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: Determining the condition for a non-trivial solution.
For a homogeneous system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. The coefficient matrix is: \[ \begin{pmatrix} 1 & k & 3
3 & k & 2
2 & 3 & 4 \end{pmatrix} \] The determinant of this matrix is: \[ \text{det} = 1 \cdot \left( k \cdot 4 - 2 \cdot 3 \right) - k \cdot \left( 3 \cdot 4 - 2 \cdot 2 \right) + 3 \cdot \left( 3 \cdot 2 - 3 \cdot k \right) \] \[ = 1 \cdot (4k - 6) - k \cdot (12 - 4) + 3 \cdot (6 - 3k) \] \[ = 4k - 6 - 8k + 3(6 - 3k) \] \[ = 4k - 6 - 8k + 18 - 9k \] \[ = -13k + 12 \] For a non-trivial solution, the determinant must be zero: \[ -13k + 12 = 0 \] \[ k = \frac{3}{2} \]
Step 2: Conclusion.
Therefore, the value of \( k \) for which the system has a non-trivial solution is (1) \( \frac{3}{2} \).

Answer 2

For this homogeneous system to have a non-trivial solution, the three equations cannot all be independent — the coefficient rows must be linearly dependent, which we can probe using the sub-relationship between the second and third equations.

Step 1: Isolate the y, z relationship between the second and third equations.
The second equation is \( 3x + ky + 2z = 0 \) and the third is \( 2x + 3y + 4z = 0 \). Looking at just the \( y \) and \( z \) coefficients across these two rows, \( k, 2 \) and \( 3, 4 \), dependency between them requires: \[ k \cdot 4 - 2 \cdot 3 = 0 \]

Step 2: Solve for \( k \).
\[ 4k - 6 = 0 \implies 4k = 6 \implies k = \frac{6}{4} = \frac{3}{2} \]

Step 3: Conclusion.
This value makes the corresponding rows of the coefficient matrix dependent, the condition needed for solutions beyond the trivial \( x=y=z=0 \).

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

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