Question

The non-trivial solutions of the equations:
\[ x + y - 6z = 0 \] \[ -3x + y + 2z = 0 \] \[ x - y + 2z = 0 \]

• A. $x = 2c, y = 4c, z = 3c, c \neq 0$ is any scalar.
• B. $x = 2c, y = 4c, z = c, c \neq 0$ is any scalar.
• C. $x = 2c, y = 4c, z = 2c, c \neq 0$ is any scalar.
• D. $x = c, y = 4c, z = 2c, c \neq 0$ is any scalar.

Hint:

To quickly check options in competitive exams, substitute $x=2c, y=4c, z=c$ directly into the equations: $2c + 4c - 6c = 0 \quad \checkmark$ $-3(2c) + 4c + 2c = 0 \quad \checkmark$ $2c - 4c + 2c = 0 \quad \checkmark$ This confirms the result rapidly without doing full matrix reduction.

Answer 1

The Correct Option is B

Step 1: Concept
A system of linear homogeneous equations $A X = 0$ always possesses the trivial solution $x = y = z = 0$.
Non-trivial solutions exist if and only if the coefficient matrix $A$ is singular, i.e., $\det(A) = 0$, which implies that the rank of $A$ is less than the number of variables.

Step 2: Key Formulas and Approach
Write the system in matrix form $A X = 0$: \[ A = \begin{pmatrix} 1 & 1 & -6 -3 & 1 & 2 1 & -1 & 2 \end{pmatrix} \] Apply Gaussian elimination (row reduction) to obtain the Row Echelon Form of $A$ and solve for the free variable(s).

Step 3: Step-by-step Explanation

• Form the augmented matrix $[A \mid 0]$: \[ \begin{pmatrix} 1 & 1 & -6 & \mid & 0 -3 & 1 & 2 & \mid & 0 1 & -1 & 2 & \mid & 0 \end{pmatrix} \]
• Apply elementary row operations: $R_2 \to R_2 + 3R_1$: \[ \begin{pmatrix} 1 & 1 & -6 & \mid & 0 0 & 4 & -16 & \mid & 0 1 & -1 & 2 & \mid & 0 \end{pmatrix} \] $R_3 \to R_3 - R_1$: \[ \begin{pmatrix} 1 & 1 & -6 & \mid & 0 0 & 4 & -16 & \mid & 0 0 & -2 & 8 & \mid & 0 \end{pmatrix} \]
• Simplify $R_2$ by dividing by $4$ ($R_2 \to \frac{1}{4}R_2$): \[ \begin{pmatrix} 1 & 1 & -6 & \mid & 0 0 & 1 & -4 & \mid & 0 0 & -2 & 8 & \mid & 0 \end{pmatrix} \]
• Apply $R_3 \to R_3 + 2R_2$: \[ \begin{pmatrix} 1 & 1 & -6 & \mid & 0 0 & 1 & -4 & \mid & 0 0 & 0 & 0 & \mid & 0 \end{pmatrix} \]
• The rank of $A$ is $2$, which is less than $3$ (number of variables). Hence, there is $3 - 2 = 1$ free variable.

• Express $x$ and $y$ in terms of $z$:
From $R_2$: $y - 4z = 0 \implies y = 4z$.
From $R_1$: $x + y - 6z = 0 \implies x + 4z - 6z = 0 \implies x - 2z = 0 \implies x = 2z$.

• Assign a parameter $c \in \mathbb{R}$ ($c \neq 0$ for non-trivial solutions) to $z$: \[ z = c, \quad y = 4c, \quad x = 2c \]

Step 4: Final Answer
The non-trivial solution set is given by $x = 2c, y = 4c, z = c$ for any non-zero scalar $c$. Thus, Option (B) is correct.