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 given system of equations is: \[ 2x + \lambda y + 3z = 5 3x + 2y - z = 7 \\4x + 5y + \mu z = 9 \] To check for infinitely many solutions, we use the determinant of the coefficient matrix. The coefficient matrix is: \[ \Delta = \begin{vmatrix} 2 & \lambda & 3 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} \] \[ \Delta = 0 \quad \text{(for infinitely many solutions)} \] Expanding the determinant: \[ \Delta = 2 \begin{vmatrix} 2 & -1 5 & \mu \end{vmatrix} - \lambda \begin{vmatrix} 3 & -1 \\ 4 & \mu \end{vmatrix} + 3 \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} \] \[ = 2(\lambda \mu - (-5)) - \lambda (3 \mu - (-4)) + 3(15 - 8) \] \[ = 2(\lambda \mu + 5) - \lambda (3 \mu + 4) + 3(7) \] \[ = 2\lambda \mu + 10 - \lambda (3 \mu + 4) + 21 \] \[ = 2\lambda \mu + 10 - \lambda 3 \mu - 4 \lambda + 21 \] After solving the system, we find: \[ \Delta_3 = 0 \quad \text{and} \quad 2(7) + \lambda(1) + 5(7) = 0 \] Solving for \( \lambda \) and \( \mu \), we find \( \lambda = -1 \) and \( \mu = -5 \). Hence, \[ \lambda^2 + \mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26 \]

Answer 2

The problem provides a system of three linear equations in three variables, \(x, y, z\), and asks for the value of \( \lambda^2 + \mu^2 \) given that the system has infinitely many solutions.

Concept Used:

For a system of linear equations of the form \( AX = B \), the condition for having infinitely many solutions is that the determinant of the coefficient matrix, \( \Delta \), must be zero, and the determinants \( \Delta_x, \Delta_y, \text{and } \Delta_z \) must also be zero. These determinants are defined as:

\[ \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}, \quad \Delta_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}, \quad \Delta_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}, \quad \Delta_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix} \]

For infinitely many solutions, we must have \( \Delta = \Delta_x = \Delta_y = \Delta_z = 0 \).

Step-by-Step Solution:

Step 1: Write down the determinant of the coefficient matrix, \( \Delta \), and set it to zero.

The given system of equations is:

• \( 2x + \lambda y + 3z = 5 \)
• \( 3x + 2y - z = 7 \)
• \( 4x + 5y + \mu z = 9 \)

The coefficient matrix is \( A = \begin{pmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{pmatrix} \). Its determinant is:

\[ \Delta = \begin{vmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} = 0 \]

Expanding along the first row:

\[ 2(2\mu - (-5)) - \lambda(3\mu - (-4)) + 3(15 - 8) = 0 \] \[ 2(2\mu + 5) - \lambda(3\mu + 4) + 3(7) = 0 \] \[ 4\mu + 10 - 3\lambda\mu - 4\lambda + 21 = 0 \] \[ 4\mu - 4\lambda - 3\lambda\mu + 31 = 0 \quad \cdots (1) \]

Step 2: Calculate the determinant \( \Delta_z \) and set it to zero to find \( \lambda \).

We replace the third column (coefficients of z) with the constant terms:

\[ \Delta_z = \begin{vmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{vmatrix} = 0 \]

Expanding along the first row:

\[ 2(2 \cdot 9 - 7 \cdot 5) - \lambda(3 \cdot 9 - 7 \cdot 4) + 5(3 \cdot 5 - 2 \cdot 4) = 0 \] \[ 2(18 - 35) - \lambda(27 - 28) + 5(15 - 8) = 0 \] \[ 2(-17) - \lambda(-1) + 5(7) = 0 \] \[ -34 + \lambda + 35 = 0 \] \[ \lambda + 1 = 0 \implies \lambda = -1 \]

Step 3: Substitute the value of \( \lambda \) into equation (1) to find \( \mu \).

Using \( \lambda = -1 \) in the equation \( 4\mu - 4\lambda - 3\lambda\mu + 31 = 0 \):

\[ 4\mu - 4(-1) - 3(-1)\mu + 31 = 0 \] \[ 4\mu + 4 + 3\mu + 31 = 0 \] \[ 7\mu + 35 = 0 \] \[ 7\mu = -35 \] \[ \mu = -5 \]

Step 4: Calculate the final required value, \( \lambda^2 + \mu^2 \).

We have found \( \lambda = -1 \) and \( \mu = -5 \).

\[ \lambda^2 + \mu^2 = (-1)^2 + (-5)^2 \] \[ = 1 + 25 \] \[ = 26 \]

Thus, for the system to have infinitely many solutions, the value of \( \lambda^2 + \mu^2 \) is 26.