Question

If \[ \begin{bmatrix} 1 & 3 & 1 \\ k & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix} \] has a determinant of \( \pm 6 \), then the value of \( k \) is:

• A. \( 1 \)
• B. \( -2 \)
• C. \( 2 \)
• D. \( \pm 2 \)

Hint:

To solve determinants, expand using the first row for simpler calculations.

Answer 1

The Correct Option is D

The determinant is:

\[ \det = 1 \cdot \begin{vmatrix} 0 & 1 \\ 0 & 1 \end{vmatrix} - 3 \cdot \begin{vmatrix} k & 1 \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} k & 0 \\ 1 & 0 \end{vmatrix}. \]

Simplify:

\[ \det = 0 - 3(k - 1) + k = -3k + 3 + k = -2k + 3. \]

Given \( |\det| = 6 \), solve:

\[ -2k + 3 = \pm 6 \quad \Rightarrow \quad k = \pm 2. \] Final Answer: \( \boxed{\pm 2} \)