ข้อ 30
\[\begin{array}{l}
from\quad {A^3} = 2I\\
and\quad \left| {{A^3}} \right| = \left| {2I} \right| = {2^2}\left| I \right| = {2^3}\\
then\quad \left| A \right| = 2\\
from\quad \left| {{C^{ - 1}}} \right| = 4 \Rightarrow \left| C \right| = \frac{1}{4}\\
from\quad \left| {{B^t}C} \right| = \left| {\begin{array}{*{20}{c}}
{ - 3g}&{ - 3h}&{ - 3i}\\
{ - a}&{ - b}&{ - c}\\
{2d}&{2e}&{2f}
\end{array}} \right| = \left( { - 3} \right)\left( { - 1} \right)\left( 2 \right)\left| {\begin{array}{*{20}{c}}
g&h&i\\
a&b&c\\
d&e&f
\end{array}} \right|\\
and\quad \left| B \right|\left| C \right| = 6\left| {\begin{array}{*{20}{c}}
a&b&c\\
d&e&f\\
g&h&i
\end{array}} \right| = 6\left| A \right|\\
then\quad \left| B \right| = 6 \times 2 \times 4
\end{array}\]
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