Question

2．已知點P（3，1）在矩陣A=$[\begin{array}{l}{a}&{2}\\&{-1}\end{array}]$ 變換下得到點P′（5，-1）．試求矩陣A和它的逆矩陣A^-1．

Answer 1

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

解答 解：依題意得$[\begin{array}{l}{a}&{2}\\&{-1}\end{array}]$ $[\begin{array}{l}{3}\\{1}\end{array}]$=$[\begin{array}{l}{3a+2}\\{2b-1}\end{array}]$=$[\begin{array}{l}{5}\\{-1}\end{array}]$，
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=$[\begin{array}{l}{1}&{2}\\{0}&{-1}\end{array}]$．

點評 本題考查矩陣的變換，逆矩陣的求法，考查計算能力，屬于基礎(chǔ)題．