Question

If the system of linear equations \[ x + ky + 3z = 0, \quad 3x + ky - 2z = 0, \quad 2x + 4y - 3z = 0 \] has a non-zero solution  \( (x, y, z) \), then \( \frac{xz}{y^2} \) is equal to:

• A. -10
• B. -30
• C. 30
• D. 10

Hint:

For systems of linear equations with non-zero solutions, check the determinant. If the determinant is zero, the system has a non-zero solution.

Answer 1

The Correct Option is A

For the system of equations to have a non-zero solution, the determinant of the coefficient matrix must be zero. The given system of equations is:

\[ \begin{bmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{bmatrix} \]

Step 1: Compute the determinant of the matrix.

\[ \text{Determinant} = \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{vmatrix} \]

Expanding along the first row:

\[ = 1 \begin{vmatrix} k & -2 \\ 4 & -3 \end{vmatrix} - k \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} + 3 \begin{vmatrix} 3 & k \\ 2 & 4 \end{vmatrix} \]

Computing the 2×2 determinants:

\[ \begin{vmatrix} k & -2 \\ 4 & -3 \end{vmatrix} = k(-3) - (-2)(4) = -3k + 8 \] \[ \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} = 3(-3) - (-2)(2) = -9 + 4 = -5 \] \[ \begin{vmatrix} 3 & k \\ 2 & 4 \end{vmatrix} = 3(4) - k(2) = 12 - 2k \]

Substituting back:

\[ \text{Determinant} = 1(-3k + 8) - k(-5) + 3(12 - 2k) \] \[ = -3k + 8 + 5k + 36 - 6k \] \[ = -4k + 44 \]

Step 2: Solve for \( k \).

\[ -4k + 44 = 0 \quad \Rightarrow \quad k = 11 \]

Substituting \( k = 11 \) into the equations, we get:

\[ x + 11y + 3z = 0, \quad 3x + 11y - 2z = 0, \quad 2x + 4y - 3z = 0 \]

Solving this system, we find:

\[ x = 10, \quad y = 1, \quad z = -1 \]

Step 3: Compute \( \frac{xz}{y^2} \).

\[ \frac{xz}{y^2} = \frac{10 \times (-1)}{1^2} = -10 \]

Final Answer: \( \mathbf{-10} \).