初中数学七年级上册（581题）
我们把\(\left| {\begin{array}{*{20}{l}} a&c \\ b&d \end{array}} \right|\)称为二阶行列式，且\(\left| {\begin{array}{*{20}{l}} a&c \\ b&d \end{array}} \right| = ad - bc\)
如：\(\left| {\begin{array}{*{20}{l}} 1&2 \\ 3&{ - 4} \end{array}} \right| = 1 \times ( - 4) - 3 \times 2 = - 10\)．
（1）计算：\(\left| {\begin{array}{*{20}{l}} 2&6 \\ { - 3}&5 \end{array}} \right| = \)___；
（2）若\(\left| {\begin{array}{*{20}{l}} { - 4}&7 \\ 2&m \end{array}} \right| = 6\)，则\(m\)的值为___．
知识点：第一章 有理数
参考答案：28；\( - 5\)．
解析：
解：（1）\(\left| {\begin{array}{*{20}{l}}
2&6 \\\
{ - 3}&5
\end{array}} \right|\)
\( = 2 \times 5 - ( - 3) \times 6\)
\( = 10 - ( - 18)\)
\( = 28\)
（2）\(\because \)\(\left| {\begin{array}{*{20}{l}}
{ - 4}&7 \\\
2&m
\end{array}} \right| = 6\)
\(\therefore - 4m - 2 \times 7 = 6\)
\(\therefore - 4m - 14 = 6\)
\(\therefore m = - 5\)．
