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13-(a)

図 2: (a)
\includegraphics[width=5.00truecm,scale=1.1]{a.eps}

13-(a)解答

$\displaystyle (q_j,\vr_j)=\left(q,1,0\right),\,\left(-q,-1,0\right) $

$\displaystyle {\bf d}= \begin{pmatrix}q+q \\ 0 \end{pmatrix}= \begin{pmatrix}2... ...q-q & 0 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$


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著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp