Question

If the system of equations \(x + 4y - z = \lambda\), \(7x + 9y + \mu z = -3\), \(5x + y + 2z = -1\) has infinitely many solutions, then \((2\mu + 3\lambda)\) is equal to:

• A. 2
• B. -3
• C. 3
• D. -2

Answer 1

The Correct Option is B

Given the system of equations:

\[ x + 4y - z = \lambda, \quad 7x + 9y + \mu z = -3, \quad 5x + y + 2z = -1, \] 

we form the coefficient matrix: \[ \Delta = \begin{vmatrix} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{vmatrix}. \] 

For the system to have infinitely many solutions, the determinant of this matrix must be zero: \[ \Delta = 1 \cdot \begin{vmatrix} 9 & \mu \\ 1 & 2 \end{vmatrix} - 4 \cdot \begin{vmatrix} 7 & \mu \\ 5 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 7 & 9 \\ 5 & 1 \end{vmatrix}. \] 

Calculating each term: \[ \Delta = 1 \cdot (18 - \mu) - 4 \cdot (14 - 5\mu) - (7 - 45). \] 

Simplifying: \[ \Delta = 18 - \mu - 4(14 - 5\mu) - (-38). \] \[ \Delta = 18 - \mu - 56 + 20\mu + 38. \] \[ \Delta = 19\mu + 0. \] 

For the determinant to be zero (infinitely many solutions): \[ 19\mu = 0 \implies \mu = 0. \] Substituting \( \mu = 0 \) into the augmented matrix and setting \( \Delta_x = \Delta_y = \Delta_z = 0 \), 

we find: \[ \Delta_x = \begin{vmatrix} \lambda & 4 & -1 \\ -3 & 9 & 0 \\ -1 & 1 & 2 \end{vmatrix} = 0. \] 

Expanding: \[ \lambda \cdot \begin{vmatrix} 9 & 0 \\ 1 & 2 \end{vmatrix} - 4 \cdot \begin{vmatrix} -3 & 0 \\ -1 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} -3 & 9 \\ -1 & 1 \end{vmatrix}. \] 

Calculating each term: \[ \lambda (18) - 4(-6) - (-12) = 0. \] Simplifying: \[ 18\lambda + 24 + 12 = 0. \] \[ 18\lambda = -36 \implies \lambda = -1. \] 

Finally, calculating \( 2\mu + 3\lambda \): \[ 2\mu + 3\lambda = 2(0) + 3(-1) = -3. \] 

Therefore: \[ -3. \]