Question

1．有關(guān)行列式展開：
（1）分別按第一行以及第一列展開行列式$|\begin{array}{l}{2}&{1}&{3}\\{0}&{4}&{2}\\{0}&{1}&{1}\end{array}|$；
（2）試將展開式a$|\begin{array}{l}{1}&{2}\\{0}&{4}\end{array}|$+b$|\begin{array}{l}{-1}&{3}\\{0}&{4}\end{array}|$+c$|\begin{array}{l}{-1}&{3}\\{1}&{2}\end{array}|$寫成一個三階行列式．

Answer 1

分析 （1）利用行列式展開的方法，即可得出結(jié)論；
（2）根據(jù)行列式的展開規(guī)律，將展開式還原．

解答 解：（1）按第一行展開：2×$|\begin{array}{l}{4}&{2}\\{1}&{1}\end{array}|$-1×$|\begin{array}{l}{0}&{2}\\{0}&{1}\end{array}|$+3×$|\begin{array}{l}{0}&{4}\\{0}&{1}\end{array}|$，
按第一列展開：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）$|\begin{array}{l}{a}&{-1}&{3}\\{-b}&{1}&{2}\\{c}&{0}&{4}\end{array}|$按第一列展開可得：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$|\begin{array}{l}{-1}&{3}\\{1}&{2}\end{array}|$寫成一個三階行列式$|\begin{array}{l}{a}&{-1}&{3}\\{-b}&{1}&{2}\\{c}&{0}&{4}\end{array}|$．

點評 本題考查行列式展開的方法，考查學(xué)生的計算能力，屬于基礎(chǔ)題．