Question

15．已知變換T：$[\begin{array}{l}{x}\\{y}\end{array}]$→$[\begin{array}{l}{{x}^{′}}\\{y′}\end{array}]$=$[\begin{array}{l}{x+2y}\\{y}\end{array}]$，試寫出變換T對應(yīng)的矩陣A，并求出其逆矩陣A^-1．

Answer 1

分析 由題意求得變換矩陣T，根據(jù)二階矩陣的求法，求得行列式丨A丨及其伴隨矩陣，即可求得逆矩陣A^-1．

解答 解：由題意可知設(shè)變換矩陣T=$[\begin{array}{l}{a}&\\{c}&8ajq9k6\end{array}]$，
∴$[\begin{array}{l}{a}&\\{c}&k9ryf3z\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}]$，
丨A丨=1
∴逆矩陣A^-1=$[\begin{array}{l}{1}&{-2}\\{0}&{1}\end{array}]$．

點(diǎn)評 本題考查矩陣的變換，考查逆變換與逆矩陣，矩陣變換是附加題中�？嫉�，屬于基礎(chǔ)題．