Question

The system of equations $2x + y = 2z$, $y + z = 0$, $3y + \alpha z = x$ has:

• A. No solution for any value of $\alpha$
• B. Infinite number of solutions for $\alpha = \frac{7}{2}$
• C. Only trivial solution for $\alpha = \frac{9}{2}$
• D. Infinite solutions of the form $(3k, -2k, 2k)$

Hint:

When options give a specific parametric solution form like $(3k, -2k, 2k)$, directly plug it back into the easiest equations to verify if it satisfies them identically. This saves the time needed to compute complex determinants.

Answer 1

The Correct Option is D

Concept: First, let us rewrite the given system of linear equations in standard homogeneous form $AX = 0$: 1) $2x + y - 2z = 0$ 2) $0x + y + z = 0$ 3) $-x + 3y + \alpha z = 0$ A homogeneous system always admits the trivial solution $(0, 0, 0)$. For non-trivial (infinite) solutions to exist, the determinant of the coefficient matrix $A$ must equal zero ($\left|A\right| = 0$). Alternatively, we can analyze the structure of the solutions directly from the simplest equations to check the validity of the given option forms.

Step 1: Analyze the relationship between variables using the first two equations.
From equation (2): \[ y + z = 0 \implies y = -z \] Let us substitute $y = -z$ into equation (1): \[ 2x + (-z) - 2z = 0 \implies 2x - 3z = 0 \implies 2x = 3z \implies x = \frac{3}{2}z \]

Step 2: Express the solution in parameterized form.
Let $z = 2k$, where $k$ is any real number parameter. Then: \[ y = -z = -2k \] \[ x = \frac{3}{2}(2k) = 3k \] Thus, any solution to the first two equations must have the general structure: \[ (x, y, z) = (3k, -2k, 2k) \]

Step 3: Substitute this parametric form into the third equation to analyze consistency and $\alpha$.
The third equation is $-x + 3y + \alpha z = 0$. Substituting $x = 3k$, $y = -2k$, and $z = 2k$: \[ -(3k) + 3(-2k) + \alpha(2k) = 0 \] \[ -3k - 6k + 2\alpha k = 0 \implies -9k + 2\alpha k = 0 \implies k(-9 + 2\alpha) = 0 \] For non-trivial solutions to exist for non-zero $k$, we must have: \[ -9 + 2\alpha = 0 \implies \alpha = \frac{9}{2} \] Let us check the given choices: - Option (B) states infinite solutions for $\alpha = \frac{7}{2}$, which is incorrect because $\alpha$ must be $\frac{9}{2}$ for non-trivial solutions. - Option (C) says "Only trivial solution for $\alpha = \frac{9}{2}$", which contradicts the fact that $\alpha = \frac{9}{2}$ creates infinite non-trivial solutions. - Option (D) asserts that the system has infinite solutions of the form $(3k, -2k, 2k)$. This completely matches our structural parametric derivation regardless of checking specific constraints on option boundaries first.