Question

The number of solution(s) of the system of linear equations
\[ x + y + z = 0
x - y + z = 0
x + 2y - z = 0 \]

• A. 1
• B. 0
• C. Infinite
• D. More than one but finite

Hint:

When solving systems of linear equations, try elimination or substitution methods to find the values of the variables step-by-step.

Answer 1

The Correct Option is A

We are given a system of 3 linear equations: \[ x + y + z = 0 \tag{1} \] \[ x - y + z = 0 \tag{2} \] \[ x + 2y - z = 0 \tag{3} \] Let’s solve this system using elimination or substitution.
Step 1: Subtract equation (2) from equation (1): \[ (x + y + z) - (x - y + z) = 0 \] \[ 2y = 0 \] Thus, \( y = 0 \).
Step 2: Substitute \( y = 0 \) into equations (1) and (2): Substituting into equation (1): \[ x + 0 + z = 0 \implies x + z = 0 \implies z = -x \tag{4} \] Substituting into equation (2): \[ x - 0 + z = 0 \implies x + z = 0 \implies z = -x \tag{5} \] Since both equations give the same relationship, we now know that \( z = -x \).
Step 3: Substitute \( y = 0 \) and \( z = -x \) into equation (3): Substituting into equation (3): \[ x + 2(0) - (-x) = 0 \implies x + x = 0 \implies 2x = 0 \implies x = 0. \] Thus, \( x = 0 \) and \( z = -x = 0 \). Therefore, \( y = 0 \), and the unique solution is \( x = y = z = 0 \).
Thus, the system has only one solution. The correct answer is (A) 1.