1. Key Concepts and Formulas

• Conditions for Infinitely Many Solutions: For a system of linear equations $AX=B$ with $n$ variables, it has infinitely many solutions if and only if:
  1. The determinant of the coefficient matrix, $|A|$, is zero.
  2. The rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A|B]$, and this common rank is less than the number of variables, $n$. ($\text{rank}(A) = \text{rank}([A|B]) < n$).
• Linear Dependence of Rows: When $|A|=0$ for a $3 \times 3$ matrix, it implies that the rows (or columns) of $A$ are linearly dependent. For infinitely many solutions, the rows of the augmented matrix $[A|B]$ must also be linearly dependent in the same way. That is, if one row of $A$ (e.g., $R_3$) is a linear combination of the other rows (e.g., $R_3 = pR_1 + qR_2$), then the corresponding element in the constant matrix $B$ (e.g., $B_3$) must also be the same linear combination of the other elements of $B$ (e.g., $B_3 = pB_1 + qB_2$).
• Determinant of a $3 \times 3$ Matrix: For $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$, the determinant is $|A| = a(ei-fh) - b(di-fg) + c(dh-eg)$.

2. Step-by-Step Solution

Step 1: Represent the system in matrix form. The given system of linear equations is: $\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$ This can be written in the matrix form $AX = B$, where: $A = \begin{bmatrix} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 2 \\ \mu \end{bmatrix}$

Step 2: Apply the condition $|A|=0$ to find $\lambda$. For the system to have infinitely many solutions, the determinant of the coefficient matrix $A$ must be zero. $|A| = \left| \begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array} \right| = 0$ Expanding the determinant along the first row: $3 \left| \begin{array}{cc} 11 & -9 \\ 155 & -189 \end{array} \right| - 5 \left| \begin{array}{cc} 7 & -9 \\ 97 & -189 \end{array} \right| + \lambda \left| \begin{array}{cc} 7 & 11 \\ 97 & 155 \end{array} \right| = 0$ Now, we calculate the $2 \times 2$ determinants:

• $11(-189) - (-9)(155) = -2079 + 1395 = -684$
• $7(-189) - (-9)(97) = -1323 + 873 = -450$
• $7(155) - 11(97) = 1085 - 1067 = 18$ Substitute these values back into the determinant equation: $3(-684) - 5(-450) + \lambda(18) = 0$ $-2052 + 2250 + 18\lambda = 0$ $198 + 18\lambda = 0$ $18\lambda = -198$ $\lambda = \frac{-198}{18} = -11$ So, the value of $\lambda$ is $-11$.

Step 3: Apply the condition of linear dependence to find $\mu$. Since $|A|=0$, the rows of $A$ are linearly dependent. For the system to have infinitely many solutions, the third row of the augmented matrix $[A|B]$ must be a linear combination of the first two rows. Let's assume $R_3 = p R_1 + q R_2$ for some scalars $p$ and $q$. We compare the coefficients of $x, y, z$ and the constant terms:

1. For $x$-coefficients: $97 = 3p + 7q$
2. For $y$-coefficients: $155 = 5p + 11q$
3. For $z$-coefficients: $-189 = \lambda p - 9q$
4. For constant terms: $\mu = 3p + 2q$

First, we solve the system of equations (1) and (2) for $p$ and $q$: Multiply equation (1) by 5: $15p + 35q = 485$ Multiply equation (2) by 3: $15p + 33q = 465$ Subtracting the second modified equation from the first: $(15p + 35q) - (15p + 33q) = 485 - 465$ $2q = 20 \implies q = 10$ Substitute $q=10$ back into equation (1): $3p + 7(10) = 97$ $3p + 70 = 97$ $3p = 27 \implies p = 9$ So, the linear combination coefficients are $p=9$ and $q=10$.

Now, we verify this linear combination with the $z$-coefficients using our calculated $\lambda = -11$: $-189 = \lambda p - 9q$ $-189 = (-11)(9) - 9(10)$ $-189 = -99 - 90$ $-189 = -189$ This confirms that our values for $p, q,$ and $\lambda$ are consistent.

Finally, we use these values of $p$ and $q$ to find $\mu$ from the constant terms (equation 4): $\mu = 3p + 2q$ $\mu = 3(9) + 2(10)$ $\mu = 27 + 20$ $\mu = 47$ Thus, the value of $\mu$ is $47$.

Step 4: Calculate $\mu + 2\lambda$. We need to find the value of $\mu + 2\lambda$. Substitute the values $\lambda = -11$ and $\mu = 47$: $\mu + 2\lambda = 47 + 2(-11)$ $\mu + 2\lambda = 47 - 22$ $\mu + 2\lambda = 25$

3. Common Mistakes & Tips

• Arithmetic Errors: Determinant calculations and solving for $p$ and $q$ involve several arithmetic operations. A single mistake can lead to an incorrect final answer. Always double-check your calculations.
• Incomplete Conditions: Remember that $|A|=0$ alone is not sufficient for infinitely many solutions; it could also lead to no solution. The additional condition of consistent linear dependence across the augmented matrix is crucial.
• Systematic Approach: Break down the problem into smaller, manageable steps (finding $\lambda$, then $p$ and $q$, then $\mu$). This reduces the chances of error and makes the solution easier to follow.

4. Summary

To solve this problem, we first used the condition that for a system to have infinitely many solutions, the determinant of its coefficient matrix must be zero. This allowed us to calculate $\lambda = -11$. Next, we applied the condition that the rows of the augmented matrix must be linearly dependent in the same way as the coefficient matrix. By expressing the third row as a linear combination of the first two rows ($R_3 = 9R_1 + 10R_2$), we found the values of $p=9$ and $q=10$. Using these values, we then determined $\mu = 47$. Finally, we computed the required expression $\mu + 2\lambda = 47 + 2(-11) = 25$.

5. Final Answer

The final answer is $\boxed{25}$.