题目内容
1.有关行列式展开:(1)分别按第一行以及第一列展开行列式$|\begin{array}{l}{2}&{1}&{3}\\{0}&{4}&{2}\\{0}&{1}&{1}\end{array}|$;
(2)试将展开式a$|\begin{array}{l}{1}&{2}\\{0}&{4}\end{array}|$+b$|\begin{array}{l}{-1}&{3}\\{0}&{4}\end{array}|$+c$|\begin{array}{l}{-1}&{3}\\{1}&{2}\end{array}|$写成一个三阶行列式.
分析 (1)利用行列式展开的方法,即可得出结论;
(2)根据行列式的展开规律,将展开式还原.
解答 解:(1)按第一行展开:2×$|\begin{array}{l}{4}&{2}\\{1}&{1}\end{array}|$-1×$|\begin{array}{l}{0}&{2}\\{0}&{1}\end{array}|$+3×$|\begin{array}{l}{0}&{4}\\{0}&{1}\end{array}|$,
按第一列展开:2×$|\begin{array}{l}{4}&{2}\\{1}&{1}\end{array}|$-0$|\begin{array}{l}{1}&{3}\\{1}&{1}\end{array}|$+0×$|\begin{array}{l}{1}&{3}\\{4}&{2}\end{array}|$;
(2)$|\begin{array}{l}{a}&{-1}&{3}\\{-b}&{1}&{2}\\{c}&{0}&{4}\end{array}|$按第一列展开可得:a$|\begin{array}{l}{1}&{2}\\{0}&{4}\end{array}|$+b$|\begin{array}{l}{-1}&{3}\\{0}&{4}\end{array}|$+c$|\begin{array}{l}{-1}&{3}\\{1}&{2}\end{array}|$,
a$|\begin{array}{l}{1}&{2}\\{0}&{4}\end{array}|$+b$|\begin{array}{l}{-1}&{3}\\{0}&{4}\end{array}|$+c$|\begin{array}{l}{-1}&{3}\\{1}&{2}\end{array}|$写成一个三阶行列式$|\begin{array}{l}{a}&{-1}&{3}\\{-b}&{1}&{2}\\{c}&{0}&{4}\end{array}|$.
点评 本题考查行列式展开的方法,考查学生的计算能力,属于基础题.
| A. | 17 | B. | 16 | C. | 15 | D. | 14 |
| A. | $\left\{{\begin{array}{l}{x=3x'}\\{y=\frac{1}{2}y'}\end{array}}\right.$ | B. | $\left\{{\begin{array}{l}{x'=3x}\\{y'=\frac{1}{2}y}\end{array}}\right.$ | C. | $\left\{{\begin{array}{l}{x=3x'}\\{y=2y'}\end{array}}\right.$ | D. | $\left\{{\begin{array}{l}{x'=3x}\\{y'=2y}\end{array}}\right.$ |