Question

If the system of linear equations \( x+y+z=2 \), \( 2x+y-z=3 \) and \( 3x+2y+kz=4 \) has a unique solution, then:

• A. \( k = 0 \)
• B. \( -1 \lt k \lt 1 \)
• C. \( 0 \lt k \lt 1 \)
• D. \( k \neq 0 \)

Hint:

Cramer's Rule states that if \( \Delta \neq 0 \), a unique solution exists. Always start by finding the determinant of the coefficients.

Answer 1

The Correct Option is D

Concept: A system of linear equations \( AX = B \) has a unique solution based on the determinant.

• The system has a unique solution if the coefficient matrix is non-singular.

• This requires the determinant \( |A| \) to be non-zero.

• If \( |A| = 0 \), the system has either no solution or infinite solutions.

Step 1: Setting up the coefficient matrix.
Extract the coefficients of \( x, y, \) and \( z \) from the equations. \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{bmatrix} \]

Step 2: Calculating the determinant of \( A \).
Expand the determinant along the first row. \[ |A| = 1(1(k) - (-1)(2)) - 1(2(k) - (-1)(3)) + 1(2(2) - 1(3)) \] Simplify the terms inside the parentheses. \[ |A| = (k + 2) - (2k + 3) + (4 - 3) \] Combine the algebraic terms. \[ |A| = k + 2 - 2k - 3 + 1 = -k \]

Step 3: Applying the uniqueness condition.
For a unique solution, we must have \( |A| \neq 0 \). Substituting the calculated value: \( -k \neq 0 \). This simplifies to \( k \neq 0 \).