Question:
If the system of simultaneous linear equations \( x + \lambda y - 2z = 1 \), \( x - y + \lambda z = 2 \) and \( x - 2y + 3z = 3 \) is inconsistent for \( \lambda = \lambda_1 \) and \( \lambda_2 \), then \( \lambda_1 + \lambda_2 = \)
If the system of simultaneous linear equations \( x + \lambda y - 2z = 1 \), \( x - y + \lambda z = 2 \) and \( x - 2y + 3z = 3 \) is inconsistent for \( \lambda = \lambda_1 \) and \( \lambda_2 \), then \( \lambda_1 + \lambda_2 = \)
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When a question asks for the sum or product of parameters that make a system inconsistent, don't waste time solving for the exact roots. Form the quadratic equation for \( \Delta = 0 \) and use Vieta's formulas (\( \text{Sum} = -b/a \)) to get the answer instantly.
Updated On: Jun 3, 2026
- \( 5 \)
- \( \sqrt{5} \)
- \( 1 \)
- \( -1 \)
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The Correct Option is C
Solution and Explanation
Concept:
A system of non-homogeneous linear equations is inconsistent if the determinant of the coefficient matrix (\( \Delta \)) is zero, but at least one of the determinants \( \Delta_x, \Delta_y, \) or \( \Delta_z \) is non-zero. For this problem, we first find the values of \( \lambda \) that make \( \Delta = 0 \).
Step 1: Find the values of \( \lambda \) for which the determinant \( \Delta = 0 \).
The coefficient matrix is: \[ \Delta = \begin{vmatrix} 1 & \lambda & -2 1 & -1 & \lambda 1 & -2 & 3 \end{vmatrix} \] Expanding along the first row: \[ \Delta = 1(-3 + 2\lambda) - \lambda(3 - \lambda) - 2(-2 + 1) = 0 \] \[ -3 + 2\lambda - 3\lambda + \lambda^2 + 2 = 0 \] \[ \lambda^2 - \lambda - 1 = 0 \]
Step 2: Apply the properties of quadratic equations.
The values \( \lambda_1 \) and \( \lambda_2 \) are the roots of the quadratic equation \( \lambda^2 - \lambda - 1 = 0 \). For any quadratic equation \( a\lambda^2 + b\lambda + c = 0 \), the sum of the roots is given by \( -\frac{b}{a} \). Here, \( a = 1 \) and \( b = -1 \). \[ \lambda_1 + \lambda_2 = -\frac{-1}{1} = 1 \]
Step 3: Verify inconsistency.
For inconsistency, we must ensure \( \Delta_x, \Delta_y, \text{ or } \Delta_z \neq 0 \) for these roots. \[ \Delta_x = \begin{vmatrix} 1 & \lambda & -2 2 & -1 & \lambda 3 & -2 & 3 \end{vmatrix} = 1(-3 + 2\lambda) - \lambda(6 - 3\lambda) - 2(-4 + 3) \] \[ \Delta_x = -3 + 2\lambda - 6\lambda + 3\lambda^2 + 2 = 3\lambda^2 - 4\lambda - 1 \] Substituting \( \lambda^2 = \lambda + 1 \) from our previous equation: \[ \Delta_x = 3(\lambda + 1) - 4\lambda - 1 = 3\lambda + 3 - 4\lambda - 1 = 2 - \lambda \] Since the roots of \( \lambda^2 - \lambda - 1 = 0 \) are \( \frac{1 \pm \sqrt{5}}{2} \), \( \Delta_x \neq 0 \). Thus, the system is inconsistent for these values.
Step 1: Find the values of \( \lambda \) for which the determinant \( \Delta = 0 \).
The coefficient matrix is: \[ \Delta = \begin{vmatrix} 1 & \lambda & -2 1 & -1 & \lambda 1 & -2 & 3 \end{vmatrix} \] Expanding along the first row: \[ \Delta = 1(-3 + 2\lambda) - \lambda(3 - \lambda) - 2(-2 + 1) = 0 \] \[ -3 + 2\lambda - 3\lambda + \lambda^2 + 2 = 0 \] \[ \lambda^2 - \lambda - 1 = 0 \]
Step 2: Apply the properties of quadratic equations.
The values \( \lambda_1 \) and \( \lambda_2 \) are the roots of the quadratic equation \( \lambda^2 - \lambda - 1 = 0 \). For any quadratic equation \( a\lambda^2 + b\lambda + c = 0 \), the sum of the roots is given by \( -\frac{b}{a} \). Here, \( a = 1 \) and \( b = -1 \). \[ \lambda_1 + \lambda_2 = -\frac{-1}{1} = 1 \]
Step 3: Verify inconsistency.
For inconsistency, we must ensure \( \Delta_x, \Delta_y, \text{ or } \Delta_z \neq 0 \) for these roots. \[ \Delta_x = \begin{vmatrix} 1 & \lambda & -2 2 & -1 & \lambda 3 & -2 & 3 \end{vmatrix} = 1(-3 + 2\lambda) - \lambda(6 - 3\lambda) - 2(-4 + 3) \] \[ \Delta_x = -3 + 2\lambda - 6\lambda + 3\lambda^2 + 2 = 3\lambda^2 - 4\lambda - 1 \] Substituting \( \lambda^2 = \lambda + 1 \) from our previous equation: \[ \Delta_x = 3(\lambda + 1) - 4\lambda - 1 = 3\lambda + 3 - 4\lambda - 1 = 2 - \lambda \] Since the roots of \( \lambda^2 - \lambda - 1 = 0 \) are \( \frac{1 \pm \sqrt{5}}{2} \), \( \Delta_x \neq 0 \). Thus, the system is inconsistent for these values.
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