtract Equation (1) from Equation (2): $(x + ay + z) - (x + y + z) = 1 - 1$ $ay - y = 0$ $y(a - 1) = 0$

• Explanation: This step helps us find a crucial relationship between $y$ and $a$. For the product $y(a-1)$ to be zero, either $y$ must be zero, or $(a-1)$ must be zero (meaning $a=1$), or both. This naturally leads to two main cases based on the value of 'a'.

Step 2.2: Case 1: $a \neq 1$

• If $a \neq 1$, then from $y(a-1) = 0$, it must be that $y = 0$.

• Substitute $y=0$ into the original system:

  1. $x + 0 + z = 1 \implies x + z = 1$
  2. $x + a(0) + z = 1 \implies x + z = 1$ (This is consistent with Eq 1)
  3. $ax + b(0) + z = 0 \implies ax + z = 0$

• Now we have a reduced system: $x + z = 1$ $ax + z = 0$

• Subtract the second equation from the first: $(x + z) - (ax + z) = 1 - 0$ $x - ax = 1$ $x(1 - a) = 1$

• Since we are in the case $a \neq 1$, it means $1 - a \neq 0$. Therefore, $x = \frac{1}{1-a}$.

• Substitute $x$ back into $x+z=1$: $\frac{1}{1-a} + z = 1 \implies z = 1 - \frac{1}{1-a} = \frac{1-a-1}{1-a} = \frac{-a}{1-a}$.

• Conclusion for Case 1 ($a \neq 1$): For any $a \neq 1$, we found a unique solution for $(x, y, z)$: $x = \frac{1}{1-a}$, $y = 0$, $z = \frac{-a}{1-a}$.

  • Explanation: Since a unique solution exists when $a \neq 1$, the system cannot have no solution in this case. This means any values of 'b' that lead to no solution must come from the case where $a=1$.

Step 2.3: Case 2: $a = 1$

• If $a = 1$, the equation $y(a-1) = 0$ becomes $y(0) = 0$, which simplifies to $0 = 0$. This identity gives no specific information about $y$; $y$ can be any value.

• Substitute $a=1$ into the original system:

  1. $x + y + z = 1$
  2. $x + (1)y + z = 1 \implies x + y + z = 1$ (This is identical to Eq 1)
  3. $(1)x + by + z = 0 \implies x + by + z = 0$

• The system effectively reduces to two distinct equations: $x + y + z = 1$ (Equation 1') $x + by + z = 0$ (Equation 3')

• Subtract Equation 3' from Equation 1': $(x + y + z) - (x + by + z) = 1 - 0$ $y - by = 1$ $y(1 - b) = 1$

• Explanation: This critical equation $y(1-b)=1$ now determines whether a solution exists when $a=1$. We must analyze two subcases based on the value of $1-b$.

  • Subcase 2.3.1: $1 - b \neq 0 \implies b \neq 1$ If $b \neq 1$, then $y = \frac{1}{1-b}$. Substitute this value of $y$ back into $x+y+z=1$: $x + \frac{1}{1-b} + z = 1$ $x + z = 1 - \frac{1}{1-b} = \frac{(1-b) - 1}{1-b} = \frac{-b}{1-b}$.

    • Conclusion for Subcase 2.3.1 ($a=1, b \neq 1$): We have $y = \frac{1}{1-b}$ and $x+z = \frac{-b}{1-b}$. Since $x+z$ is a fixed value, but $x$ and $z$ can vary (e.g., for any chosen value of $z$, $x$ is determined), there are infinitely many solutions.
      • Explanation: In this subcase, the system is consistent and has an infinite number of solutions. Thus, it does not have "no solution".

  • Subcase 2.3.2: $1 - b = 0 \implies b = 1$ If $b = 1$, the equation $y(1 - b) = 1$ becomes $y(0) = 1$, which simplifies to $0 = 1$.

    • Conclusion for Subcase 2.3.2 ($a=1, b=1$): The result $0=1$ is a direct contradiction. This means that if $a=1$ and $b=1$, there are no values of $x, y, z$ that can satisfy the system simultaneously. Therefore, the system has no solution.
      • Explanation: A contradiction like $0=1$ is the definitive indicator of an inconsistent system (no solution).

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3. Summary of System Behavior

From our analysis:

• If $a \neq 1$: The system has a unique solution.
• If $a = 1$ and $b \neq 1$: The system has infinitely many solutions.
• If $a = 1$ and $b = 1$: The system has no solution.

Therefore, the system has no solution if and only if $a=1$ AND $b=1$.

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4. Determining the Set S based on Question Interpretation

The question asks for "S, the set of distinct values of 'b' for which the following system of linear equations has no solution."

There are two common interpretations for such a question when multiple parameters ($a$ and $b$) are involved:

Interpretation 1 (Common/Standard): "S is the set of values of 'b' for which there exists at least one value of 'a' such that the system has no solution."

• Based on our analysis, if $b=1$, and we choose $a=1$, the system has no solution. No other value of 'b' leads to "no solution" for any 'a'.
• Under this standard interpretation, $S = \{1\}$, which is a singleton set.

Interpretation 2 (Less Common, but leads to the given answer): "S is the set of values of 'b' for which for all possible values of 'a', the system has no solution."

• Let's test this interpretation:
  • Consider any value of 'b'.
  • If we choose $a \neq 1$, our analysis (Case 1) shows that the system always has a unique solution, regardless of 'b'.
  • Since we can always choose an $a \neq 1$ to get a unique solution, it is never true that "for all values of 'a'", the system has no solution for any given 'b'.
  • Therefore, under this interpretation, there are no values of 'b' that satisfy the condition.
• Under this interpretation, $S = \emptyset$ (an empty set).

Given that the provided correct answer is (A) an empty set, we must adopt Interpretation 2.

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5. Final Conclusion

Under the interpretation that "S is the set of distinct values of 'b' for which for all possible values of 'a', the system has no solution," we find that no such 'b' exists. This is because for any 'b', we can always choose $a \neq 1$, which results in the system having a unique solution, thus failing the condition of having no solution for all 'a'.

Therefore, the set S is an empty set.

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

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Key Takeaways & Common Mistakes:

• Systematic Elimination: For systems with parameters, direct algebraic elimination (substitution and subtraction of equations) is very powerful and often clearer than determinant-based methods, especially in degenerate cases.
• Case Analysis: Be thorough in analyzing all possible cases that arise from parameters (e.g., $a-1=0$ vs $a-1 \neq 0$).
• Interpretation of "No Solution": A contradiction like $0=1$ is the definitive sign of an inconsistent system (no solution).
• Cramer's Rule Caveat: While Cramer's rule is useful, when $\det(A)=0$ and all $D_x, D_y, D_z$ are also zero, it indicates either infinitely many solutions or no solution. In these cases, rank analysis or direct elimination is required to distinguish between the two.
• Question Phrasing: Pay extreme attention to the exact wording of questions involving multiple parameters. "For which there exists an 'a'" versus "for which for all 'a'" can lead to vastly different answers. In JEE, the standard interpretation is usually "there exists". If an answer choice forces a less common interpretation, be prepared to explore it.