1. Key Concepts and Formulas

• System of Linear Equations: A system of $n$ linear equations with $n$ 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 Infinitely Many Solutions (Cramer's Rule Extension): For a system of three linear equations in three variables to have infinitely many solutions, the following conditions must be met:
  1. The determinant of the coefficient matrix, denoted as $\Delta$ (or $det(A)$), must be zero ($\Delta = 0$).
  2. All the determinants obtained by replacing a column of the coefficient matrix with the constant terms ($\Delta_x, \Delta_y, \Delta_z$) must also be zero ($\Delta_x = 0, \Delta_y = 0, \Delta_z = 0$).
• Determinant Expansion: For a $3 \times 3$ matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, its determinant is $a(ei - fh) - b(di - fg) + c(dh - eg)$.

2. Step-by-Step Solution

Step 1: Represent the System in Matrix Form First, we express the given system of linear equations in the standard matrix form $AX=B$.

The given system is:

1. $x + y + z = 6$
2. $4x + \lambda y - \lambda z = \lambda - 2$
3. $3x + 2y - 4z = - 5$

From this, we identify the coefficient matrix $A$ and the constant matrix $B$:

$$A = \begin{pmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 6 \\ \lambda - 2 \\ -5 \end{pmatrix}$$

Step 2: Calculate $\Delta$ and Find Potential Values of $\lambda$ For the system to have infinitely many solutions, the determinant of the coefficient matrix, $\Delta$, must be zero. We calculate $\Delta$ and set it to zero to find the possible values of $\lambda$.

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{vmatrix}$$

We expand the determinant along the first row ($R_1$):

$$\Delta = 1 \cdot (\lambda(-4) - (-\lambda)(2)) - 1 \cdot (4(-4) - (-\lambda)(3)) + 1 \cdot (4(2) - \lambda(3))$$ $$\Delta = 1 \cdot (-4\lambda + 2\lambda) - 1 \cdot (-16 + 3\lambda) + 1 \cdot (8 - 3\lambda)$$ $$\Delta = -2\lambda + 16 - 3\lambda + 8 - 3\lambda$$

Combine like terms:

$$\Delta = (-2 - 3 - 3)\lambda + (16 + 8)$$ $$\Delta = -8\lambda + 24$$

Now, set $\Delta = 0$ to find the value(s) of $\lambda$:

$$-8\lambda + 24 = 0 \\ -8\lambda = -24 \\ \lambda = 3$$

This means $\lambda = 3$ is the only value for which the system could potentially have infinitely many solutions or no solution.

Step 3: Verify Conditions for Infinitely Many Solutions (Check $\Delta_x, \Delta_y, \Delta_z$ for $\lambda=3$) For infinitely many solutions, not only must $\Delta=0$, but also $\Delta_x=0$, $\Delta_y=0$, and $\Delta_z=0$. We substitute $\lambda=3$ into the expressions for $\Delta_x, \Delta_y, \Delta_z$ and verify they are all zero.

• Calculate $\Delta_x$ for $\lambda=3$: Replace the first column of $A$ with $B$:

  $$\Delta_x = \begin{vmatrix} 6 & 1 & 1 \\ \lambda - 2 & \lambda & -\lambda \\ -5 & 2 & -4 \end{vmatrix}$$

  Substitute $\lambda = 3$:

  $$\Delta_x = \begin{vmatrix} 6 & 1 & 1 \\ 3 - 2 & 3 & -3 \\ -5 & 2 & -4 \end{vmatrix} = \begin{vmatrix} 6 & 1 & 1 \\ 1 & 3 & -3 \\ -5 & 2 & -4 \end{vmatrix}$$

  Expand $\Delta_x$ along the first row:

  $$\Delta_x = 6(3(-4) - (-3)(2)) - 1(1(-4) - (-3)(-5)) + 1(1(2) - 3(-5)) \\ \Delta_x = 6(-12 + 6) - 1(-4 - 15) + 1(2 + 15) \\ \Delta_x = 6(-6) - 1(-19) + 1(17) \\ \Delta_x = -36 + 19 + 17 \\ \Delta_x = -36 + 36 = 0$$

  This condition is met.

• Calculate $\Delta_y$ for $\lambda=3$: Replace the second column of $A$ with $B$:

  $$\Delta_y = \begin{vmatrix} 1 & 6 & 1 \\ 4 & \lambda - 2 & -\lambda \\ 3 & -5 & -4 \end{vmatrix}$$

  Substitute $\lambda = 3$:

  $$\Delta_y = \begin{vmatrix} 1 & 6 & 1 \\ 4 & 3 - 2 & -3 \\ 3 & -5 & -4 \end{vmatrix} = \begin{vmatrix} 1 & 6 & 1 \\ 4 & 1 & -3 \\ 3 & -5 & -4 \end{vmatrix}$$

  Expand $\Delta_y$ along the first row:

  $$\Delta_y = 1(1(-4) - (-3)(-5)) - 6(4(-4) - (-3)(3)) + 1(4(-5) - 1(3)) \\ \Delta_y = 1(-4 - 15) - 6(-16 + 9) + 1(-20 - 3) \\ \Delta_y = -19 - 6(-7) + (-23) \\ \Delta_y = -19 + 42 - 23 \\ \Delta_y = 42 - 42 = 0$$

  This condition is also met.

• Calculate $\Delta_z$ for $\lambda=3$ (Optional, but good for completeness): Replace the third column of $A$ with $B$:

  $$\Delta_z = \begin{vmatrix} 1 & 1 & 6 \\ 4 & \lambda & \lambda - 2 \\ 3 & 2 & -5 \end{vmatrix}$$

  Substitute $\lambda = 3$:

  $$\Delta_z = \begin{vmatrix} 1 & 1 & 6 \\ 4 & 3 & 3 - 2 \\ 3 & 2 & -5 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 6 \\ 4 & 3 & 1 \\ 3 & 2 & -5 \end{vmatrix}$$

  Expand $\Delta_z$ along the first row:

  $$\Delta_z = 1(3(-5) - 1(2)) - 1(4(-5) - 1(3)) + 6(4(2) - 3(3)) \\ \Delta_z = 1(-15 - 2) - 1(-20 - 3) + 6(8 - 9) \\ \Delta_z = -17 - 1(-23) + 6(-1) \\ \Delta_z = -17 + 23 - 6 \\ \Delta_z = 6 - 6 = 0$$

  All conditions ($\Delta = 0, \Delta_x = 0, \Delta_y = 0, \Delta_z = 0$) are met for $\lambda = 3$. Therefore, the system has infinitely many solutions when $\lambda = 3$.

Step 4: Identify the Quadratic Equation Now, we need to find which of the given quadratic equations has $\lambda = 3$ as a root. We substitute $\lambda=3$ into each option:

• (A) $\lambda^2 + \lambda - 6 = 0$ $(3)^2 + (3) - 6 = 9 + 3 - 6 = 12 - 6 = 6 \neq 0$. So, (A) is not the answer.

• (B) $\lambda^2 - \lambda - 6 = 0$ $(3)^2 - (3) - 6 = 9 - 3 - 6 = 6 - 6 = 0$. This equation is satisfied by $\lambda=3$.

• (C) $\lambda^2 - 3\lambda - 4 = 0$ $(3)^2 - 3(3) - 4 = 9 - 9 - 4 = -4 \neq 0$. So, (C) is not the answer.

• (D) $\lambda^2 + 3\lambda - 4 = 0$ $(3)^2 + 3(3) - 4 = 9 + 9 - 4 = 18 - 4 = 14 \neq 0$. So, (D) is not the answer.

Thus, $\lambda=3$ is a root of the quadratic equation $\lambda^2 - \lambda - 6 = 0$.

3. Common Mistakes & Tips

• Determinant Calculation Errors: Carefully expand determinants, paying close attention to signs, especially when terms involve negative $\lambda$ or subtraction of negative products. A small arithmetic error can lead to an incorrect value of $\lambda$.
• Incomplete Conditions: Remember that $\Delta=0$ alone is not sufficient for infinitely many solutions; it could also lead to no solution. All $\Delta_x, \Delta_y, \Delta_z$ must also be zero. If $\Delta=0$ but any $\Delta_i \neq 0$, the system has no solution.
• Alternative Method (Gaussian Elimination): For verification or if determinants are complex, converting the system into an augmented matrix and performing row operations (Gaussian elimination) can also determine the nature of solutions. For infinitely many solutions, this method will result in at least one row of zeros (e.g., $0 \ 0 \ 0 \ | \ 0$).

4. Summary

To find the value of $\lambda$ for which the system of linear equations has infinitely many solutions, we first calculated the determinant of the coefficient matrix ($\Delta$) and set it to zero. This yielded $\lambda=3$. Next, we verified that for $\lambda=3$, the determinants $\Delta_x, \Delta_y, \Delta_z$ were also all zero, confirming that $\lambda=3$ leads to infinitely many solutions. Finally, we checked which of the given quadratic equations has $\lambda=3$ as a root, finding that $\lambda^2 - \lambda - 6 = 0$ is the correct equation.

The final answer is $\boxed{\text{(B)} \lambda^2 - \lambda - 6 = 0}$.