Key Concepts and Formulas

• System of Linear Equations: 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.
• Conditions for Consistency and Inconsistency:
  • Unique Solution: If $\det(A) \neq 0$, the system has a unique solution.
  • No Solution (Inconsistent): If $\det(A) = 0$ AND at least one of the determinants $D_x, D_y, D_z, \dots$ (obtained by replacing a column of $A$ with $B$, as per Cramer's Rule) is non-zero, the system is inconsistent. This means there is no set of values for the variables that satisfies all equations simultaneously.
  • Infinitely Many Solutions: If $\det(A) = 0$ AND all $D_x, D_y, D_z, \dots$ are zero, the system has infinitely many solutions.
• Identifying Inconsistency: When $\det(A)=0$, substituting the value of the parameter back into the original system and observing a contradiction (e.g., $0 = \text{non-zero constant}$) is a direct way to confirm inconsistency.

Step-by-Step Solution

Step 1: Formulate the coefficient matrix and apply the condition for non-unique solutions. The given system of linear equations is: $\lambda x+y+z=1$ $x+\lambda y+z=1$ $x+y+\lambda z=1$ We can represent this system in matrix form $AX=B$, where the coefficient matrix $A$ is: $A = \begin{pmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{pmatrix}$ For the system to be either inconsistent (no solution) or to have infinitely many solutions, the determinant of the coefficient matrix must be zero. This is a necessary condition for the system to not have a unique solution. So, we set $\det(A) = 0$: $\det(A) = \left| \begin{array}{ccc} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{array} \right| = 0$

Step 2: Expand the determinant and solve for $\lambda$. We expand the determinant of $A$ along the first row: $\lambda \left| \begin{array}{cc} \lambda & 1 \\ 1 & \lambda \end{array} \right| - 1 \left| \begin{array}{cc} 1 & 1 \\ 1 & \lambda \end{array} \right| + 1 \left| \begin{array}{cc} 1 & \lambda \\ 1 & 1 \end{array} \right| = 0$ $\lambda(\lambda^2 - 1) - 1(\lambda - 1) + 1(1 - \lambda) = 0$ $\lambda(\lambda-1)(\lambda+1) - (\lambda-1) - (\lambda-1) = 0$ Factor out $(\lambda-1)$: $(\lambda-1)[\lambda(\lambda+1) - 1 - 1] = 0$ $(\lambda-1)[\lambda^2 + \lambda - 2] = 0$ Now, factor the quadratic term $\lambda^2 + \lambda - 2$: $(\lambda-1)(\lambda+2)(\lambda-1) = 0$ This gives us the values of $\lambda$ for which $\det(A)=0$: $\lambda = 1, 1, -2$ So, the potential values of $\lambda$ are $1$ and $-2$.

Step 3: Analyze each value of $\lambda$ to determine inconsistency. We must check each value of $\lambda$ found in Step 2 to see if it leads to an inconsistent system (no solution) or a system with infinitely many solutions.

• Case 1: $\lambda = 1$ Substitute $\lambda=1$ into the original system of equations: $1x+y+z=1 \quad \Rightarrow \quad x+y+z=1$ $x+1y+z=1 \quad \Rightarrow \quad x+y+z=1$ $x+y+1z=1 \quad \Rightarrow \quad x+y+z=1$ All three equations reduce to the single equation $x+y+z=1$. This is a single linear equation with three variables, which has infinitely many solutions (e.g., we can choose any values for $x$ and $y$, and $z$ will be determined as $1-x-y$). Therefore, for $\lambda=1$, the system has infinitely many solutions, and thus it is not inconsistent.

• Case 2: $\lambda = -2$ Substitute $\lambda=-2$ into the original system of equations: $-2x+y+z=1 \quad \text{(Equation 1)}$ $x-2y+z=1 \quad \text{(Equation 2)}$ $x+y-2z=1 \quad \text{(Equation 3)}$ To check for inconsistency, we can try to find a contradiction by manipulating these equations. Let's add all three equations together: $(-2x+y+z) + (x-2y+z) + (x+y-2z) = 1+1+1$ Combine like terms: $(-2+1+1)x + (1-2+1)y + (1+1-2)z = 3$ $0x + 0y + 0z = 3$ $0 = 3$ This is a false statement, a clear contradiction. This means there are no values of $x, y, z$ that can satisfy all three equations simultaneously. Therefore, for $\lambda=-2$, the system has no solution, and thus it is inconsistent.

From this analysis, the set $S$ of all real values of $\lambda$ for which the system is inconsistent is $S = \{-2\}$.

Step 4: Calculate the required sum. We need to calculate $\sum_{\lambda \in S}\left(|\lambda|^{2}+|\lambda|\right)$. Since $S = \{-2\}$, we substitute $\lambda = -2$ into the expression: $|\lambda|^2 + |\lambda| = |-2|^2 + |-2|$ $= (2)^2 + 2$ $= 4 + 2$ $= 6$

Common Mistakes & Tips

• Don't stop at $\det(A)=0$: A common error is to assume that any $\lambda$ for which $\det(A)=0$ automatically implies an inconsistent system. As demonstrated in Case 1, $\det(A)=0$ can also lead to infinitely many solutions. Always perform the crucial second step of checking the system for each such $\lambda$.
• Methods for checking inconsistency: When $\det(A)=0$, you can check for inconsistency by:
  • Direct substitution and algebraic manipulation: Try to combine equations to find a contradiction (e.g., $0 = k$ where $k \neq 0$).
  • Cramer's Rule extension: If $\det(A)=0$ and at least one of $D_x, D_y, D_z$ (determinants of matrices where a column of $A$ is replaced by $B$) is non-zero, the system is inconsistent.
  • Rank method: If $rank(A) \neq rank([A|B])$, the system is inconsistent.
• Factorizing cubic polynomials: For equations like $\lambda^3 - 3\lambda + 2 = 0$, try testing small integer divisors of the constant term (e.g., $\pm 1, \pm 2$) to find initial roots. Once a root $r$ is found, $(\lambda-r)$ is a factor, and polynomial division can help find the remaining quadratic factor.

Summary

To find the values of $\lambda$ for which the system of equations is inconsistent, we first set the determinant of the coefficient matrix to zero and solve for $\lambda$. This gives potential values for inconsistency or infinitely many solutions. Then, we substitute each of these $\lambda$ values back into the original system. For $\lambda=1$, the system reduces to a single equation, implying infinitely many solutions. For $\lambda=-2$, adding the equations leads to a contradiction ($0=3$), indicating no solution (inconsistency). Thus, $S = \{-2\}$. Finally, we calculate the required sum $|\lambda|^2 + |\lambda|$ for $\lambda \in S$, which yields $|-2|^2 + |-2| = 4+2=6$.

The final answer is $\boxed{6}$, which corresponds to option (A).