Let A be a 3 times 3 matrix such that X^{T}AX | vedclass

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(D) Given $X^{T}AX = 0$ for all $X$,$A$ must be a skew-symmetric matrix. Let $A = \begin{bmatrix} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{bmatrix}

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$44$

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(D) Given $X^{T}AX = 0$ for all $X$,$A$ must be a skew-symmetric matrix. Let $A = \begin{bmatrix} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{bmatrix}$.From $A \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \ 4 \ -5 \end{bmatrix}$,we get $a+b=1$,$-a+c=4$,and $-b-c=-5 \Rightarrow b+c=5$.Solving these,$a=-1, b=2, c=3$. Thus $A = \begin{bmatrix} 0 & -1 & 2 \ 1 & 0 & 3 \ -2 & -3 & 0 \end{bmatrix}$.Then $A+I = \begin{bmatrix} 1 & -1 & 2 \ 1 & 1 & 3 \ -2 & -3 & 1 \end{bmatrix}$. $\det(A+I) = 1(1+9) + 1(1+6) + 2(-3+2) = 10+7-2 = 15$.$2(A+I)$ is a $3 	imes 3$ matrix,so $\det(2(A+I)) = 2^3 \det(A+I) = 8 	imes 15 = 120$.Using $\det(\operatorname{adj}(M)) = (\det M)^{n-1}$,$\det(\operatorname{adj}(2(A+I))) = (120)^{3-1} = 120^2 = (2^3 \cdot 3 \cdot 5)^2 = 2^6 \cdot 3^2 \cdot 5^2$.Thus $\alpha=6, \beta=2, \gamma=2$. Therefore,$\alpha^2+\beta^2+\gamma^2 = 36+4+4 = 44$.

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