Question

The equations \( \lambda x - y = 2 \), \( 2x - 3y = -\lambda \), and \( 3x - 2y = -1 \) are consistent for:

• A. \( \lambda = -4 \)
• B. \( \lambda = 1, 4 \)
• C. \( \lambda = 1, -4 \)
• D. \( \lambda = -1, 4 \)
• E. \( \lambda = -1 \)

Hint:

Another way to solve this is to find the intersection point \((x, y)\) of the two equations that do not contain \(\lambda\), then substitute that point into the third equation to solve for \(\lambda\).

Answer 1

The Correct Option is C

Concept: A system of three linear equations in two variables is said to be consistent if all three lines intersect at a common point. For this to happen, the determinant of the augmented matrix (using the coefficients and the constants) must be equal to zero. If the lines are represented as \( a_i x + b_i y + c_i = 0 \), the condition for consistency is: \[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \]

Step 1: Set up the determinant for consistency.
Rewrite the equations in the form \( ax + by + c = 0 \): 1. \( \lambda x - y - 2 = 0 \) 2. \( 2x - 3y + \lambda = 0 \) 3. \( 3x - 2y + 1 = 0 \) The determinant is: \[ \begin{vmatrix} \lambda & -1 & -2 \\ 2 & -3 & \lambda 3 & -2 & 1 \end{vmatrix} = 0 \]

Step 2: Expand the determinant to form a quadratic equation in \( \lambda \).
Expanding along the first row: \[ \lambda[(-3)(1) - (\lambda)(-2)] - (-1)[(2)(1) - (\lambda)(3)] + (-2)[(2)(-2) - (-3)(3)] = 0 \] \[ \lambda[-3 + 2\lambda] + 1[2 - 3\lambda] - 2[-4 + 9] = 0 \] \[ 2\lambda^2 - 3\lambda + 2 - 3\lambda - 2(5) = 0 \] \[ 2\lambda^2 - 6\lambda - 8 = 0 \]

Step 3: Solve for \( \lambda \).
Divide the entire equation by 2: \[ \lambda^2 - 3\lambda - 4 = 0 \] Factoring the quadratic: \[ (\lambda - 4)(\lambda + 1) = 0 \] \[ \lambda = 4, \, -1 \]