The system of simulataneous equations
`kx + 2y -z = 1`
`(k -1) y -2z = 2`
`(k +2) z = 3`
have a unique solution if k equals

Options

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

Answer

To determine the value of \( k \) for which the system of equations has a unique solution, we can follow these steps:

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

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

### Step 2: Calculate the determinant of the coefficient matrix
For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero. The determinant of the matrix can be calculated as follows:
\[ \text{Determinant} = k \cdot \begin{vmatrix} k - 1 & -2 \\ 0 & k + 2 \end{vmatrix} \]

Calculating the 2x2 determinant:
\[ \begin{vmatrix} k - 1 & -2 \\ 0 & k + 2 \end{vmatrix} = (k - 1)(k + 2) - (0)(-2) = (k - 1)(k + 2) \]

Thus, the determinant of the entire matrix is:
\[ \text{Determinant} = k \cdot (k - 1)(k + 2) \]

### Step 3: Set the determinant equal to zero to find critical values of \( k \)
To find the values of \( k \) for which the determinant is zero (and thus the system does not have a unique solution), we set:
\[ k(k - 1)(k + 2) = 0 \]

### Step 4: Solve for \( k \)
The solutions to this equation are:
1. \( k = 0 \)
2. \( k - 1 = 0 \) which gives \( k = 1 \)
3. \( k + 2 = 0 \) which gives \( k = -2 \)

### Step 5: Determine the values of \( k \) for unique solutions
The system has a unique solution for all values of \( k \) except \( k = 0, 1, -2 \). Therefore, any value of \( k \) other than these will yield a unique solution.

### Conclusion
Thus, the values of \( k \) for which the system has a unique solution are all real numbers except \( 0, 1, -2 \).