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 \( \lambda^4 - \mu \): \[ \lambda^4 - \mu = (-2)^4 - (-31) = 16 + 31 = 47 \]
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to:
If \[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \] find \( A^{-1} \).
Using \( A^{-1} \), solve the following system of equations:
\[ \begin{aligned} 2x - 3y + 5z &= 11 \quad \text{(1)} \\ 3x + 2y - 4z &= -5 \quad \text{(2)} \\ x + y - 2z &= -3 \quad \text{(3)} \end{aligned} \]
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).