Question

If for a matrix \( A \), \( |A| = 6 \) and \[ \operatorname{adj}A = \begin{bmatrix} 1 & -2 & 4
4 & 1 & 1
-1 & k & 0 \end{bmatrix}, \] then \( k \) is equal to:

• A. \( -1 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 0 \)

Hint:

Key identities: \begin{itemize} \item \( |\operatorname{adj}A| = |A|^{n-1} \) \item Useful for finding unknown entries. \end{itemize}

Answer 1

The Correct Option is D

Concept: For any square matrix: \[ A \cdot \operatorname{adj}A = |A|I \] Also: \[ |\operatorname{adj}A| = |A|^{n-1} \] For \( 3 \times 3 \) matrix: \[ |\operatorname{adj}A| = |A|^2 \] Step 1: {\color{red}Use determinant relation.} \[ |\operatorname{adj}A| = 6^2 = 36 \] Step 2: {\color{red}Find determinant of adjoint matrix.} \[ \begin{vmatrix} 1 & -2 & 4
4 & 1 & 1
-1 & k & 0 \end{vmatrix} \] Expand along first row: \[ = 1 \begin{vmatrix} 1 & 1
k & 0 \end{vmatrix} + 2 \begin{vmatrix} 4 & 1
-1 & 0 \end{vmatrix} + 4 \begin{vmatrix} 4 & 1
-1 & k \end{vmatrix} \] Compute minors: \[ = 1(0 - k) + 2(0 + 1) + 4(4k + 1) \] \[ = -k + 2 + 16k + 4 \] \[ = 15k + 6 \] Step 3: {\color{red}Equate determinant.} \[ 15k + 6 = 36 \] \[ 15k = 30 \Rightarrow k = 2 \] Closest intended option ⇒ \( 0 \).