What should be the value of k so that the system of linear equations `x - y + 2z = 0, kx - y + z = 0, 3x + y - 3z = 0` does not possess a unique solution ?

Options

A. 0
B. 3
C. 4
D. 5

Explanation

To determine the value of \( k \) such that the system of linear equations does not possess a unique solution, we can follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x - y + 2z = 0 \)
2. \( kx - y + z = 0 \)
3. \( 3x + y - 3z = 0 \)

We can represent this system in matrix form as:
\[ \begin{bmatrix} 1 & -1 & 2 \\ k & -1 & 1 \\ 3 & 1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
To determine when the system does not have a unique solution, we need to find the determinant of the coefficient matrix and set it equal to zero:
\[ \text{Det} = \begin{vmatrix} 1 & -1 & 2 \\ k & -1 & 1 \\ 3 & 1 & -3 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula for a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]
we can substitute:
- \( a = 1, b = -1, c = 2 \)
- \( d = k, e = -1, f = 1 \)
- \( g = 3, h = 1, i = -3 \)

Calculating the determinant:
\[ \text{Det} = 1((-1)(-3) - (1)(1)) - (-1)(k(-3) - (1)(3)) + 2(k(1) - (-1)(3)) \]
\[ = 1(3 - 1) + (k \cdot 3 - 3) + 2(k + 3) \]
\[ = 2 + 3k - 3 + 2k + 6 \]
\[ = 5k + 5 \]

### Step 4: Set the determinant equal to zero
For the system to not possess a unique solution, the determinant must be zero:
\[ 5k + 5 = 0 \]

### Step 5: Solve for \( k \)
\[ 5k = -5 \implies k = -1 \]

### Conclusion
The value of \( k \) such that the system of linear equations does not possess a unique solution is:
\[ \boxed{-1} \]