Question

The number of real values $\lambda$, such that the system of linear equations $2 x-3 y+5 z=9$ $x+3 y-z=-18$ $3 x-y+\left(\lambda^2-|\lambda|\right) z=16$ has no solution, is :-

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A)

0

Answer 2

2x−3y+5z=9
x+3y−z=−18
3x−y+(λ2−∣λ∣)z=16
D=∣∣​213​−33−1​5−1λ2−∣λ​∣∣​=0
⇒3λ2−3∣λ∣−11=0
Clearly one negative and one positive root since ∣λ∣ is there so negative not possible and two values of λ corresponding to positive value
D3​=∣∣​213​−33−1​9−1816​∣∣​=0 so no solution.