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

We are given the system:

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

Step 1: Solve (1) for $x$:

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

Step 2: Substitute $x$ in (2):

$2(-4+2y-z) - y + 2z = 2 \Rightarrow -8 + 4y -2z - y +2z = 2$.

Simplify:

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

Step 3: Substitute $x$ and $y$ into (3):

$(-4+2y-z) + y + \lambda z = 4 \Rightarrow -4 +3y + (\lambda -1)z =4$.

Plug $y=\frac{10}{3}$:

$-4 + 10 + (\lambda-1)z = 4 \Rightarrow 6 + (\lambda-1)z = 4$.

Thus:

$(\lambda-1)z = -2$.

For the system to have no solution, the coefficient of $z$ must vanish while the constant term remains nonzero, i.e.,

$\lambda - 1 = 0$ and $-2 \neq 0$.

This gives:

$\lambda = 1$.