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:
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:
Show Hint
To solve determinants, expand using the first row for simpler calculations.
Updated On: Jan 13, 2026
- \( 1 \)
- \( -2 \)
- \( 2 \)
- \( \pm 2 \)
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The Correct Option is D
Solution and Explanation
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: \( \boxed{\pm 2} \)
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