Question

The value of $\lambda$ such that the system $x - 2y + z = -4, 2x - y + 2z = 2$ and $x + y + \lambda z = 4$ has no solution is

• A. 3
• B. 1
• C. 0
• D. -3

Answer 1

Answer: (B)

1

Answer 2

Given the system

$x - 2y + z = -4$, $2x - y + 2z = 2$, $x + y + \lambda z = 4$,

express $x$ from the first equation:

$x = -4 + 2y - z$.

Substitute into the second equation:

$2(-4+2y-z) - y + 2z = 2 \implies -8 + 4y - 2z - y + 2z = 2$,

which simplifies to

$3y = 10 \implies y = \frac{10}{3}$.

Then,

$x = -4 + 2(\frac{10}{3}) - z = \frac{8}{3} - z$.

Substitute $x$ and $y$ into the third equation:

$(\frac{8}{3}-z) + \frac{10}{3} + \lambda z = 4 \implies \frac{18}{3} + (\lambda -1)z = 4$,

i.e.,

$6 + (\lambda -1)z = 4 \implies (\lambda -1)z = -2$.

For a unique solution there exists a value of $z$ given by $z = \dfrac{-2}{\lambda -1}$ provided $\lambda \neq 1$. If $\lambda = 1$, we get

$0 \cdot z = -2$,

which is impossible. Hence, the system has no solution for

$\lambda = 1$.