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 determine 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. \( x - 3y + z = 0 \)

### Step 1: Write the system of equations in matrix form
We can represent the above equations in the form of a matrix \( A \) as follows:

\[ A = \begin{bmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{bmatrix} \]

### Step 2: Set up the determinant
For the system to have a non-zero solution, the determinant of the coefficient matrix must be equal to zero:

\[ \text{det}(A) = 0 \]

### Step 3: Calculate the determinant
We can calculate the determinant of matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -k & -1 \\ -3 & 1 \end{vmatrix} - k \cdot \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -k \\ 1 & -3 \end{vmatrix} \]

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

Substituting back into the determinant equation:

\[ \text{det}(A) = 1(-k - 3) - k(4) - 1(k - 9) \]

Expanding this gives:

\[ -k - 3 - 4k - k + 9 = 0 \]

Combining like terms:

\[ -6k + 6 = 0 \]

### Step 4: Solve for \( k \)
Now, we can solve for \( k \):

\[ -6k + 6 = 0 \implies -6k = -6 \implies k = 1 \]

### Conclusion
The value of \( k \) for which the system of equations has a non-zero solution is:

\[ \boxed{1} \]