Question

The value of \( k \) for which the system of equations \( x + ky + 3z = 0 \); \( 3x + ky + 2z = 0 \); \( 2x + 3y + 4z = 0 \) has a non-trivial solution is

• A. \( \frac{3}{2} \)
• B. \( \frac{33}{2} \)
• C. 15
• D. -27

Hint:

For a homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

Answer 1

The Correct Option is A

Step 1: Eliminate \( x \) from equations 2 and 3 using equation 1 (\( x+ky+3z=0 \)): from equation 3, \( R_3 - 2R_1 \) gives \( (3-2k)y - 2z = 0 \).
Step 2: For a non-trivial solution to exist, this reduced relation between \( y \) and \( z \) must itself be degenerate, requiring \( 3 - 2k = 0 \).
Step 3: Solving, \( 2k = 3 \), so \( k = \frac{3}{2} \).
\[ \boxed{k = \frac{3}{2}} \]

Answer 2

Another way to approach this is to notice that a homogeneous system has a non-trivial solution when some pair of rows in the coefficient matrix becomes proportional. Comparing the \( x \) and \( y \) coefficients of the first equation \( (1, k) \) with those of the third equation \( (2, 3) \):

\[ \frac{1}{2} = \frac{k}{3} \]

Cross-multiplying gives \( 3 = 2k \), so \( k = \frac{3}{2} \). This proportionality between the first and third rows' \( x, y \) entries is exactly the kind of degeneracy that permits a non-trivial solution for the system.

Therefore, the correct answer is \( \frac{3}{2} \).