Question

If the system of linear equations
$2x+3y−z=−2$
$x+y+z=4$
$x−y+âˆ\pounds λâˆ\pounds z=4λ−4$
where $λ∈R$, has no solution, then

• A. $λ=7$
• B. $λ=−7$
• C. $λ=8$
• D. ${λ}^2=1$

Answer 1

Answer: (B)

$â\begin{array}{ccc}2& 3& −1\\ 1& 1& 1\\ 1& −1& âˆ\pounds λâˆ\pounds <br/>\end{array}â=0$
$⇒âˆ\pounds λâˆ\pounds =7⇒λ=Â\pm 7$ ...(1)
System :
$2x+3y−z=−2$ ...(2)
$x+y+z=4$ ...(3)
$x−y+âˆ\pounds λâˆ\pounds z=4λ−4$ ...(4)
Eliminating y from equal (2) & (3) we get
$x+4z=14$ ...(5)
$(3)+(4)⇒x+\left(\frac{âˆ\pounds λâˆ\pounds +1}{2}\right)z=2λ$ ...(6)
Clearly for $λ=−7$, system is inconsistent.