Question

19．設(shè)全集A={$[\begin{array}{l}{x}&{3}\\{4}&{-2}\end{array}]$，$|\begin{array}{l}{1}&{tanα}\\{sinβ}&{-2}\end{array}|$}，B={$[\begin{array}{l}{1}&{y}\\{z}&{-2}\end{array}]$}，且∁_AB={$[\begin{array}{l}{1}&{1}\\{-\frac{1}{2}}&{-2}\end{array}]$}，試求x，y，z，α，β

Answer 1

分析 利用補(bǔ)集性質(zhì)得$[\begin{array}{l}{x}&{3}\\{4}&{-2}\end{array}]$=$[\begin{array}{l}{1}&{y}\\{z}&{-2}\end{array}]$，且$[\begin{array}{l}{1}&{tanα}\\{sinβ}&{-2}\end{array}]$=$[\begin{array}{l}{1}&{1}\\{-\frac{1}{2}}&{-2}\end{array}]$，再由二階矩陣的性質(zhì)能求出x，y，z，α，β的值．

解答 解：∵全集A={$[\begin{array}{l}{x}&{3}\\{4}&{-2}\end{array}]$，$[\begin{array}{l}{1}&{tanα}\\{sinβ}&{-2}\end{array}]$}，B={$[\begin{array}{l}{1}&{y}\\{z}&{-2}\end{array}]$}，且∁_AB={$[\begin{array}{l}{1}&{1}\\{-\frac{1}{2}}&{-2}\end{array}]$}，
∴$[\begin{array}{l}{x}&{3}\\{4}&{-2}\end{array}]$=$[\begin{array}{l}{1}&{y}\\{z}&{-2}\end{array}]$，且$[\begin{array}{l}{1}&{tanα}\\{sinβ}&{-2}\end{array}]$=$[\begin{array}{l}{1}&{1}\\{-\frac{1}{2}}&{-2}\end{array}]$，
∴$\left\{\begin{array}{l}{x=1}\\{y=3}\\{z=4}\end{array}\right.$，且$\left\{\begin{array}{l}{tanα=1}\\{sinβ=-\frac{1}{2}}\end{array}\right.$，
∴x=1，y=3，z=4，α=$\frac{π}{4}$+kπ，k∈Z，β=$\frac{7π}{6}+2kπ$或$β=\frac{11π}{6}+2kπ$，k∈Z．

點(diǎn)評 本題考查滿足條件的實(shí)數(shù)值的求法，是基礎(chǔ)題，解題時要認(rèn)真審題，注意補(bǔ)集性質(zhì)和二階矩陣的性質(zhì)的合理運(yùn)用．