Question

1．在平面直角坐標(biāo)系xOy中，設(shè)點(diǎn)A（-1，2）在矩陣$M=[{\begin{array}{l}{-1}&0\\ 0&1\end{array}}]$對(duì)應(yīng)的變換作用下得到點(diǎn)A′，將點(diǎn)B（3，4）繞點(diǎn)A′逆時(shí)針旋轉(zhuǎn)90°得到點(diǎn)B′，求點(diǎn)B′的坐標(biāo)．

Answer 1

分析 設(shè)B′（x，y），$[\begin{array}{l}{-1}&{0}\\{0}&{1}\end{array}]$$[\begin{array}{l}{-1}\\{2}\end{array}]$=$[\begin{array}{l}{1}\\{2}\end{array}]$，求得A′的坐標(biāo)，寫(xiě)出向量$\overrightarrow{A′B}$，$\overrightarrow{A′B′}$，$[\begin{array}{l}{0}&{-1}\\{1}&{0}\end{array}]$$[\begin{array}{l}{2}\\{2}\end{array}]$=$[\begin{array}{l}{x-1}\\{y-2}\end{array}]$，即可求得x和y，求得點(diǎn)B′的坐標(biāo)．

解答 解：設(shè)B′（x，y），
由題意可知：$[\begin{array}{l}{-1}&{0}\\{0}&{1}\end{array}]$$[\begin{array}{l}{-1}\\{2}\end{array}]$=$[\begin{array}{l}{1}\\{2}\end{array}]$，得A′（1，2），
則$\overrightarrow{A′B}$=（2，2），$\overrightarrow{A′B′}$=（x-1，y-2），
即旋轉(zhuǎn)矩陣N=$[\begin{array}{l}{0}&{-1}\\{1}&{0}\end{array}]$，
則$[\begin{array}{l}{0}&{-1}\\{1}&{0}\end{array}]$$[\begin{array}{l}{2}\\{2}\end{array}]$=$[\begin{array}{l}{x-1}\\{y-2}\end{array}]$，
即$[\begin{array}{l}{-2}\\{2}\end{array}]$=$[\begin{array}{l}{x-1}\\{y-2}\end{array}]$，解得：$\left\{\begin{array}{l}{x=-1}\\{y=4}\end{array}\right.$，
所以B′的坐標(biāo)為（-1，4）．

點(diǎn)評(píng) 本題考查矩陣的變換，考查矩陣的乘法，考查點(diǎn)在變換下點(diǎn)的坐標(biāo)的求法，屬于中檔題．