1. Key Concepts and Formulas

• System of Linear Equations: A system of linear equations $AX = B$ (where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix) can have a unique solution, no solution, or infinitely many solutions.
• Cramer's Rule Conditions:
  • Unique Solution: The system has a unique solution if and only if the determinant of the coefficient matrix, $\Delta = \det(A)$, is non-zero ($\Delta \neq 0$).
  • No Solution (Inconsistent System): The system has no solution if and only if $\Delta = \det(A) = 0$ AND at least one of the determinants $\Delta_x, \Delta_y, \Delta_z$ is non-zero. ($\Delta_x$ is the determinant of the matrix formed by replacing the x-coefficient column of $A$ with the constant matrix $B$, and similarly for $\Delta_y$ and $\Delta_z$).
  • Infinitely Many Solutions (Consistent System): The system has infinitely many solutions if and only if $\Delta = \det(A) = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$.
• Absolute Value Property: For any real number $k$, if $|k| = a$ (where $a \ge 0$), then $k = a$ or $k = -a$. Both possibilities must be considered.

2. Step-by-Step Solution

Step 1: Write down the system of equations and identify the coefficient matrix and constant matrix.

The given system of linear equations is: $2x + 3y - z = - 2 \quad (E_1)$ $x + y + z = 4 \quad (E_2)$ $x - y + |\lambda |z = 4\lambda - 4 \quad (E_3)$

From these equations, we can form the coefficient matrix $A$ and the constant matrix $B$: $A = \begin{pmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & |\lambda| \end{pmatrix}, \quad B = \begin{pmatrix} -2 \\ 4 \\ 4\lambda - 4 \end{pmatrix}$

Step 2: Calculate the determinant of the coefficient matrix, $\Delta$, and set it to zero.

For the system to have no solution or infinitely many solutions, the determinant of the coefficient matrix must be zero ($\Delta = 0$). This is the initial necessary condition to investigate these two cases.

Let's calculate $\Delta$: $\Delta = \left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & |\lambda| \end{matrix} \right|$

Expanding the determinant along the first row: $\Delta = 2(1 \cdot |\lambda| - 1 \cdot (-1)) - 3(1 \cdot |\lambda| - 1 \cdot 1) + (-1)(1 \cdot (-1) - 1 \cdot 1)$ $\Delta = 2(|\lambda| + 1) - 3(|\lambda| - 1) - 1(-1 - 1)$ $\Delta = 2|\lambda| + 2 - 3|\lambda| + 3 + 2$ $\Delta = -|\lambda| + 7$

Now, we set $\Delta = 0$ to find the values of $\lambda$ for which the system does not have a unique solution: $-|\lambda| + 7 = 0$ $|\lambda| = 7$

This equation implies two possible values for $\lambda$: $\lambda = 7$ or $\lambda = -7$. We must analyze each of these values further using Cramer's Rule conditions to determine which one leads to "no solution".

Step 3: Analyze the case when $\lambda = 7$.

Substitute $\lambda = 7$ into the original system of equations. Note that $|\lambda| = |7| = 7$. The equations become: $2x + 3y - z = - 2 \quad (E_1)$ $x + y + z = 4 \quad (E_2)$ $x - y + 7z = 4(7) - 4 \Rightarrow x - y + 7z = 24 \quad (E_3')$

Since we have $\Delta = 0$, we need to calculate $\Delta_x, \Delta_y, \Delta_z$. If all of them are zero, there are infinitely many solutions. If at least one is non-zero, there is no solution.

Let's calculate $\Delta_x$: $\Delta_x = \left| \begin{matrix} -2 & 3 & -1 \\ 4 & 1 & 1 \\ 24 & -1 & 7 \end{matrix} \right|$ Expanding along the first row: $= -2(1 \cdot 7 - 1 \cdot (-1)) - 3(4 \cdot 7 - 1 \cdot 24) + (-1)(4 \cdot (-1) - 1 \cdot 24)$ $= -2(7+1) - 3(28-24) - 1(-4-24)$ $= -2(8) - 3(4) - 1(-28)$ $= -16 - 12 + 28 = 0$

Since $\Delta = 0$ and $\Delta_x = 0$, we must also check $\Delta_y$ and $\Delta_z$. Let's calculate $\Delta_y$: $\Delta_y = \left| \begin{matrix} 2 & -2 & -1 \\ 1 & 4 & 1 \\ 1 & 24 & 7 \end{matrix} \right|$ $= 2(4 \cdot 7 - 1 \cdot 24) - (-2)(1 \cdot 7 - 1 \cdot 1) + (-1)(1 \cdot 24 - 4 \cdot 1)$ $= 2(28 - 24) + 2(7 - 1) - 1(24 - 4)$ $= 2(4) + 2(6) - 1(20)$ $= 8 + 12 - 20 = 0$

Let's calculate $\Delta_z$: $\Delta_z = \left| \begin{matrix} 2 & 3 & -2 \\ 1 & 1 & 4 \\ 1 & -1 & 24 \end{matrix} \right|$ $= 2(1 \cdot 24 - 4 \cdot (-1)) - 3(1 \cdot 24 - 4 \cdot 1) + (-2)(1 \cdot (-1) - 1 \cdot 1)$ $= 2(24 + 4) - 3(24 - 4) - 2(-1 - 1)$ $= 2(28) - 3(20) - 2(-2)$ $= 56 - 60 + 4 = 0$

Since $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$, the system of equations for $\lambda=7$ has infinitely many solutions.

Step 4: Analyze the case when $\lambda = -7$.

Substitute $\lambda = -7$ into the original system of equations. Note that $|\lambda| = |-7| = 7$. The equations become: $2x + 3y - z = - 2 \quad (E_1)$ $x + y + z = 4 \quad (E_2)$ $x - y + 7z = 4(-7) - 4 \Rightarrow x - y + 7z = -32 \quad (E_3'')$

Again, we have $\Delta = 0$ for this case (since $|\lambda|=7$). We need to check for consistency by calculating $\Delta_x, \Delta_y, \Delta_z$.

Let's calculate $\Delta_x$: $\Delta_x = \left| \begin{matrix} -2 & 3 & -1 \\ 4 & 1 & 1 \\ -32 & -1 & 7 \end{matrix} \right|$ Expanding along the first row: $= -2(1 \cdot 7 - 1 \cdot (-1)) - 3(4 \cdot 7 - 1 \cdot (-32)) + (-1)(4 \cdot (-1) - 1 \cdot (-32))$ $= -2(7+1) - 3(28+32) - 1(-4+32)$ $= -2(8) - 3(60) - 1(28)$ $= -16 - 180 - 28 = -224$

Since $\Delta = 0$ and $\Delta_x = -224 \neq 0$, the system of equations has no solution when $\lambda = -7$.

3. Common Mistakes & Tips

• Don't stop at $\Delta = 0$! This is the most common mistake. $\Delta = 0$ only tells you that there is not a unique solution. You must further check the consistency using $\Delta_x, \Delta_y, \Delta_z$ to distinguish between "no solution" and "infinitely many solutions".
• Be careful with absolute values: Remember that $|\lambda| = 7$ implies $\lambda = 7$ or $\lambda = -7$. Both cases must be checked because the value of $\lambda$ itself, not just $|\lambda|$, affects the constant term $4\lambda-4$.
• Linear Combination Method: When $\Delta = 0$, an alternative to calculating all $\Delta_x, \Delta_y, \Delta_z$ is to check for linear dependency of the equations. If one equation (say, $E_k$) is a linear combination of others ($E_k = a_1 E_1 + a_2 E_2 + \dots$), check if the constant term of $E_k$ also matches the linear combination of the other constant terms ($C_k = a_1 C_1 + a_2 C_2 + \dots$). If the constant terms match, it's infinitely many solutions. If they don't match, it's no solution. This method confirms our results:
  • For $\lambda=7$, $E_3' = x-y+7z=24$. We find $E_3' = -2E_1 + 5E_2$ for coefficients AND constant terms ($24 = -2(-2) + 5(4) = 24$). Thus, infinitely many solutions.
  • For $\lambda=-7$, $E_3'' = x-y+7z=-32$. We find $E_3'' = -2E_1 + 5E_2$ for coefficients, but NOT constant terms ($-32 \neq -2(-2) + 5(4) = 24$). Thus, no solution.

4. Summary

We first calculated the determinant of the coefficient matrix, $\Delta$, and found that $\Delta = 0$ when $|\lambda|=7$, which means $\lambda=7$ or $\lambda=-7$. We then analyzed each case separately. For $\lambda=7$, all determinants $\Delta_x, \Delta_y, \Delta_z$ were found to be zero, indicating infinitely many solutions. For $\lambda=-7$, while $\Delta=0$, the determinant $\Delta_x$ was non-zero, indicating no solution. Therefore, the system of linear equations has no solution only when $\lambda = -7$.

The final answer is $\boxed{\lambda = -7}$, which corresponds to option (B).