1. Key Concepts and Formulas

For a system of three linear equations in three variables: $\begin{aligned} a_1 x + b_1 y + c_1 z &= d_1 \\ a_2 x + b_2 y + c_2 z &= d_2 \\ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned}$ to have infinitely many solutions, the following conditions must be met (based on Cramer's Rule or the concept of matrix ranks):

• Determinant of the Coefficient Matrix ($D$) must be zero: $D = \left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right| = 0$
• Determinants $D_x, D_y,$ and $D_z$ must also be zero: These are obtained by replacing the coefficients of $x, y,$ and $z$ respectively with the constant terms $d_1, d_2, d_3$. $D_x = \left|\begin{array}{ccc} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{array}\right| = 0$ $D_y = \left|\begin{array}{ccc} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{array}\right| = 0$ $D_z = \left|\begin{array}{ccc} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{array}\right| = 0$ All four determinants ($D, D_x, D_y, D_z$) must be simultaneously zero for the system to have infinitely many solutions. If $D=0$ but any of $D_x, D_y, D_z$ is non-zero, the system has no solution.

2. Step-by-Step Solution

Step 1: Identify Coefficients and Formulate Determinants

The given system of equations is: $\begin{aligned} x+2 y-3 z&=2 \\ 2 x+\lambda y+5 z&=5 \\ 14 x+3 y+\mu z&=33 \end{aligned}$ The coefficients and constant terms are: $a_1=1, b_1=2, c_1=-3, d_1=2$ $a_2=2, b_2=\lambda, c_2=5, d_2=5$ $a_3=14, b_3=3, c_3=\mu, d_3=33$

We formulate the determinant of the coefficient matrix, $D$, and $D_y$ (chosen first because it involves only one unknown, $\mu$, simplifying the initial calculation):

$D = \left|\begin{array}{ccc} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{array}\right|$

$D_y = \left|\begin{array}{ccc} 1 & 2 & -3 \\ 2 & 5 & 5 \\ 14 & 33 & \mu \end{array}\right|$

Step 2: Calculate $D_y$ and Solve for $\mu$

We expand the determinant $D_y$ along the first row and set it to zero, as required for infinitely many solutions. $D_y = 1 \cdot \left|\begin{array}{cc} 5 & 5 \\ 33 & \mu \end{array}\right| - 2 \cdot \left|\begin{array}{cc} 2 & 5 \\ 14 & \mu \end{array}\right| + (-3) \cdot \left|\begin{array}{cc} 2 & 5 \\ 14 & 33 \end{array}\right|$ Explanation: We compute the $2 \times 2$ determinants and combine the terms. $D_y = 1(5\mu - 5 \times 33) - 2(2\mu - 5 \times 14) - 3(2 \times 33 - 5 \times 14)$ $D_y = (5\mu - 165) - 2(2\mu - 70) - 3(66 - 70)$ $D_y = 5\mu - 165 - 4\mu + 140 - 3(-4)$ $D_y = \mu - 25 + 12$ $D_y = \mu - 13$ For infinitely many solutions, $D_y$ must be zero: $\mu - 13 = 0 \implies \mu = 13$ Explanation: Setting $D_y=0$ allows us to directly solve for the value of $\mu$.

Step 3: Calculate $D$ and Solve for $\lambda$

Now that we have $\mu=13$, we substitute this value into the determinant $D$ and set it to zero to find $\lambda$. $D = \left|\begin{array}{ccc} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & 13 \end{array}\right|$ Explanation: We substitute the known value of $\mu$ into $D$ to reduce the number of unknowns. Expand the determinant $D$ along the first row: $D = 1 \cdot \left|\begin{array}{cc} \lambda & 5 \\ 3 & 13 \end{array}\right| - 2 \cdot \left|\begin{array}{cc} 2 & 5 \\ 14 & 13 \end{array}\right| + (-3) \cdot \left|\begin{array}{cc} 2 & \lambda \\ 14 & 3 \end{array}\right|$ $D = 1(13\lambda - 5 \times 3) - 2(2 \times 13 - 5 \times 14) - 3(2 \times 3 - \lambda \times 14)$ $D = (13\lambda - 15) - 2(26 - 70) - 3(6 - 14\lambda)$ $D = 13\lambda - 15 - 2(-44) - 18 + 42\lambda$ $D = 13\lambda - 15 + 88 - 18 + 42\lambda$ Combine like terms: $D = (13\lambda + 42\lambda) + (-15 + 88 - 18)$ $D = 55\lambda + 55$ For infinitely many solutions, $D$ must be zero: $55\lambda + 55 = 0$ $55\lambda = -55$ $\lambda = -1$ Explanation: Setting $D=0$ provides an equation that we can solve for $\lambda$.

Step 4: Verify with $D_z=0$ (Optional but Recommended)

Although we have found $\lambda$ and $\mu$, it's good practice to verify that other determinants ($D_x$ or $D_z$) also become zero with these values. Let's check $D_z=0$. $D_z = \left|\begin{array}{ccc} 1 & 2 & 2 \\ 2 & \lambda & 5 \\ 14 & 3 & 33 \end{array}\right|$ Substitute $\lambda = -1$: $D_z = \left|\begin{array}{ccc} 1 & 2 & 2 \\ 2 & -1 & 5 \\ 14 & 3 & 33 \end{array}\right|$ Expand along the first row: $D_z = 1(-1 \times 33 - 5 \times 3) - 2(2 \times 33 - 5 \times 14) + 2(2 \times 3 - (-1) \times 14)$ $D_z = 1(-33 - 15) - 2(66 - 70) + 2(6 + 14)$ $D_z = -48 - 2(-4) + 2(20)$ $D_z = -48 + 8 + 40$ $D_z = -48 + 48 = 0$ Explanation: This verification confirms that our values of $\lambda=-1$ and $\mu=13$ are consistent with all conditions for infinitely many solutions.

Step 5: Calculate $\lambda + \mu$

We found $\lambda = -1$ and $\mu = 13$. $\lambda + \mu = -1 + 13 = 12$

3. Common Mistakes & Tips

• All Determinants Must Be Zero: A frequent error is to only set $D=0$. While $D=0$ is necessary, it indicates either no solution or infinitely many solutions. To confirm infinitely many solutions, all of $D_x, D_y, D_z$ must also be zero.
• Arithmetic Errors: Determinant calculations can be lengthy and prone to calculation mistakes, especially with signs. Always double-check your expansion and arithmetic.
• Strategic Choice: When solving for multiple unknown variables, strategically choose which determinant ($D_x, D_y,$ or $D_z$) to evaluate first. Often, one determinant will contain fewer unknowns or simpler numbers, allowing you to solve for one variable more quickly. In this problem, $D_y$ was a good starting point as it only contained $\mu$.

4. Summary

To determine the values of $\lambda$ and $\mu$ for which the given system of linear equations has infinitely many solutions, we used the condition that the determinant of the coefficient matrix ($D$) and the determinants $D_x, D_y, D_z$ must all be zero. By first calculating $D_y=0$, we found $\mu=13$. Substituting this value into $D=0$, we then solved for $\lambda=-1$. Finally, we calculated the sum $\lambda+\mu = -1+13 = 12$.

5. Final Answer

The final answer is $\boxed{12}$, which corresponds to option (C).