Here's a clear, educational, and well-structured solution to the problem, focusing on mathematical rigor and common JEE approaches.

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1. Key Concepts and Formulas

For a system of linear equations in three variables, say: $a_1x + b_1y + c_1z = d_1$ $a_2x + b_2y + c_2z = d_2$ $a_3x + b_3y + c_3z = d_3$

to have infinitely many solutions, the following conditions must be met:

• Using Determinants (Cramer's Rule extended):

  1. The determinant of the coefficient matrix ($D$) must be zero.
  2. All the determinants $D_x$, $D_y$, and $D_z$ (obtained by replacing the respective column of coefficients with the constant terms) must also be zero.
     • $D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$
     • $D_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}$, and similarly for $D_y$ and $D_z$.

• Using Rank (Gaussian Elimination): For a system $AX=B$, represented by the augmented matrix $[A|B]$, infinitely many solutions exist if and only if:

  1. The rank of the coefficient matrix ($A$) is equal to the rank of the augmented matrix ($[A|B]$).
  2. This common rank is less than the number of variables (in this case, 3). Practically, this means that when the augmented matrix is reduced to row echelon form, there must be at least one row of zeros in the coefficient part, and the corresponding constant term in that row must also be zero.

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2. Step-by-Step Solution

We are given the system of equations:

1. $x+y+z=2$
2. $2x+4y–z=6$
3. $3x+2y+\lambda z=\mu$

We will use Gaussian elimination on the augmented matrix to find the conditions on $\lambda$ and $\mu$.

Step 1: Set up the Augmented Matrix First, we write the system in its augmented matrix form $[A|B]$: $\begin{pmatrix} 1 & 1 & 1 & | & 2 \\ 2 & 4 & -1 & | & 6 \\ 3 & 2 & \lambda & | & \mu \end{pmatrix}$

Step 2: Perform Row Operations to Achieve Row Echelon Form Our goal is to transform this matrix into a row echelon form, making the entries below the main diagonal zero.

• Step 2.1: Eliminate $x$ from the second and third equations. Apply the row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$. This uses the first row to create zeros in the first column of the second and third rows. $\begin{pmatrix} 1 & 1 & 1 & | & 2 \\ 2 - 2(1) & 4 - 2(1) & -1 - 2(1) & | & 6 - 2(2) \\ 3 - 3(1) & 2 - 3(1) & \lambda - 3(1) & | & \mu - 3(2) \end{pmatrix}$ Simplifying these operations, we get: $\begin{pmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 2 & -3 & | & 2 \\ 0 & -1 & \lambda - 3 & | & \mu - 6 \end{pmatrix}$

• Step 2.2: Eliminate $y$ from the third equation. Apply the row operation $R_3 \to 2R_3 + R_2$. This eliminates the $y$ term in the third row, using the second row, without affecting the $x$ terms in the first column. $\begin{pmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 2 & -3 & | & 2 \\ 2(0) + 0 & 2(-1) + 2 & 2(\lambda - 3) + (-3) & | & 2(\mu - 6) + 2 \end{pmatrix}$ Simplifying the entries:

  • Third row, third column: $2\lambda - 6 - 3 = 2\lambda - 9$
  • Third row, constant term: $2\mu - 12 + 2 = 2\mu - 10$ The matrix now is in row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 2 & -3 & | & 2 \\ 0 & 0 & 2\lambda - 9 & | & 2\mu - 10 \end{pmatrix}$

Step 3: Apply Conditions for Infinitely Many Solutions For the system to have infinitely many solutions, the rank of the coefficient matrix must be 2 (one less than the number of variables, 3), and the rank of the augmented matrix must also be 2. This implies that the last row of the row-reduced augmented matrix must consist entirely of zeros.

Therefore, we must have:

• $2\lambda - 9 = 0$
• $2\mu - 10 = 0$

Solving these equations for $\lambda$ and $\mu$: From $2\lambda - 9 = 0 \implies 2\lambda = 9 \implies \lambda = \frac{9}{2}$. From $2\mu - 10 = 0 \implies 2\mu = 10 \implies \mu = 5$.

Step 4: Determine the Relationship between $\lambda$ and $\mu$ from the Options Now we substitute these unique values of $\lambda$ and $\mu$ into each of the given options to find the correct relationship. Given $\lambda = \frac{9}{2}$ and $\mu = 5$:

• (A) $2\lambda - \mu = 5$ Substitute values: $2\left(\frac{9}{2}\right) - 5 = 9 - 5 = 4$. Since $4 \neq 5$, option (A) is not satisfied.

• (B) $\lambda - 2\mu = -5$ Substitute values: $\frac{9}{2} - 2(5) = \frac{9}{2} - 10 = \frac{9}{2} - \frac{20}{2} = -\frac{11}{2}$. Since $-\frac{11}{2} \neq -5$, option (B) is not satisfied.

• (C) $2\lambda + \mu = 14$ Substitute values: $2\left(\frac{9}{2}\right) + 5 = 9 + 5 = 14$. This relationship is satisfied by our derived values.

• (D) $\lambda + 2\mu = 14$ Substitute values: $\frac{9}{2} + 2(5) = \frac{9}{2} + 10 = \frac{9}{2} + \frac{20}{2} = \frac{29}{2}$. Since $\frac{29}{2} \neq 14$, option (D) is not satisfied.

Thus, the values of $\lambda$ and $\mu$ that ensure infinitely many solutions satisfy the relationship $2\lambda + \mu = 14$.

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3. Common Mistakes & Tips

• Arithmetic Errors: Row operations and determinant calculations are prone to small arithmetic mistakes, especially with negative signs. Always double-check each step.
• Conditions for Solutions: Clearly understand and differentiate the conditions for a unique solution ($D \neq 0$), no solution ($D=0$ and at least one of $D_x, D_y, D_z \neq 0$), and infinitely many solutions ($D=0$ and $D_x=D_y=D_z=0$, or using rank conditions as shown).
• Consistency: When using Cramer's rule, if $D=0$, you must check all $D_x, D_y, D_z$. If even one of them is non-zero, there is no solution. If all are zero, there are infinitely many solutions. Gaussian elimination naturally reveals these conditions.

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4. Summary

To determine the conditions for a system of linear equations to have infinitely many solutions, we can use Gaussian elimination to reduce the augmented matrix to row echelon form. For infinitely many solutions, the rank of the coefficient matrix must equal the rank of the augmented matrix, and this common rank must be less than the number of variables. This implies that the last row of the row-reduced augmented matrix must consist entirely of zeros. By applying this to the given system, we found that $\lambda = \frac{9}{2}$ and $\mu = 5$. Substituting these values into the given options, we found that they satisfy the relation $2\lambda + \mu = 14$.

5. Final Answer

The final answer is $\boxed{2\lambda + \mu = 14}$, which corresponds to option (C).