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(B) Given $A+A^T=O$, $A$ is a skew-symmetric matrix. Let $A = \begin{bmatrix} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{bmatrix}$.From $A\begin{bmat

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$18$

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(B) Given $A+A^T=O$, $A$ is a skew-symmetric matrix. Let $A = \begin{bmatrix} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{bmatrix}$.From $A\begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} = \begin{bmatrix} 3 \ 3 \ 2 \end{bmatrix}$, we get:$-a = 3 \Rightarrow a = -3$$-b+c = 2$$3a + 2b = -3 \Rightarrow 3(-3) + 2b = -3 \Rightarrow 2b = 6 \Rightarrow b = 3$.Then $c = 2+b = 5$.So, $A = \begin{bmatrix} 0 & -3 & 3 \ 3 & 0 & 5 \ -3 & -5 & 0 \end{bmatrix}$.Then $A+I = \begin{bmatrix} 1 & -3 & 3 \ 3 & 1 & 5 \ -3 & -5 & 1 \end{bmatrix}$.$|A+I| = 1(1+25) + 3(3+15) + 3(-15+3) = 26 + 54 - 36 = 44$.We need $\det(	ext{adj}(2	ext{adj}(A+I)))$.Since $A+I$ is a $3 	imes 3$ matrix, $	ext{adj}(A+I)$ is also $3 	imes 3$.$\det(2	ext{adj}(A+I)) = 2^3 |	ext{adj}(A+I)| = 8 |A+I|^2 = 8(44)^2$.Then $\det(	ext{adj}(2	ext{adj}(A+I))) = (8 \cdot 44^2)^2 = (2^3 \cdot (2^2 \cdot 11)^2)^2 = (2^3 \cdot 2^4 \cdot 11^2)^2 = (2^7 \cdot 11^2)^2 = 2^{14} \cdot 11^4$.Comparing with $(2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, we have $\alpha=14, \beta=0, \gamma=4$.Thus, $\alpha+\beta+\gamma = 14+0+4 = 18$.

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