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`

Answer

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 formed by the equations. The equations given are:

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 system can be written as:

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

### Step 2: Calculate the Determinant of the Coefficient 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 \( \text{det}(A) \) can be calculated using the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

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

### Step 3: Simplify the Determinant
Calculating each term:

1. The first term: \( 1 \cdot (k + 2) = k + 2 \)
2. The second term: \( -1 \cdot (2k + 3) = -2k - 3 \)
3. The third term: \( 1 \cdot (4 - 3) = 1 \)

Combining these, we have:

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

Simplifying this:

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

### Step 4: Set the Determinant Not Equal to Zero
For the system to have a unique solution, we require:

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

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