1.If the system of linear equations2x+3y−z=−2x+y+z=4x−y+λz=4λ−4\begin{aligned} 2x + 3y - z &= -2 \\ x + y + z &= 4 \\ x - y + \lambda z &= 4\lambda - 4 \end{aligned}2x+3y−zx+y+zx−y+λz=−2=4=4λ−4where λ∈R\lambda \in \mathbb{R}λ∈R, has no solution, then:a.λ=7\lambda = 7λ=7b.λ=−7\lambda = -7λ=−7c.λ=8\lambda = 8λ=8d.λ2=1\lambda^2 = 1λ2=1Login to continueOnly logged in users canattempt or see the solution.
