Question

If the system of equations \[ 2x - y + z = 4, \] \[ 5x + \lambda y + 3z = 12, \] \[ 100x - 47y + \mu z = 212, \] has infinitely many solutions, then \( \mu - 2\lambda \) is equal to:

• A. 56
• B. 59
• C. 55
• D. 57

Hint:

To find the condition for infinitely many solutions, compute the determinant of the coefficient matrix and set it equal to zero. This ensures the system is consistent and has infinitely many solutions.

Answer 1

The Correct Option is D

To find the value of \( \mu - 2\lambda \) for the given system of equations that has infinitely many solutions, the system must be consistent and dependent, implying at least one redundant equation. The process is as follows:

1. Consider the system of equations:
   • \( 2x - y + z = 4 \)
   • \( 5x + \lambda y + 3z = 12 \)
   • \( 100x - 47y + \mu z = 212 \)
2. For infinite solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix is:

\[\begin{bmatrix} 2 & -1 & 1 \\ 5 & \lambda & 3 \\ 100 & -47 & \mu \end{bmatrix}\]

1. Set the determinant of this matrix to zero:

\[\text{Det} = 2 \begin{vmatrix} \lambda & 3 \\ -47 & \mu \end{vmatrix} + 1 \begin{vmatrix} 5 & 3 \\ 100 & \mu \end{vmatrix} + 1 \begin{vmatrix} 5 & \lambda \\ 100 & -47 \end{vmatrix}\]

1. Calculate the 2x2 determinants:
   • First determinant:

\[\begin{vmatrix} \lambda & 3 \\ -47 & \mu \end{vmatrix} = \lambda \mu - (-47)(3) = \lambda \mu + 141\]

• Second determinant:

\[\begin{vmatrix} 5 & 3 \\ 100 & \mu \end{vmatrix} = 5\mu - 300\]

• Third determinant:

\[\begin{vmatrix} 5 & \lambda \\ 100 & -47 \end{vmatrix} = 5(-47) - 100\lambda = -235 - 100\lambda\]

1. Substitute these values into the determinant equation:

\[2(\lambda \mu + 141) + (5\mu - 300) + (-235 - 100\lambda) = 0 \] \[ 2\lambda \mu + 282 + 5\mu - 300 - 235 - 100\lambda = 0\]

1. Simplify the equation:

\[2\lambda \mu + 5\mu - 100\lambda + 282 - 300 - 235 = 0 \] \[ 2\lambda \mu + 5\mu - 100\lambda - 253 = 0\]

1. Rearrange the terms:

\[2\lambda \mu + 5\mu - 100\lambda = 253\]

1. Express the equation in terms of \( \mu \):

\[\mu(2\lambda + 5) = 100\lambda + 253\]

1. By comparing coefficients, we find \( \mu = 100 \) and \( 2\lambda = 43 \). Therefore, \( \mu - 2\lambda = 100 - 43 = 57 \).

The result is 57.