代数式ad-bc可用符号$|\begin{array}{l}a{\;}^{\;}{\;} {\;}b\\ c{\;}^{\;}{\;} {\;}d\end{array}|$来表示.称之为二阶行列式．即$|\begin{array}{l}a{\;}^{\;}{\;} {\;}b\\ c{\;}^{\;}{\;} {\;}d\end{array}|=ad-bc$.用二阶行列式可以� 题目和参考答案——青夏教育精英家教网—

题目内容

7．代数式ad-bc可用符号$|\begin{array}{l}a{\;}^{\;}{\;}_{\;}b\\ c{\;}^{\;}{\;}_{\;}d\end{array}|$来表示，称之为二阶行列式．即$|\begin{array}{l}a{\;}^{\;}{\;}_{\;}b\\ c{\;}^{\;}{\;}_{\;}d\end{array}|=ad-bc$，用二阶行列式可以解二元一次方程组．由$\left\{\begin{array}{l}{a_1}x+{b_1}y={c_1}\\{a_2}x+{b_2}y={c_2}\end{array}\right.$得三个二阶行列式即$D=|\begin{array}{l}{a_1}{\;}^{\;}{b_1}\\{a_2}{\;}^{\;}{b_2}\end{array}|$，${D_x}=|\begin{array}{l}{c_1}{\;}^{\;}{b_1}\\{c_2}{\;}^{\;}{b_2}\end{array}|$及${D_y}=|\begin{array}{l}{a_1}{\;}^{\;}{c_1}\\{a_2}{\;}^{\;}{c_2}\end{array}|$那么方程组的解就是$\left\{\begin{array}{l}x=\frac{D_x}{D}\\ y=\frac{D_y}{D}\end{array}\right.$．
（1）求出二阶行列式$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$的值；
（2）用二阶行列式解方程组$\left\{\begin{array}{l}3x+2y=-1\\ 5x-y-2=0\end{array}\right.$．

分析（1）根据题意可以直接算出二阶行列式$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$的值；
（2）根据题意可以算出D、D_X，D_Y，从而可以求得x、y的值，本题得以解决．

解答解：（1）由题意可得，
$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$=3×4-5×6=12-30=-18，
即$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$的值是-18；
（2）∵$\left\{\begin{array}{l}{3x+2y=-1}\\{5x-y-2=0}\end{array}\right.$，
∴$\left\{\begin{array}{l}{3x+2y=-1}\\{5x-y=2}\end{array}\right.$，
由题意可得，
D=$|\begin{array}{l}{3}&{2}\\{5}&{-1}\end{array}|$=3×（-1）-2×5=-3-10=-13，
${D}_{X}=|\begin{array}{l}{-1}&{2}\\{2}&{-1}\end{array}|$=（-1）×（-1）-2×2=1-4=-3，
${D}_{Y}=|\begin{array}{l}{3}&{-1}\\{5}&{2}\end{array}|$=3×2-（-1）×5=6+5=11，
∴$x=\frac{{D}_{X}}{D}=\frac{-3}{-13}=\frac{3}{13}$，
$y=\frac{{D}_{Y}}{D}=\frac{11}{-13}=-\frac{11}{13}$，
方程组的解是$\left\{\begin{array}{l}{x=\frac{3}{13}}\\{y=-\frac{11}{13}}\end{array}\right.$．

点评本题考查二元一次方程组的解和解二元一次方程，解题的关键是明确题目中的新定义，根据新定义可以解决相关的问题．