For what values of k, does the system of linear equations `x+y+z=2, 2x+y-z=3, 3x+2y+kz=4` have a unique solution ?

Options

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

Explanation

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

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

### Step 1: Write the coefficient matrix
The coefficient matrix \( A \) for the system of equations is:

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

### Step 2: Calculate the determinant of the matrix
To find the values of \( k \) for which the system has a unique solution, we need to calculate the determinant of matrix \( A \) and set it not equal to zero. The determinant \( |A| \) can be calculated as follows:

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

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ |A| = 1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -1 \\ 3 & k \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} \]

### Step 3: Calculate the minors
Now we calculate each of the minors:

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

### Step 4: Substitute back into the determinant
Substituting these minors back into the determinant expression:

\[ |A| = 1(k + 2) - 1(2k + 3) + 1(1) \]
\[ = k + 2 - 2k - 3 + 1 \]
\[ = -k + 0 \]
\[ = -k \]

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

\[ -k \neq 0 \]

This implies:

\[ k \neq 0 \]

### Conclusion
The system of linear equations has a unique solution for all values of \( k \) except \( k = 0 \).