Question

If the system of linear equations
2x + 3y – z = –2
x + y + z = 4
x – y + |λ|z = 4λ – 4
where λ ∈ R, has no solution, then

• A. λ = 7
• B. λ = -7
• C. λ = 8
• D. λ² = 1

Answer 1

The Correct Option is B

 To determine the value of \(\lambda\) for which the given system of linear equations has no solution, we will analyze the equations:

1. The system of equations is: \(2x + 3y - z = -2\)
   \(x + y + z = 4\)
   \(x - y + |\lambda|z = 4\lambda - 4\)
2. First, observe that a system of linear equations has no solution if the equations are inconsistent. This usually happens when two planes (equations) are parallel and the third one does not intersect at the same point.
3. Substituting each equation to find any inconsistencies:
   • From the second equation, express \(z\) as \(z = 4 - x - y\).
   • Substitute this into the first equation: \(2x + 3y - (4 - x - y) = -2\)
     Simplify: \(2x + 3y - 4 + x + y = -2\)
     \(3x + 4y - 4 = -2\)
     \(3x + 4y = 2\) ...(i)
   • From the third equation: \(x - y + |\lambda|z = 4\lambda - 4\)
     Substitute for \(z\): \(x - y + |\lambda|(4 - x - y) = 4\lambda - 4\)
     Simplify: \(x - y + 4|\lambda| - |\lambda|x + |\lambda|y = 4\lambda - 4\)
     Rearrange to get: \((1 - |\lambda|)x + (|\lambda| - 1)y = 4\lambda - 4 - 4|\lambda|\)
     Further simplification gives us: \((1 - |\lambda|)x + (|\lambda| - 1)y = -4\beta\) where \(\beta = \lambda\)
4. For the system to be inconsistent, identify conditions where equations (i) and the derived equation cannot be satisfied together:
   • Equations become identical when \(|\lambda| = 1\), leading to contradiction under different constants.
   • Thus, solve for \(\lambda\), setting\)
   • For exact inconsistency \(\lambda = -7\) proves the system cannot be solved as proposed.

Thus, the correct option making the system inconsistent is \(\lambda = -7\). When checked with the equations, \(\lambda \neq 1\) proves valid elsewhere.