题目内容
15.已知直线l:2x-y=3,若矩阵A=$(\begin{array}{l}{-1}&{a}\\{b}&{3}\end{array})$a,b∈R所对应的变换σ把直线l变换为它自身.(Ⅰ)求矩阵A;
(Ⅱ)求矩阵A的逆矩阵.
分析 (Ⅰ)通过设直线2x-y=3上任意一点P(x,y),利用其在A的作用下变为(x′,y′),可用x、y表示出x′、y′,代入2x′-y′=3,计算即可;
(Ⅱ)直接计算即可.
解答 解:(Ⅰ)设P(x,y)为直线2x-y=3上任意一点,
其在A的作用下变为(x′,y′),
则$[\begin{array}{l}{-1}&{a}\\{b}&{3}\end{array}]$$[\begin{array}{l}{x}\\{y}\end{array}]$=$[\begin{array}{l}{-x+ay}\\{bx+3y}\end{array}]$=$[\begin{array}{l}{x′}\\{y′}\end{array}]$,∴$\left\{\begin{array}{l}{x′=-x+ay}\\{y′=bx+3y}\end{array}\right.$,
代入2x′-y′=3得:-(b+2)x+(2a-3)y=3,
∵其与2x-y=3完全一样,∴$\left\{\begin{array}{l}{-b-2=2}\\{2a-3=-1}\end{array}\right.$,即$\left\{\begin{array}{l}{a=1}\\{b=-4}\end{array}\right.$,
∴矩阵A=$[\begin{array}{l}{-1}&{1}\\{-4}&{3}\end{array}]$;
(Ⅱ)∵$|\begin{array}{l}{-1}&{1}\\{-4}&{3}\end{array}|$=1,∴矩阵M的逆矩阵为A-1=$[\begin{array}{l}{3}&{-1}\\{4}&{-1}\end{array}]$.
点评 本题主要考查矩阵变换的性质,考查二阶矩阵的乘法,注意解题方法的积累,属于基础题.
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