Question

Let \( A \) be a \( 3 \times 3 \) matrix such that \( A + A^{T} = O \). If

\[ A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}, \quad A^{2} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix} \] and

\[ \det\!\big(\operatorname{adj}(2\,\operatorname{adj}(A+I))\big) = 2^{\alpha}\,3^{\beta}\,11^{\gamma}, \] where \( \alpha, \beta, \gamma \) are non-negative integers, then the value of
\( \alpha + \beta + \gamma \) is ________.

Hint:

For any skew-symmetric matrix of odd order, the determinant is always zero, which simplifies determinant calculations significantly.

Answer 1

Correct Answer: 6

Since \( A + A^{T} = O \), the matrix \( A \) is skew-symmetric.
For a skew-symmetric matrix of odd order,
\[ \det(A) = 0. \] Step 1: Determinant of \( A+I \).
Since \( \det(A)=0 \) and \( A \) is skew-symmetric of order 3,
\[ \det(A+I) = 1. \] Step 2: Determinant of adjoint.
For an \( n \times n \) matrix \( M \),
\[ \det(\operatorname{adj}(M)) = (\det M)^{n-1}. \] Here \( n = 3 \), hence
\[ \det(\operatorname{adj}(A+I)) = (\det(A+I))^{2} = 1. \] Step 3: Scalar multiplication inside adjoint.
For a \( 3 \times 3 \) matrix,
\[ \operatorname{adj}(2M) = 2^{2}\operatorname{adj}(M). \] Therefore,
\[ \det(\operatorname{adj}(2\,\operatorname{adj}(A+I))) = 2^{6}\det(\operatorname{adj}(\operatorname{adj}(A+I))). \] Step 4: Adjoint of adjoint.
\[ \det(\operatorname{adj}(\operatorname{adj}(M))) = (\det M)^{(n-1)^{2}}. \] Thus,
\[ \det(\operatorname{adj}(\operatorname{adj}(A+I))) = 1. \] Hence,
\[ \det(\operatorname{adj}(2\,\operatorname{adj}(A+I))) = 2^{6}. \] Comparing with \( 2^{\alpha} 3^{\beta} 11^{\gamma} \), we get
\[ \alpha = 6,\quad \beta = 0,\quad \gamma = 0. \] Final Answer:
\[ \boxed{6} \]