1. Key Concepts and Formulas

• System of Linear Equations: A system of linear equations $AX=B$ (where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix) has infinitely many solutions if and only if:
  1. The determinant of the coefficient matrix, $\Delta = \det(A)$, is zero.
  2. The determinants formed by replacing each column of the coefficient matrix with the constant terms, $\Delta_x$, $\Delta_y$, and $\Delta_z$ (also denoted as $\Delta_1, \Delta_2, \Delta_3$), are all zero.
• Determinant of a 3x3 Matrix: For a 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: Formulate the coefficient matrix and the constant matrix. The given system of equations is: $2x + 7y + \lambda z = 3$ $3x + 2y + 5z = 4$ $x + \mu y + 32z = -1$

The coefficient matrix $A$ is: $A = \begin{pmatrix} 2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & \mu & 32 \end{pmatrix}$ The constant matrix $B$ is: $B = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix}$

Step 2: Apply the condition $\Delta = 0$. For infinitely many solutions, the determinant of the coefficient matrix must be zero. $\Delta = \begin{vmatrix} 2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & \mu & 32 \end{vmatrix} = 0$ Expanding the determinant along the first row: $2(2 \times 32 - 5 \times \mu) - 7(3 \times 32 - 5 \times 1) + \lambda(3 \times \mu - 2 \times 1) = 0$ $2(64 - 5\mu) - 7(96 - 5) + \lambda(3\mu - 2) = 0$ $128 - 10\mu - 7(91) + 3\lambda\mu - 2\lambda = 0$ $128 - 10\mu - 637 + 3\lambda\mu - 2\lambda = 0$ $3\lambda\mu - 10\mu - 2\lambda - 509 = 0 \quad \text{(Equation 1)}$

Step 3: Apply the condition $\Delta_y = 0$ to find $\lambda$. $\Delta_y$ is the determinant formed by replacing the second column of $A$ with the constant terms from $B$. $\Delta_y = \begin{vmatrix} 2 & 3 & \lambda \\ 3 & 4 & 5 \\ 1 & -1 & 32 \end{vmatrix} = 0$ Expanding the determinant along the first row: $2(4 \times 32 - 5 \times (-1)) - 3(3 \times 32 - 5 \times 1) + \lambda(3 \times (-1) - 4 \times 1) = 0$ $2(128 + 5) - 3(96 - 5) + \lambda(-3 - 4) = 0$ $2(133) - 3(91) + \lambda(-7) = 0$ $266 - 273 - 7\lambda = 0$ $-7 - 7\lambda = 0$ $7\lambda = -7$ $\lambda = -1$

Step 4: Apply the condition $\Delta_z = 0$ to find $\mu$. $\Delta_z$ is the determinant formed by replacing the third column of $A$ with the constant terms from $B$. $\Delta_z = \begin{vmatrix} 2 & 7 & 3 \\ 3 & 2 & 4 \\ 1 & \mu & -1 \end{vmatrix} = 0$ Expanding the determinant along the first row: $2(2 \times (-1) - 4 \times \mu) - 7(3 \times (-1) - 4 \times 1) + 3(3 \times \mu - 2 \times 1) = 0$ $2(-2 - 4\mu) - 7(-3 - 4) + 3(3\mu - 2) = 0$ $-4 - 8\mu - 7(-7) + 9\mu - 6 = 0$ $-4 - 8\mu + 49 + 9\mu - 6 = 0$ $\mu + 39 = 0$ $\mu = -39$

Step 5: Verify the values with $\Delta_x = 0$ (optional but good practice). $\Delta_x$ is the determinant formed by replacing the first column of $A$ with the constant terms from $B$. $\Delta_x = \begin{vmatrix} 3 & 7 & \lambda \\ 4 & 2 & 5 \\ -1 & \mu & 32 \end{vmatrix} = 0$ Substitute $\lambda = -1$ and $\mu = -39$: $\Delta_x = \begin{vmatrix} 3 & 7 & -1 \\ 4 & 2 & 5 \\ -1 & -39 & 32 \end{vmatrix}$ Expanding along the first row: $3(2 \times 32 - 5 \times (-39)) - 7(4 \times 32 - 5 \times (-1)) + (-1)(4 \times (-39) - 2 \times (-1)) = 0$ $3(64 + 195) - 7(128 + 5) - 1(-156 + 2) = 0$ $3(259) - 7(133) - 1(-154) = 0$ $777 - 931 + 154 = 0$ $931 - 931 = 0$ $0 = 0$. The values $\lambda = -1$ and $\mu = -39$ are consistent with all conditions.

Step 6: Calculate $(\lambda - \mu)$. We have $\lambda = -1$ and $\mu = -39$. $(\lambda - \mu) = -1 - (-39) = -1 + 39 = 38$.

3. Common Mistakes & Tips

• Forgetting all conditions: A common mistake is to only check $\Delta = 0$. For non-homogeneous systems, $\Delta = 0$ alone is not sufficient; all $\Delta_x, \Delta_y, \Delta_z$ must also be zero for infinitely many solutions. If $\Delta=0$ but at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.
• Arithmetic errors in determinants: Determinant calculations are prone to sign errors or multiplication mistakes. Double-check each step, especially when dealing with negative numbers. Expanding along different rows/columns can help verify results.
• Understanding the implications of $\Delta=0$: If $\Delta \neq 0$, the system has a unique solution (Cramer's Rule applies directly). If $\Delta = 0$, then further investigation using $\Delta_x, \Delta_y, \Delta_z$ is required to distinguish between infinitely many solutions and no solution.

4. Summary

To determine the values of $\lambda$ and $\mu$ for which the given system of linear equations has infinitely many solutions, we applied Cramer's Rule conditions. This involved setting the determinant of the coefficient matrix ($\Delta$) and the determinants obtained by replacing each column with the constant terms ($\Delta_x, \Delta_y, \Delta_z$) to zero. Solving $\Delta_y = 0$ yielded $\lambda = -1$, and solving $\Delta_z = 0$ yielded $\mu = -39$. These values were then verified to satisfy $\Delta = 0$ and $\Delta_x = 0$. Finally, the required expression $(\lambda - \mu)$ was calculated using these values.

The final answer is $\boxed{2}$.