Question

15．已知变换T：$[\begin{array}{l}{x}\\{y}\end{array}]$→$[\begin{array}{l}{{x}^{′}}\\{y′}\end{array}]$=$[\begin{array}{l}{x+2y}\\{y}\end{array}]$，试写出变换T对应的矩阵A，并求出其逆矩阵A^-1．

Answer 1

分析 由题意求得变换矩阵T，根据二阶矩阵的求法，求得行列式丨A丨及其伴随矩阵，即可求得逆矩阵A^-1．

解答 解：由题意可知设变换矩阵T=$[\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}]$，
∴$[\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}]$$[\begin{array}{l}{x}\\{y}\end{array}]$=$[\begin{array}{l}{x+2y}\\{y}\end{array}]$，
∴$\left\{\begin{array}{l}{ax+by=x+2y}\\{cx+dy=y}\end{array}\right.$，解得：$\left\{\begin{array}{l}{a=1}\\{b=2}\\{c=0}\\{d=1}\end{array}\right.$，
∴A=$[\begin{array}{l}{1}&{2}\\{0}&{1}\end{array}]$，
丨A丨=1
∴逆矩阵A^-1=$[\begin{array}{l}{1}&{-2}\\{0}&{1}\end{array}]$．

点评 本题考查矩阵的变换，考查逆变换与逆矩阵，矩阵变换是附加题中常考的，属于基础题．