Question

2．已知点P（3，1）在矩阵A=$[\begin{array}{l}{a}&{2}\\{b}&{-1}\end{array}]$ 变换下得到点P′（5，-1）．试求矩阵A和它的逆矩阵A^-1．

Answer 1

分析 由$[\begin{array}{l}{a}&{2}\\{b}&{-1}\end{array}]$ $[\begin{array}{l}{3}\\{1}\end{array}]$=$[\begin{array}{l}{3a+2}\\{2b-1}\end{array}]$=$[\begin{array}{l}{5}\\{-1}\end{array}]$，列方程求得a和b的值，求得矩阵A，丨A丨及A*，由A^-1=$\frac{1}{丨A丨}$•A*，即可求得A^-1．

解答 解：依题意得$[\begin{array}{l}{a}&{2}\\{b}&{-1}\end{array}]$ $[\begin{array}{l}{3}\\{1}\end{array}]$=$[\begin{array}{l}{3a+2}\\{2b-1}\end{array}]$=$[\begin{array}{l}{5}\\{-1}\end{array}]$，
所以$\left\{\begin{array}{l}3a+2=5\\ 3b-1=-1\end{array}$，解得：$\left\{\begin{array}{l}a=1\\ b=0\end{array}$，
A=$[\begin{array}{l}{1}&{2}\\{0}&{-1}\end{array}]$，
丨A丨=$|\begin{array}{l}{1}&{2}\\{0}&{-1}\end{array}|$=1×（-1）-0×2=-1，
A*=$[\begin{array}{l}{-1}&{-2}\\{0}&{1}\end{array}]$，
∴A^-1=$\frac{1}{丨A丨}$•A*=$[\begin{array}{l}{1}&{2}\\{0}&{-1}\end{array}]$，
∴A^-1=$[\begin{array}{l}{1}&{2}\\{0}&{-1}\end{array}]$．

点评 本题考查矩阵的变换，逆矩阵的求法，考查计算能力，属于基础题．