The system of linear equations
`x + y + z = 2`
`2x + y -z = 3`
`3x + 2y + kz = 4`
has a unique solution, if

Options

A. `k != 0`
B. `-1 lt k lt 1`
C. `-2 lt k lt 2`
D. `k = 0`

Answer

To determine the value of \( k \) for which the system of linear equations has a unique solution, we need to analyze the determinant of the coefficient matrix. The system of equations is given as:

1. \( x + y + z = 2 \)
2. \( 2x + y - z = 3 \)
3. \( 3x + 2y + kz = 4 \)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) for the above system of equations is:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{bmatrix} \]

### Step 2: Calculate the Determinant
To find the condition for a unique solution, we need to calculate the determinant of matrix \( A \) and set it not equal to zero.

\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix} \]

### Step 3: Expand the Determinant
We can expand the determinant using the first row:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -1 \\ 3 & k \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} = 1 \cdot k - (-1) \cdot 2 = k + 2 \)

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

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

Putting it all together:

\[ \text{det}(A) = (k + 2) - (2k + 3) + 1 \]

### Step 4: Simplify the Determinant
Now, simplify the expression:

\[ \text{det}(A) = k + 2 - 2k - 3 + 1 = -k + 0 = -k \]

### Step 5: Set the Determinant Not Equal to Zero
For the system to have a unique solution, the determinant must not equal zero:

\[ -k \neq 0 \implies k \neq 0 \]

### Conclusion
The system of linear equations has a unique solution if \( k \neq 0 \).