1. Understanding the Condition for Infinitely Many Solutions

For a system of linear equations in three variables: $\begin{array}{l} a_1 x+b_1 y+c_1 z=d_1 \\ a_2 x+b_2 y+c_2 z=d_2 \\ a_3 x+b_3 y+c_3 z=d_3 \end{array}$ This system can be represented in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix. $A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}$ According to Cramer's Rule, for this system to have infinitely many solutions, two crucial conditions must be met simultaneously:

1. The determinant of the coefficient matrix, denoted as $\Delta = \det(A)$, must be zero.
2. All the determinants obtained by replacing a column of $A$ with the constant terms from $B$, denoted as $\Delta_1, \Delta_2, \Delta_3$, must also be zero.

Let's define these determinants explicitly: $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ $\Delta_1 = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}, \quad \Delta_2 = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}, \quad \Delta_3 = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$ The condition for infinitely many solutions is: $\Delta = 0 \quad \text{AND} \quad \Delta_1 = 0 \quad \text{AND} \quad \Delta_2 = 0 \quad \text{AND} \quad \Delta_3 = 0$ Why this condition? If $\Delta \neq 0$, the system has a unique solution. If $\Delta = 0$, the system is either inconsistent (no solution) or has infinitely many solutions. The additional condition that all $\Delta_i$ must also be zero distinguishes the case of infinitely many solutions from the case of no solution. If $\Delta=0$ but at least one $\Delta_i \neq 0$, then the system has no solution.

2. Setting Up the System

The given system of linear equations is:

1. $11 x+y+\lambda z=-5$
2. $2 x+3 y+5 z=3$
3. $8 x-19 y-39 z=\mu$

From these equations, we can identify the coefficient matrix $A$ and the constant vector $B$: $A = \begin{pmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{pmatrix}, \quad B = \begin{pmatrix} -5 \\ 3 \\ \mu \end{pmatrix}$

3. Finding the Value of $\lambda$

To find the value of $\lambda$, we must apply the first necessary condition for infinitely many solutions: $\Delta = 0$. This condition will allow us to determine $\lambda$.

Let's calculate the determinant of the coefficient matrix $A$: $\Delta = \begin{vmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{vmatrix}$ To simplify the calculation of the determinant, we can use row operations to create zeros in a column or row. Let's perform $R_3 \to R_3 + 19R_2$ to simplify the second column: $\Delta = \begin{vmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8+19(2) & -19+19(3) & -39+19(5) \end{vmatrix}$ $\Delta = \begin{vmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 46 & 38 & 56 \end{vmatrix}$ Now, we expand the determinant along the first row: $\Delta = 11 \begin{vmatrix} 3 & 5 \\ 38 & 56 \end{vmatrix} - 1 \begin{vmatrix} 2 & 5 \\ 46 & 56 \end{vmatrix} + \lambda \begin{vmatrix} 2 & 3 \\ 46 & 38 \end{vmatrix}$ $\Delta = 11((3)(56) - (5)(38)) - 1((2)(56) - (5)(46)) + \lambda((2)(38) - (3)(46))$ $\Delta = 11(168 - 190) - 1(112 - 230) + \lambda(76 - 138)$ $\Delta = 11(-22) - 1(-118) + \lambda(-62)$ $\Delta = -242 + 118 - 62\lambda$ $\Delta = -124 - 62\lambda$ For the system to have infinitely many solutions, we must have $\Delta = 0$: $-124 - 62\lambda = 0$ $-62\lambda = 124$ $\lambda = \frac{124}{-62}$ $\lambda = -2$ Thus, we have found the value of $\lambda = -2$.

Tip for Determinant Calculation: Using row/column operations to introduce zeros before expanding can significantly simplify calculations and reduce the chance of arithmetic errors, especially for larger matrices. Remember that operations of the form $R_i \to R_i + k R_j$ (or $C_i \to C_i + k C_j$) do not change the determinant value.

4. Finding the Value of $\mu$

Now that we have $\lambda = -2$, we need to use the other conditions, $\Delta_1=0$, $\Delta_2=0$, or $\Delta_3=0$, to find $\mu$. Since the problem states that the system has infinitely many solutions, all these determinants must be zero. We only need to set one of them to zero to find $\mu$. Let's choose $\Delta_3 = 0$.

First, we form $\Delta_3$ by replacing the third column of the coefficient matrix $A$ with the constant vector $B$. We also substitute the value of $\lambda=-2$ into the matrix: $\Delta_3 = \begin{vmatrix} 11 & 1 & -5 \\ 2 & 3 & 3 \\ 8 & -19 & \mu \end{vmatrix}$ Now, we set $\Delta_3 = 0$ and solve for $\mu$. Let's expand along the third column to directly involve $\mu$: $\Delta_3 = (-5) \cdot \text{cofactor}(1,3) + (3) \cdot \text{cofactor}(2,3) + (\mu) \cdot \text{cofactor}(3,3)$ $\Delta_3 = -5 \begin{vmatrix} 2 & 3 \\ 8 & -19 \end{vmatrix} - 3 \begin{vmatrix} 11 & 1 \\ 8 & -19 \end{vmatrix} + \mu \begin{vmatrix} 11 & 1 \\ 2 & 3 \end{vmatrix}$ $\Delta_3 = -5((2)(-19) - (3)(8)) - 3((11)(-19) - (1)(8)) + \mu((11)(3) - (1)(2))$ $\Delta_3 = -5(-38 - 24) - 3(-209 - 8) + \mu(33 - 2)$ $\Delta_3 = -5(-62) - 3(-217) + \mu(31)$ $\Delta_3 = 310 + 651 + 31\mu$ $\Delta_3 = 961 + 31\mu$ For infinitely many solutions, we must have $\Delta_3 = 0$: $961 + 31\mu = 0$ $31\mu = -961$ $\mu = \frac{-961}{31}$ $\mu = -31$ So, we have found the value of $\mu = -31$.

Common Mistake to Avoid: When calculating $\Delta_i$, ensure that you substitute the value of $\lambda$ (if $\lambda$ is part of the determinant $\Delta_i$) before setting $\Delta_i=0$ to solve for $\mu$. Also, be extremely careful with arithmetic, especially managing negative signs in determinant expansions.

5. Calculating the Final Expression

We have found $\lambda = -2$ and $\mu = -31$. The question asks for the value of $\lambda^4 - \mu$. $\lambda^4 - \mu = (-2)^4 - (-31)$ $= 16 - (-31)$ $= 16 + 31$ $= 47$ Therefore, $\lambda^4 - \mu = 47$.

6. Summary and Key Takeaways

This problem is a direct application of the conditions for a system of linear equations to have infinitely many solutions, based on Cramer's Rule. Key Takeaways:

• A system $AX=B$ has infinitely many solutions if and only if $\Delta = 0$ AND $\Delta_1 = 0$ AND $\Delta_2 = 0$ AND $\Delta_3 = 0$.
• The condition $\Delta=0$ is typically used first to find parameters within the coefficient matrix (like $\lambda$ in this case).
• Subsequently, any of the $\Delta_i=0$ conditions can be used to find parameters in the constant vector (like $\mu$ here), after substituting the values found from $\Delta=0$.
• Leverage determinant properties like row/column operations to simplify calculations and reduce potential errors.
• Meticulous arithmetic and sign management are crucial when evaluating determinants.

The final answer is $\boxed{\text{47}}$.