Key Concept: Conditions for a System of Linear Equations to Have More Than One Solution

For a system of linear equations $AX = B$ where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, it has more than one solution (i.e., infinitely many solutions) if and only if two conditions are met:

1. The determinant of the coefficient matrix, $\det(A)$, is zero.
2. The system is consistent. This implies that the equations are linearly dependent, and the same linear dependency must hold for the constants on the right-hand side. In other words, if one equation can be expressed as a linear combination of the others, then the corresponding constant must also be the same linear combination of the other constants.

Step-by-step Solution:

1. Identify the Coefficient Matrix and Calculate its Determinant The given system of linear equations is: Equation 1 (P1): $2x + 2y + 3z = a$ Equation 2 (P2): $3x - y + 5z = b$ Equation 3 (P3): $x - 3y + 2z = c$

First, we form the coefficient matrix $A$: $A = \begin{pmatrix} 2 & 2 & 3 \\ 3 & -1 & 5 \\ 1 & -3 & 2 \end{pmatrix}$

Next, we calculate the determinant of $A$, $\det(A)$. This is crucial because if $\det(A) \neq 0$, there would be a unique solution, contradicting the problem statement. $\det(A) = 2 \begin{vmatrix} -1 & 5 \\ -3 & 2 \end{vmatrix} - 2 \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} + 3 \begin{vmatrix} 3 & -1 \\ 1 & -3 \end{vmatrix}$ $= 2((-1)(2) - (5)(-3)) - 2((3)(2) - (5)(1)) + 3((3)(-3) - (-1)(1))$ $= 2(-2 + 15) - 2(6 - 5) + 3(-9 + 1)$ $= 2(13) - 2(1) + 3(-8)$ $= 26 - 2 - 24$ $= 0$ Since $\det(A) = 0$, the system either has no solution or infinitely many solutions. The problem states it has "more than one solution", which implies infinitely many solutions. This confirms our determinant calculation is consistent with the problem statement.

2. Find the Linear Dependency Among the Equations For the system to have infinitely many solutions, the equations must be linearly dependent. This means one equation's left-hand side (LHS) can be expressed as a linear combination of the LHS of the other equations. We need to find this specific linear relationship.

Let's try simple combinations of the equations. A common strategy is to try adding or subtracting equations to see if they produce another equation. Consider adding Equation 1 (P1) and Equation 3 (P3): LHS of (P1) + LHS of (P3) = $(2x + 2y + 3z) + (x - 3y + 2z)$ $= (2+1)x + (2-3)y + (3+2)z$ $= 3x - y + 5z$ Observe that this result is precisely the LHS of Equation 2 (P2). So, we have found the linear dependency: LHS(P1) + LHS(P3) = LHS(P2).

3. Apply the Linear Dependency to the Constants For the system to be consistent and have infinitely many solutions, the same linear relationship that holds for the left-hand sides of the equations must also hold for the right-hand sides (the constants $a, b, c$). Since LHS(P1) + LHS(P3) = LHS(P2), it must be that RHS(P1) + RHS(P3) = RHS(P2). Therefore, we must have: $a + c = b$

4. Rearrange the Condition to Match the Options Rearranging the condition $a + c = b$, we get: $b - a - c = 0$ or, $b - c - a = 0$

5. Compare with Given Options Comparing our derived condition $b - c - a = 0$ with the given options: (A) $b - c - a = 0$ (B) $a + b + c = 0$ (C) $b - c + a = 0$ (D) $b + c - a = 0$

Our condition matches option (A).

Tips and Common Mistakes:

• Always start with $\det(A)$: This is the primary condition for non-unique solutions. If $\det(A) \neq 0$, there's a unique solution, and the problem statement would be contradicted.
• Systematic Search for Linear Dependency: If $\det(A)=0$, there must be a linear dependency among the rows (or columns) of $A$. Sometimes, it's obvious by inspection (like $R_1 + R_3 = R_2$ here). If not, you can use row operations on the augmented matrix $[A|B]$ to reduce it to row echelon form. An all-zero row in the coefficient part (A) indicates dependency.
• Consistency is Key: Just $\det(A)=0$ is not enough for infinite solutions; the system must also be consistent. If the linear dependency for the LHS does not hold for the RHS (e.g., if we found LHS(P1) + LHS(P3) = LHS(P2), but $a+c \neq b$), then the system would have no solution, even with $\det(A)=0$.
• Cramer's Rule Extension: For $\det(A)=0$, if the system has infinite solutions, then $\det(A_x) = \det(A_y) = \det(A_z) = 0$ must also hold (where $A_x$ is $A$ with the first column replaced by $B$, etc.). This is a more general way to ensure consistency, but finding the direct linear relationship is often quicker for competitive exams if it's simple.

Summary:

The problem asks for the condition under which a given system of linear equations has more than one solution. This implies infinitely many solutions. For such a system, the determinant of the coefficient matrix must be zero, and the system must be consistent. We first calculated the determinant of the coefficient matrix and found it to be 0. Then, we identified a linear dependency among the left-hand sides of the equations (Equation 1 + Equation 3 = Equation 2). For consistency, this same linear dependency must hold for the right-hand side constants, leading to $a + c = b$. Rearranging this gives the final condition $b - c - a = 0$.