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

Answer: (C)

47

Answer 2

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 $\lambda^4 - \mu$: $\lambda^4 - \mu = (-2)^4 - (-31) = 16 + 31 = 47$