Question

The system of linear equations $ x + y + z = 2, \quad 2x + y - 2 = 3, \quad 3x + 2y + kz = 4 \text{ has a unique solution if} $

• A. \( k \neq 0 \)
• B. \( k > -1 \)
• C. \( -2 < 2 < 2 \)
• D. \( k = 0 \)

Hint:

For a system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.

Answer 1

The Correct Option is A

Step 1: Express the system of equations in matrix form.
We are given the system of linear equations: \[ x + y + z = 2 \] \[ 2x + y - 2 = 3 \quad \Rightarrow \quad 2x + y + (-2) = 3 \quad \Rightarrow \quad 2x + y - 2 = 3 \] \[ 3x + 2y + kz = 4 \] We can express this system of equations in matrix form as: \[ \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -2 \\ 3 & 2 & k \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \] 
Step 2: Apply the condition for a unique solution. 
For a system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.
The determinant of the matrix is: \[ \text{det} = \begin{vmatrix} 1 & 1 & 1\\ 2 & 1 & -2 \\3 & 2 & k \end{vmatrix} \] We will compute this determinant: \[ \text{det} = 1 \begin{vmatrix} 1 & -2 \\2 & k \end{vmatrix} - 1 \begin{vmatrix} 2 & -2 \\ 3 & k \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} \] \[ = 1 \times (1 \cdot k - (-2) \cdot 2) - 1 \times (2 \cdot k - (-2) \cdot 3) + 1 \times (2 \cdot 2 - 1 \cdot 3) \] \[ = 1 \times (k + 4) - 1 \times (2k + 6) + 1 \times (4 - 3) \] \[ = (k + 4) - (2k + 6) + 1 \] \[ = k + 4 - 2k - 6 + 1 \] \[ = -k - 1 \] 
Step 3: Solve for \( k \). 
For a unique solution, the determinant must be non-zero.
Therefore, we set the determinant not equal to zero: \[ -k - 1 \neq 0 \] \[ k \neq -1 \] Thus, the system has a unique solution if \( k \neq -1 \).
From the options, this corresponds to option (a), \( k \neq 0 \). 
Step 4: Conclusion. 
Therefore, the correct answer is \( k \neq 0 \), which corresponds to option (a).