Question

3．已知點(diǎn)A（1，1）在矩陣$M=[{\begin{array}{l}1&a\\ 0&b\end{array}}]$對(duì)應(yīng)的變換作用下得到點(diǎn)B（1，2），點(diǎn)B在矩陣$N=[{\begin{array}{l}m&{-1}\\ n&0\end{array}}]$對(duì)應(yīng)的變換作用下得到點(diǎn)C（-2，1），求矩陣MN的逆矩陣．

Answer 1

分析 根據(jù)條件，先求出$M=[{\begin{array}{l}1&0\\ 0&2\end{array}}]$，再求出$N=[{\begin{array}{l}0&{-1}\\ 1&0\end{array}}]$，因此得出$MN=[{\begin{array}{l}1&0\\ 0&2\end{array}}][{\begin{array}{l}0&{-1}\\ 1&0\end{array}}]=[{\begin{array}{l}0&{-1}\\ 2&0\end{array}}]$，最后根據(jù)逆矩陣定義的得出${[{\begin{array}{l}0&{-1}\\ 2&0\end{array}}]^{-1}}=[{\begin{array}{l}0&{\frac{1}{2}}\\{-1}&0\end{array}}]$．

解答 解：由題意可得，$[{\begin{array}{l}1&a\\ 0&b\end{array}}][\begin{array}{l}1\\ 1\end{array}]=[\begin{array}{l}1+a\\ b\end{array}]=[\begin{array}{l}1\\ 2\end{array}]$，
解得a=0，b=2，所以$M=[{\begin{array}{l}1&0\\ 0&2\end{array}}]$，
又$[{\begin{array}{l}m&{-1}\\ n&0\end{array}}][\begin{array}{l}1\\ 2\end{array}]=[{\begin{array}{l}{m-2}\\ n\end{array}}]=[\begin{array}{l}-2\\ 1\end{array}]$，
解得m=0，n=1，所以$N=[{\begin{array}{l}0&{-1}\\ 1&0\end{array}}]$，
則$MN=[{\begin{array}{l}1&0\\ 0&2\end{array}}][{\begin{array}{l}0&{-1}\\ 1&0\end{array}}]=[{\begin{array}{l}0&{-1}\\ 2&0\end{array}}]$，
所以，矩陣MN的逆矩陣為${[{\begin{array}{l}0&{-1}\\ 2&0\end{array}}]^{-1}}=[{\begin{array}{l}0&{\frac{1}{2}}\\{-1}&0\end{array}}]$．

點(diǎn)評(píng) 本題主要考查了矩陣的運(yùn)算以及逆矩陣的求解，涉及矩陣的運(yùn)算法則和逆矩陣的定義，屬于中檔題．