Here's a clear, educational, and well-structured solution to the problem:

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1. Understanding the Conditions for Infinitely Many Solutions

For a system of linear equations in three variables $x, y, z$: $\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}$ We use Cramer's Rule to determine the nature of solutions, which relies on four determinants:

• Coefficient Determinant ($\Delta$): This is the determinant of the coefficients of $x, y, z$. $\Delta = \left| \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right|$
• $\Delta_x$: Formed by replacing the $x$-coefficient column in $\Delta$ with the constant terms. $\Delta_x = \left| \begin{matrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{matrix} \right|$
• $\Delta_y$: Formed by replacing the $y$-coefficient column in $\Delta$ with the constant terms. $\Delta_y = \left| \begin{matrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{matrix} \right|$
• $\Delta_z$: Formed by replacing the $z$-coefficient column in $\Delta$ with the constant terms. $\Delta_z = \left| \begin{matrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{matrix} \right|$

The conditions for the nature of solutions are:

• Unique Solution: If $\Delta \neq 0$.
• No Solution (Inconsistent): If $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
• Infinitely Many Solutions (Consistent): If $\Delta = 0$ AND $\Delta_x = 0$ AND $\Delta_y = 0$ AND $\Delta_z = 0$.

In this problem, we are given that the system has infinitely many solutions. Therefore, we must satisfy the condition that all four determinants are equal to zero: $\Delta = 0$, $\Delta_x = 0$, $\Delta_y = 0$, and $\Delta_z = 0$.

2. Setting Up the Determinants for the Given System

The given system of linear equations is:

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

From these equations, we can write down the four determinants:

• Coefficient Determinant ($\Delta$): $\Delta = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{matrix} \right|$
• $\Delta_x$: $\Delta_x = \left| \begin{matrix} 5 & 1 & 1 \\ 6 & 2 & 2 \\ \mu & 3 & \lambda \end{matrix} \right|$
• $\Delta_y$: $\Delta_y = \left| \begin{matrix} 1 & 5 & 1 \\ 1 & 6 & 2 \\ 1 & \mu & \lambda \end{matrix} \right|$
• $\Delta_z$: $\Delta_z = \left| \begin{matrix} 1 & 1 & 5 \\ 1 & 2 & 6 \\ 1 & 3 & \mu \end{matrix} \right|$

3. Calculating $\Delta$ and Finding $\lambda$

The first step for infinitely many solutions is $\Delta = 0$. Let's calculate $\Delta$: $\Delta = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{matrix} \right|$ To simplify the calculation, we can perform row operations. Subtract $R_1$ from $R_2$ and $R_1$ from $R_3$: ($R_2 \to R_2 - R_1$) and ($R_3 \to R_3 - R_1$) $\Delta = \left| \begin{matrix} 1 & 1 & 1 \\ 1-1 & 2-1 & 2-1 \\ 1-1 & 3-1 & \lambda-1 \end{matrix} \right| = \left| \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 2 & \lambda-1 \end{matrix} \right|$ Now, expand the determinant along the first column (since it has two zeros, simplifying the calculation): $\Delta = 1 \cdot \left| \begin{matrix} 1 & 1 \\ 2 & \lambda-1 \end{matrix} \right| - 0 + 0$ $\Delta = 1 \cdot ((1)(\lambda-1) - (1)(2))$ $\Delta = (\lambda - 1 - 2) = \lambda - 3$ Since the system has infinitely many solutions, we must have $\Delta = 0$: $\lambda - 3 = 0 \implies \lambda = 3$

4. Calculating $\Delta_x, \Delta_y, \Delta_z$ and Finding $\mu$

Now that we have $\lambda = 3$, we need to ensure that $\Delta_x = 0$, $\Delta_y = 0$, and $\Delta_z = 0$. We can use any one of these conditions to find $\mu$.

• Calculate $\Delta_x$ with $\lambda = 3$: $\Delta_x = \left| \begin{matrix} 5 & 1 & 1 \\ 6 & 2 & 2 \\ \mu & 3 & 3 \end{matrix} \right|$ Observe that the second column ($C_2$) and the third column ($C_3$) are identical. A property of determinants states that if any two rows or columns are identical, the determinant is zero. Therefore, $\Delta_x = 0$ for any value of $\mu$. This means $\Delta_x = 0$ does not help us find $\mu$. We must proceed to $\Delta_y$ or $\Delta_z$.

• Calculate $\Delta_y$ with $\lambda = 3$: $\Delta_y = \left| \begin{matrix} 1 & 5 & 1 \\ 1 & 6 & 2 \\ 1 & \mu & 3 \end{matrix} \right|$ Again, we perform row operations to simplify: ($R_2 \to R_2 - R_1$) and ($R_3 \to R_3 - R_1$) $\Delta_y = \left| \begin{matrix} 1 & 5 & 1 \\ 0 & 1 & 1 \\ 0 & \mu-5 & 2 \end{matrix} \right|$ Expand along the first column: $\Delta_y = 1 \cdot \left| \begin{matrix} 1 & 1 \\ \mu-5 & 2 \end{matrix} \right| - 0 + 0$ $\Delta_y = (1)(2) - (1)(\mu-5)$ $\Delta_y = 2 - \mu + 5 = 7 - \mu$ For infinitely many solutions, we must have $\Delta_y = 0$: $7 - \mu = 0 \implies \mu = 7$

• (Optional Check) Calculate $\Delta_z$ with $\lambda = 3$ and $\mu = 7$: Let's verify that $\Delta_z$ also becomes zero with these values. $\Delta_z = \left| \begin{matrix} 1 & 1 & 5 \\ 1 & 2 & 6 \\ 1 & 3 & 7 \end{matrix} \right|$ Perform row operations: ($R_2 \to R_2 - R_1$) and ($R_3 \to R_3 - R_1$) $\Delta_z = \left| \begin{matrix} 1 & 1 & 5 \\ 0 & 1 & 1 \\ 0 & 2 & 2 \end{matrix} \right|$ Observe that the third row ($R_3$) is twice the second row ($R_2$) ($R_3 = 2R_2$). When two rows or columns are proportional, the determinant is zero. Thus, $\Delta_z = 0$, which confirms our values of $\lambda = 3$ and $\mu = 7$.

5. Finding the Value of $\lambda + \mu$

We have found $\lambda = 3$ and $\mu = 7$. The question asks for the value of $\lambda + \mu$. $\lambda + \mu = 3 + 7 = 10$

Tips for Success & Common Mistakes:

• Systematic Approach: Always start by calculating $\Delta$. If $\Delta \neq 0$, it's a unique solution, and you don't need to check other determinants.
• Check All Conditions: For infinitely many solutions, all four determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$) must be zero. If one of them is non-zero (while $\Delta=0$), the system has no solution.
• Simplifying Determinants: Use row/column operations (like $R_i \to R_i - kR_j$) to introduce zeros in a row or column, which greatly simplifies expansion. Remember these operations do not change the value of the determinant.
• Identical/Proportional Rows/Columns: Keep an eye out for identical or proportional rows/columns, as this immediately implies the determinant is zero. This saved us work for $\Delta_x$ and $\Delta_z$.
• Calculation Errors: Determinant calculations can be prone to sign errors or arithmetic mistakes. Double-check your work, especially when dealing with parameters.

Summary and Key Takeaway

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix ($\Delta$) and all the auxiliary determinants ($\Delta_x, \Delta_y, \Delta_z$) must be zero. By systematically applying these conditions, we found $\lambda = 3$ and $\mu = 7$, leading to $\lambda + \mu = 10$.

The final answer is $\boxed{\text{10}}$.