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

$$\begin{cases} x-2y+z=-4\quad(1)\\[6mm] 2x-y+2z=2\quad(2)\\[6mm] x+y+\lambda z=4\quad(3) \end{cases}$$

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

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

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

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

which simplifies to

$$3y-8=2\quad\Rightarrow\quad y=\frac{10}{3}.$$

Step 3. Now substitute $x=-4+2y-z$ and $y=\frac{10}{3}$ in (3):

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

Substitute the value of $y$:

$$-4+2\left(\frac{10}{3}\right)-z+\frac{10}{3}+\lambda z=4.$$

Combine constant and $z$ terms:

$$-4+\frac{20}{3}+\frac{10}{3}+(\lambda-1)z=4.$$

Since $\frac{20+10}{3}=\frac{30}{3}=10$, the equation becomes:

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

Thus,

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

Step 4. For a solution to exist when $\lambda\neq 1$, we can solve for $z$. But when $\lambda = 1$, the equation becomes:

$$0\cdot z=-2,$$

which is inconsistent. Hence, the system has no solution when

$$\lambda=1.$$