Question

If the system of equations \[ x + ky - z = 0, \quad 3x - ky - z = 0, \quad x - 3y + z = 0, \] has a non-zero solution, then \( k \) is equal to:

• A. \( -1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

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

Answer 1

The Correct Option is C

Step 1: Writing the system of equations as a matrix.
The system of equations can be written in matrix form as follows:
\[ \begin{bmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero.
Step 2: Finding the determinant.
The determinant of the matrix is: \[ \text{det} = \begin{vmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{vmatrix} \] Expanding the determinant and solving gives the condition for \( k \).
Step 3: Solving for \( k \).
After solving the determinant equation, we find that \( k = 1 \) satisfies the condition for a non-zero solution.
Conclusion.
The correct answer is (3) \( 1 \), as \( k = 1 \) results in a non-zero solution.