Question

16．如圖，設(shè)OP與x軸的正方向的夾角為α，OP'與OP的夾角為β，現(xiàn)將OP繞O點(diǎn)旋轉(zhuǎn)到與OP'重合，旋轉(zhuǎn)角β=$\frac{π}{6}$，則這個(gè)旋轉(zhuǎn)變換對(duì)應(yīng)的矩陣為$[\begin{array}{l}{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}\\{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}\end{array}]$．

Answer 1

分析 根據(jù)矩陣的線性變換，即可求得旋轉(zhuǎn)變換對(duì)應(yīng)的矩陣．

解答 解：由二階矩陣的線性變換性質(zhì)可得：$[\begin{array}{l}{cosβ}&{-sinβ}\\{sinβ}&{cosβ}\end{array}]$$[\begin{array}{l}{x}\\{y}\end{array}]$=$[\begin{array}{l}{xcosβ-ysinβ}\\{xsinβ+ycosβ}\end{array}]$，
則旋轉(zhuǎn)變換對(duì)應(yīng)的矩陣A=$[\begin{array}{l}{cosβ}&{-sinβ}\\{sinβ}&{cosβ}\end{array}]$=$[\begin{array}{l}{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}\\{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}\end{array}]$，
故答案為：$[\begin{array}{l}{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}\\{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}\end{array}]$．

點(diǎn)評(píng) 本題考查矩陣的線性變換的簡單應(yīng)用，考查轉(zhuǎn)化思想，屬于基礎(chǔ)題．