The system of linear equations `x+y+z=2,2x+y-z=3, 3x+2y+kz=4` has a unique solution if (A) `k!=0` (B) `-1ltklt1` (C) `-2ltklt2` (D) `k=0`

Options

A. `lambda!=0`
B. `-1ltlambdalt1`
C. `lambda=0 `
D. `-2ltlambdalt2`

Answer

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

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

We can represent this system in matrix form as \( A \mathbf{x} = \mathbf{b} \), where:

\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} \]

### Step 1: Calculate the Determinant of Matrix A

The determinant of matrix \( A \) is given by:

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

Using the determinant formula for a 3x3 matrix, we calculate:

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

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

Now substituting back into the determinant formula:

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

This simplifies to:

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

### Step 2: 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

Thus, the system of equations has a unique solution if \( k \neq 0 \).

### Final Answer

The correct option is (A) \( k \neq 0 \).