The simultaneous equations `Kx+2y-z=1, (K-1)y-2z=2` and `(K+2)z=3` have only one solution, when

Options

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

Explanation

To determine the values of \( K \) for which the simultaneous equations have only one solution, we need to analyze the given equations:

1. \( Kx + 2y - z = 1 \)
2. \( (K - 1)y - 2z = 2 \)
3. \( (K + 2)z = 3 \)

### Step 1: Write the equations in matrix form

We can express the system of equations in matrix form \( A \mathbf{x} = \mathbf{b} \), where:

\[ A = \begin{bmatrix} K & 2 & -1 \\ 0 & K - 1 & -2 \\ 0 & 0 & K + 2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]

### Step 2: Find the determinant of matrix \( A \)

For the system to have a unique solution, the determinant of matrix \( A \) must be non-zero. We calculate the determinant:

\[ \text{det}(A) = K \cdot \text{det}\begin{bmatrix} K - 1 & -2 \\ 0 & K + 2 \end{bmatrix} \]

Calculating the determinant of the 2x2 matrix:

\[ \text{det}\begin{bmatrix} K - 1 & -2 \\ 0 & K + 2 \end{bmatrix} = (K - 1)(K + 2) - (0)(-2) = (K - 1)(K + 2) \]

Thus, we have:

\[ \text{det}(A) = K \cdot (K - 1)(K + 2) \]

### Step 3: Set the determinant not equal to zero

For the system to have a unique solution, we set the determinant not equal to zero:

\[ K \cdot (K - 1)(K + 2) \neq 0 \]

### Step 4: Solve the inequality

This inequality implies:

1. \( K \neq 0 \)
2. \( K - 1 \neq 0 \) → \( K \neq 1 \)
3. \( K + 2 \neq 0 \) → \( K \neq -2 \)

### Conclusion

The values of \( K \) for which the system has only one solution are:

- \( K \neq 0 \)
- \( K \neq 1 \)
- \( K \neq -2 \)