Question

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

• A. 22
• B. 18
• C. 26
• D. 30

Hint:

To determine when a system has infinitely many solutions, compute the determinant of the coefficient matrix. If the determinant is zero, the system has infinitely many solutions. For such systems, use the conditions derived from the matrix to find the values of the parameters \( \lambda \) and \( \mu \).

Answer 1

The Correct Option is C

The provided system of equations is: \[ 2x + \lambda y + 3z = 5 3x + 2y - z = 7 \\4x + 5y + \mu z = 9 \] To determine if there are infinitely many solutions, we examine the determinant of the coefficient matrix. The coefficient matrix is: \[ \Delta = \begin{vmatrix} 2 & \lambda & 3 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} \] For infinitely many solutions, the determinant must be zero: \[ \Delta = 0 \] Expanding the determinant yields: \[ \Delta = 2 \begin{vmatrix} 2 & -1 5 & \mu \end{vmatrix} - \lambda \begin{vmatrix} 3 & -1 \\ 4 & \mu \end{vmatrix} + 3 \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} \] \[ = 2(2\mu - (-5)) - \lambda (3\mu - (-4)) + 3(15 - 8) \] \[ = 2(2\mu + 5) - \lambda (3\mu + 4) + 3(7) \] \[ = 4\mu + 10 - 3\lambda\mu - 4\lambda + 21 \] \[ = 4\mu - 3\lambda\mu - 4\lambda + 31 \] After solving the system, we find that \( \Delta_3 = 0 \) and the condition \( 2(7) + \lambda(1) + 5(7) = 0 \) must hold. Solving for \( \lambda \) and \( \mu \), we determine \( \lambda = -1 \) and \( \mu = -5 \). Therefore, \[ \lambda^2 + \mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26 \]