91Ó°ÊÓ

The column vectors in the parallelogram are \(\left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}5\\2\end{array}} \right),\left( {\begin{array}{*{20}{c}}6\\4\end{array}} \right),\) and \(\left( {\begin{array}{*{20}{c}}{11}\\6\end{array}} \right)\).

Note that the parallelogram has origin as a vertex and

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}5\\2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}6\\4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{5 + 6}\\{2 + 4}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{11}\\6\end{array}} \right).\end{array}\)

Hence, this parallelogram is determined by the columns of \(A = \left( {\begin{array}{*{20}{c}}5&6\\2&4\end{array}} \right)\).

According to the first statement of Theorem 9,if A is a\(2 \times 2\)matrix, the area of the parallelogram determined by the columns of A is\(\left| {\det A} \right|\).

By the above statement,

\(\begin{array}{c}\left| {\det A} \right| = \left| {\det \left( {\begin{array}{*{20}{c}}5&6\\2&4\end{array}} \right)} \right|\\ = \left| {20 - 12} \right|\\ = \left| 8 \right|\\\left| {\det A} \right| = 8\end{array}\)

Hence, the area of the parallelogram is 8 square units.