Question

If the system of linear equations \( x+y+z=2 \), \( 2x+y-z=3 \) and \( 3x+2y+kz=4 \) has a unique solution, then:

• A. \( k = 0 \)
• B. \( -1 \lt k \lt 1 \)
• C. \( 0 \lt k \lt 1 \)
• D. \( k \neq 0 \)

Hint:

Cramer's Rule states that if \( \Delta \neq 0 \), a unique solution exists. Always start by finding the determinant of the coefficients.

Answer 1

The Correct Option is D

Step 1: Reduce the system by elimination.
Take the equations \( x+y+z=2 \), \( 2x+y-z=3 \) and \( 3x+2y+kz=4 \). Subtracting twice the first equation from the second and three times the first from the third removes \( x \) from the last two rows, giving \[ -y-3z=-1 \quad \text{and} \quad -y+(k-3)z=-2 \]

Step 2: Eliminate \( y \) as well.
Subtracting these two reduced equations removes \( y \), leaving a single equation in \( z \): \[ kz = -1 \]

Step 3: Decide when this gives one value of \( z \).
This equation has a unique solution for \( z \) only when its coefficient is non-zero. If \( k = 0 \), the equation becomes \( 0 = -1 \), which is impossible, so a unique solution fails exactly when \( k = 0 \). \[ \boxed{k \neq 0} \]