Introduction: Understanding Systems of Linear Equations

When faced with a system of linear equations, our primary goal is to determine whether it has a unique solution, no solution, or infinitely many solutions. This classification largely depends on the coefficients of the variables and the constant terms. For a system of $n$ linear equations in $n$ variables, a powerful tool to analyze the nature of solutions is Cramer's Rule, which relies on determinants.

Key Concept: Cramer's Rule for Analyzing Solutions

Consider a system of $n$ linear equations in $n$ variables, which can be represented in matrix form as $AX=B$, where $A$ is the coefficient matrix, $X$ is the column matrix of variables, and $B$ is the column matrix of constants.

Let's define the following determinants:

• $\Delta = \det(A)$: The determinant of the coefficient matrix.
• $\Delta_j$: The determinant of the matrix obtained by replacing the $j$-th column of $A$ with the constant matrix $B$. For a system with variables $x, y, z$, these are often denoted as $\Delta_x, \Delta_y, \Delta_z$.

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

1. Unique Solution: The system has a unique solution if and only if $\Delta \neq 0$. In this case, the solution is given by $x = \frac{\Delta_x}{\Delta}$, $y = \frac{\Delta_y}{\Delta}$, $z = \frac{\Delta_z}{\Delta}$.
2. No Solution (Inconsistent System): The system has no solution if $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
3. Infinitely Many Solutions (Consistent and Dependent System): The system has infinitely many solutions if $\Delta = 0$ AND all of $\Delta_x, \Delta_y, \Delta_z$ are zero.

Step-by-Step Solution

We are given the following system of linear equations: $x+y+z=6 \quad \ldots(1)$ $x+2 y+\alpha z=10 \quad \ldots(2)$ $x+3 y+5 z=\beta \quad \ldots(3)$

Step 1: Formulate the Coefficient Matrix ($A$) and Constant Matrix ($B$)

To apply Cramer's Rule, we first write the system in its matrix form $AX=B$. This helps in systematically extracting the coefficients and constants. $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 6 \\ 10 \\ \beta \end{pmatrix}$

Step 2: Calculate the Determinant of the Coefficient Matrix ($\Delta$)

The first and most critical step is to calculate $\Delta = \det(A)$. The value of $\Delta$ immediately tells us whether a unique solution is possible or if we need further analysis.

$\Delta = \left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{array}\right|$ We expand the determinant along the first row for simplicity: $\Delta = 1 \cdot \left|\begin{array}{ll} 2 & \alpha \\ 3 & 5 \end{array}\right| - 1 \cdot \left|\begin{array}{ll} 1 & \alpha \\ 1 & 5 \end{array}\right| + 1 \cdot \left|\begin{array}{ll} 1 & 2 \\ 1 & 3 \end{array}\right|$ $\Delta = 1(2 \times 5 - \alpha \times 3) - 1(1 \times 5 - \alpha \times 1) + 1(1 \times 3 - 2 \times 1)$ $\Delta = (10 - 3\alpha) - (5 - \alpha) + (3 - 2)$ $\Delta = 10 - 3\alpha - 5 + \alpha + 1$ $\Delta = 6 - 2\alpha$

Step 3: Analyze the Case for a Unique Solution ($\Delta \neq 0$)

According to Cramer's Rule, a unique solution exists if and only if $\Delta \neq 0$. This condition sets the primary criterion for the system's behavior.

$\Delta \neq 0 \implies 6 - 2\alpha \neq 0$ $\implies 2\alpha \neq 6$ $\implies \alpha \neq 3$ Thus, if $\alpha \neq 3$, the system will always have a unique solution, irrespective of the value of $\beta$.

• Evaluating Option (D): "System has a unique solution for $\alpha=-3, \beta=14$." Since $\alpha=-3$ is not equal to $3$, $\Delta \neq 0$. Therefore, a unique solution exists, and the value of $\beta=14$ is irrelevant for the existence of a unique solution (though it would affect the actual solution values). So, statement (D) is TRUE.
• Evaluating Option (A): "System has a unique solution for $\alpha=3,\beta\ne14$." This statement claims a unique solution when $\alpha=3$. However, our analysis shows that for $\alpha=3$, $\Delta=0$, which implies there cannot be a unique solution. Therefore, statement (A) is FALSE. This is likely our answer.

Step 4: Analyze the Case where $\Delta = 0$ (No Solution or Infinitely Many Solutions)

If $\Delta = 0$, the system does not have a unique solution. This occurs when: $\Delta = 0 \implies 6 - 2\alpha = 0$ $\implies 2\alpha = 6$ $\implies \alpha = 3$ When $\alpha=3$, we must further investigate by calculating $\Delta_x, \Delta_y, \Delta_z$ to distinguish between no solution and infinitely many solutions.

Step 5: Calculate $\Delta_x, \Delta_y, \Delta_z$ for $\alpha=3$

Substitute $\alpha=3$ into the original coefficient matrix $A$ and then calculate the determinants $\Delta_x, \Delta_y, \Delta_z$ by replacing the respective columns with the constant matrix $B$.

• Calculate $\Delta_x$ (or $\Delta_1$): Replace the first column of $A$ with $B$. $\Delta_x = \left|\begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ \beta & 3 & 5 \end{array}\right|$ Expand along the first row: $\Delta_x = 6(2 \times 5 - 3 \times 3) - 1(10 \times 5 - 3 \times \beta) + 1(10 \times 3 - 2 \times \beta)$ $\Delta_x = 6(10 - 9) - 1(50 - 3\beta) + 1(30 - 2\beta)$ $\Delta_x = 6(1) - 50 + 3\beta + 30 - 2\beta$ $\Delta_x = 6 - 50 + 30 + \beta$ $\Delta_x = \beta - 14$

• Calculate $\Delta_y$ (or $\Delta_2$): Replace the second column of $A$ with $B$. $\Delta_y = \left|\begin{array}{ccc} 1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \beta & 5 \end{array}\right|$ Expand along the first row: $\Delta_y = 1(10 \times 5 - 3 \times \beta) - 6(1 \times 5 - 3 \times 1) + 1(1 \times \beta - 10 \times 1)$ $\Delta_y = 1(50 - 3\beta) - 6(5 - 3) + 1(\beta - 10)$ $\Delta_y = 50 - 3\beta - 6(2) + \beta - 10$ $\Delta_y = 50 - 3\beta - 12 + \beta - 10$ $\Delta_y = 28 - 2\beta = 2(14 - \beta)$

• Calculate $\Delta_z$ (or $\Delta_3$): Replace the third column of $A$ with $B$. $\Delta_z = \left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 3 & \beta \end{array}\right|$ Expand along the first row: $\Delta_z = 1(2 \times \beta - 10 \times 3) - 1(1 \times \beta - 10 \times 1) + 6(1 \times 3 - 2 \times 1)$ $\Delta_z = 1(2\beta - 30) - 1(\beta - 10) + 6(3 - 2)$ $\Delta_z = 2\beta - 30 - \beta + 10 + 6(1)$ $\Delta_z = \beta - 14$

So, for $\alpha=3$, we have:

• $\Delta_x = \beta - 14$
• $\Delta_y = 2(14 - \beta)$
• $\Delta_z = \beta - 14$

Step 6: Determine Conditions for No Solution and Infinitely Many Solutions (when $\alpha=3$)

Now we use the calculated $\Delta_x, \Delta_y, \Delta_z$ to establish conditions for $\beta$ when $\Delta=0$ (i.e., $\alpha=3$).

• Subcase 6a: Infinitely Many Solutions This occurs when $\Delta = 0$ AND $\Delta_x = 0$ AND $\Delta_y = 0$ AND $\Delta_z = 0$. Given $\alpha=3$ (which makes $\Delta=0$):

  • $\Delta_x = 0 \implies \beta - 14 = 0 \implies \beta = 14$

  • $\Delta_y = 0 \implies 2(14 - \beta) = 0 \implies \beta = 14$

  • $\Delta_z = 0 \implies \beta - 14 = 0 \implies \beta = 14$ All conditions are satisfied when $\beta = 14$. Therefore, for $\alpha=3$ and $\beta=14$, the system has infinitely many solutions.

  • Evaluating Option (B): "System has infinitely many solutions for $\alpha=3, \beta=14$." This matches our finding. So, statement (B) is TRUE.

• Subcase 6b: No Solution This occurs when $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero. Given $\alpha=3$ (which makes $\Delta=0$): If $\beta \neq 14$:

  • $\Delta_x = \beta - 14 \neq 0$

  • $\Delta_y = 2(14 - \beta) \neq 0$

  • $\Delta_z = \beta - 14 \neq 0$ Since all of $\Delta_x, \Delta_y, \Delta_z$ are non-zero when $\beta \neq 14$, the system has no solution. Therefore, for $\alpha=3$ and $\beta \neq 14$, the system has no solution.

  • Evaluating Option (C): "System has no solution for $\alpha=3, \beta=24$." Here, $\alpha=3$ and $\beta=24$, which is indeed $\neq 14$. This matches our finding. So, statement (C) is TRUE.

Step 7: Final Evaluation of Options

Let's consolidate our findings for each option:

• (A) System has a unique solution for $\alpha=3,\beta\ne14$.
  • Our analysis: For $\alpha=3$, $\Delta=0$. A unique solution requires $\Delta \neq 0$. Therefore, this statement is FALSE.
• (B) System has infinitely many solutions for $\alpha=3, \beta=14$.
  • Our analysis: For $\alpha=3$ and $\beta=14$, $\Delta=0$ and all $\Delta_x, \Delta_y, \Delta_z$ are $0$. Therefore, this statement is TRUE.
• (C) System has no solution for $\alpha=3, \beta=24$.
  • Our analysis: For $\alpha=3$ and $\beta=24$, $\Delta=0$ but $\Delta_x = 24-14=10 \neq 0$. Therefore, this statement is TRUE.
• (D) System has a unique solution for $\alpha=-3, \beta=14$.
  • Our analysis: For $\alpha=-3$, $\Delta = 6 - 2(-3) = 12 \neq 0$. A unique solution exists regardless of $\beta$. Therefore, this statement is TRUE.

The question asks which one of the statements is NOT true. Based on our evaluation, statement (A) is the one that is NOT true.

Tips for Success / Common Pitfalls

• Careful Determinant Calculation: Errors in calculating $\Delta$ or any $\Delta_j$ can lead to incorrect conclusions. Double-check your arithmetic, especially with signs.
• Understanding the Conditions: Memorize or clearly understand the three conditions for unique, no, and infinitely many solutions. A common mistake is to confuse the conditions for no solution and infinitely many solutions when $\Delta=0$.
• Systematic Approach: Break down the problem into cases based on $\Delta \neq 0$ and $\Delta = 0$. This ensures all possibilities are covered.
• Role of $\beta$: Remember that when $\Delta \neq 0$, the value of the constant term $\beta$ does not affect whether a unique solution exists; it only affects the values of $x, y, z$. $\beta$ becomes crucial only when $\Delta = 0$.

Summary and Key Takeaway

This problem demonstrates the complete application of Cramer's Rule to analyze the nature of solutions for a system of linear equations with parameters.

• First, calculate $\Delta$. If $\Delta \neq 0$, there's a unique solution (regardless of $\beta$).
• If $\Delta = 0$, then calculate $\Delta_x, \Delta_y, \Delta_z$.
  • If all of them are also zero, there are infinitely many solutions.
  • If at least one of them is non-zero, there is no solution.

By systematically applying these rules, we found that statement (A) incorrectly claims a unique solution when $\Delta=0$, making it the untrue statement.

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