Question

For what value of α, the system of the following equations will have infinitely many solutions:

x+y+z=0 x-y+z=0 3x + y + αz = 0

• A. α = -2
• B. a = 6
• C. α = -3
• D. a = 4

Answer 1

The system has infinitely many solutions only when α = 3. (None of the provided options is correct.)

Answer 2

We are given the system:

$$\begin{aligned} x+y+z &= 0\quad\quad(1)\\[6pt] x-y+z &= 0\quad\quad(2)\\[6pt] 3x+y+\alpha z &= 0\quad(3) \end{aligned}$$

Step 1: Subtract (2) from (1):

$$(x+y+z) - (x-y+z) = 0 \quad\Rightarrow\quad 2y = 0 \quad\Rightarrow\quad y = 0$$

Step 2: Substitute $y = 0$ in (1):

$$x + z = 0 \quad\Rightarrow\quad x = -z$$

Step 3: Substitute $x = -z$ and $y = 0$ in (3):

$$3(-z)+0+\alpha z = (-3+\alpha)z = 0$$

For the system to have infinitely many solutions, this equation must hold for all $z$ (i.e. be an identity). This requires:

$$\alpha - 3 = 0 \quad\Rightarrow\quad \alpha = 3.$$

However, none of the given options ($\alpha = -2, -3, 6, 4$) match $\alpha = 3$.

Minimal Explanation:
Subtract (2) from (1) to get $y=0$; then $x=-z$. In (3), substitute to get $(\alpha-3)z=0$; for infinitely many solutions, $\alpha-3=0$, so $\alpha=3$.

Final Answer:
The system has infinitely many solutions only when $\alpha = 3$. (None of the provided options is correct.)