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:
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:
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- 22
- 18
- 26
- 30
The Correct Option is C
Approach Solution - 1
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, \[ \lambda^2 + \mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26 \]
Approach Solution -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.
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