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Skew-symmetric 3x3 matrix found from its products with given vectors; determinant of an adjoint of 2(A + I) as powers of primes

The same question, asked 2 times: 24 Jan 2025*, 22 Jan 2026 S1. Every wording below, oldest first.

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  1. Question 1

    24 Jan 2025

    74
    Let be a 3×33 \times 3 matrix such that XTAX=OX^{T}AX = O for all nonzero 3×13 \times 1 matrices X=[xyz]X =\begin{bmatrix} x \\ y \\ z \end{bmatrix}. If A[111]=[14−5],A[121]=[04−8]A\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}=\begin{bmatrix} 1 \\ 4 \\ - 5 \end{bmatrix},A\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=\begin{bmatrix} 0 \\ 4 \\ - 8 \end{bmatrix}, and det(adj(2( A+I)))=2α3β5γ,α,β,γ,∈Ndet(adj(2(\text{ }A + I))) =2^{\alpha}3^{\beta}5^{\gamma},\alpha,\beta,\gamma, \in N, then α2+β2+γ2\alpha^{2}+\beta^{2}+\gamma^{2} is

  2. Question 2

    22 Jan 2026 S1

    72
    Let A be a 3×33 \times 3 matrix such that A+AT=OA + A^{T} = O. If A[1−10]=[332],A2[1−10]=[−319−24]A\begin{bmatrix} 1 \\ - 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix},A^{2}\begin{bmatrix} 1 \\ - 1 \\ 0 \end{bmatrix} = \begin{bmatrix} - 3 \\ 19 \\ - 24 \end{bmatrix} and det(adj(2adj(A+I)))=(2)α⋅(3)β⋅(11)γ,α,β,γdet(adj(2adj(A + I))) = (2)^{\alpha} \cdot (3)^{\beta} \cdot (11)^{\gamma},\alpha,\beta,\gamma are non-negative integers, then α+β+γ\alpha + \beta + \gamma is equal to ____\_\_\_\_