Question

If the system of equations \(11x + y + \lambda z = -5,\) \(2x + 3y + 5z = 3,\) \(8x - 19y - 39z = \mu\) has infinitely many solutions, then \( \lambda^4 - \mu \) is equal to:

• A. 49
• B. 45
• C. 47
• D. 51

Answer 1

The Correct Option is C

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. Set up the determinant \( D \) as follows:

\[ D = \begin{vmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{vmatrix} = 0 \]

Expanding the determinant:

\[ = 11(3 \cdot (-39) - 5 \cdot (-19)) - 1(2 \cdot (-39) - 5 \cdot 8) + \lambda(2 \cdot (-19) - 3 \cdot 8) \]

\[ = 11(-117 + 95) - 1(-78 - 40) + \lambda(-38 - 24) \]

\[ = 11(-22) + 118 + \lambda(-62) = 0 \]

Solving for \( \lambda \):

\[ -242 + 118 - 62\lambda = 0 \]

\[ 62\lambda = -124 \implies \lambda = -2 \]

Now substitute \( \lambda = -2 \) and calculate \( \mu \) by setting up the augmented determinant \( D_1 \):

\[ D_1 = \begin{vmatrix} -5 & 1 & -2 \\ 3 & 3 & 5 \\ \mu & -19 & -39 \end{vmatrix} = 0 \]

Expanding \( D_1 \):

\[ = -5(3 \cdot (-39) - 5 \cdot (-19)) - 1(3 \cdot (-39) - 5\mu) - 2(3 \cdot (-19) + 3\mu) \]

\[ = -5(-117 + 95) + (-117 + 5\mu) + 6\mu = 0 \]

\[ -341 + 6\mu = 0 \implies \mu = -31 \]

Finally, calculate \( \lambda^4 - \mu \): \[ \lambda^4 - \mu = (-2)^4 - (-31) = 16 + 31 = 47 \]

Answer 2

Step 1: Write the system and the condition for infinitely many solutions
We have \[ \begin{cases} 11x + y + \lambda z = -5,\\ 2x + 3y + 5z = 3,\\ 8x - 19y - 39z = \mu. \end{cases} \] A linear \(3\times 3\) system has infinitely many solutions if and only if the coefficient matrix has determinant \(0\) (so its rank is less than \(3\)) and the augmented matrix has the same rank as the coefficient matrix (consistency).

Step 2: Compute the determinant of the coefficient matrix
Let \[ A= \begin{pmatrix} 11 & 1 & \lambda\\ 2 & 3 & 5\\ 8 & -19 & -39 \end{pmatrix}. \] Compute \(\det A\) (expansion or row-reduction): \[ \det A=-62(\lambda+2). \] Thus \(\det A=0\) if and only if \(\lambda=-2\). For any other \(\lambda\), the system has a unique solution, not infinitely many.

Step 3: Impose \(\lambda=-2\) and analyze consistency
With \(\lambda=-2\), the coefficient matrix becomes \[ A_{0}= \begin{pmatrix} 11 & 1 & -2\\ 2 & 3 & 5\\ 8 & -19 & -39 \end{pmatrix},\qquad \mathbf{b}= \begin{pmatrix} -5\\ 3\\ \mu \end{pmatrix}. \] We now check whether the third row of \(A_{0}\) is a linear combination of the first two (row dependence), and then enforce the same relation on the right-hand sides for consistency.

Step 4: Find the linear dependence between rows
Seek constants \(\alpha,\beta\) such that \[ \alpha(11,1,-2)+\beta(2,3,5)=(8,-19,-39). \] Solving componentwise: \[ \begin{cases} 11\alpha+2\beta=8,\\ \alpha+3\beta=-19,\\ -2\alpha+5\beta=-39, \end{cases} \quad\Longrightarrow\quad \alpha=2,\ \beta=-7. \] Therefore, \[ (8,-19,-39)=2\cdot(11,1,-2)-7\cdot(2,3,5), \] so the third row is \(2\) times the first row minus \(7\) times the second row. Hence \(\operatorname{rank}(A_{0})=2\).

Step 5: Enforce the same relation on the constants for consistency
For the augmented matrix to have the same rank \(2\), the right-hand sides must satisfy the same linear relation: \[ \mu=2\cdot(-5)-7\cdot(3)=-10-21=-31. \] Thus, the system has infinitely many solutions precisely when \[ \lambda=-2,\qquad \mu=-31. \]

Step 6: Compute the required value
\[ \lambda^4-\mu=(-2)^4-(-31)=16+31=47. \]

Final answer

47