x+ky-z=0 , 3x-ky-z=0 and x-3y+z=0 has non-zero solution for k=

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find the value of \( k \) for which the system of equations has a non-zero solution, we need to set up the equations in matrix form and find the determinant. The equations given are:

1. \( x + ky - z = 0 \)
2. \( 3x - ky - z = 0 \)
3. \( x - 3y + z = 0 \)

We can represent this system 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 this system to have a non-zero solution, the determinant of the coefficient matrix must be zero. We will calculate the determinant of the matrix:

\[ D = \begin{vmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{vmatrix} \]

Calculating the determinant using the formula for a 3x3 matrix:

\[ D = 1 \cdot \begin{vmatrix} -k & -1 \\ -3 & 1 \end{vmatrix} - k \cdot \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -k \\ 1 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -k & -1 \\ -3 & 1 \end{vmatrix} = (-k)(1) - (-1)(-3) = -k - 3 \)

2. \( \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} = (3)(1) - (-1)(1) = 3 + 1 = 4 \)

3. \( \begin{vmatrix} 3 & -k \\ 1 & -3 \end{vmatrix} = (3)(-3) - (-k)(1) = -9 + k = k - 9 \)

Now substituting these back into the determinant expression:

\[ D = 1(-k - 3) - k(4) - 1(k - 9) \]

Simplifying this expression:

\[ D = -k - 3 - 4k - k + 9 = -6k + 6 \]

Setting the determinant equal to zero for a non-zero solution:

\[ -6k + 6 = 0 \]

Solving for \( k \):

\[ -6k = -6 \implies k = 1 \]

Thus, the value of \( k \) for which the system has a non-zero solution is:

\[ \boxed{1} \]