Question

11．已知矩阵A=$[{\begin{array}{l}{\frac{3}{2}}&{\frac{1}{2}}\\ 2&1\end{array}}]$
（1）求A^-1；
（2）满足AX=A^-1二阶矩阵X．

Answer 1

分析 （1）通过变换计算即可；
（2）通过AX=A^-1可得X=A^-1A^-1，计算即可．

解答 解：（1）∵A=$[{\begin{array}{l}{\frac{3}{2}}&{\frac{1}{2}}\\ 2&1\end{array}}]$，
∴$[\begin{array}{l}{\frac{3}{2}}&{\frac{1}{2}}&{1}&{0}\\{2}&{1}&{0}&{1}\end{array}]$$\stackrel{第一行×2}{→}$$[\begin{array}{l}{3}&{1}&{2}&{0}\\{2}&{1}&{0}&{1}\end{array}]$$\stackrel{第二行+第一行×（-\frac{2}{3}）}{→}$$[\begin{array}{l}{3}&{1}&{2}&{0}\\{0}&{\frac{1}{3}}&{-\frac{4}{3}}&{1}\end{array}]$$\stackrel{第二行×3}{→}$
$[\begin{array}{l}{3}&{1}&{2}&{0}\\{0}&{1}&{-4}&{3}\end{array}]$$\stackrel{第一行×\frac{1}{3}}{→}$$[\begin{array}{l}{1}&{\frac{1}{3}}&{\frac{2}{3}}&{0}\\{0}&{1}&{-4}&{3}\end{array}]$$\stackrel{第一行+第二行×（-\frac{1}{3}）}{→}$$[\begin{array}{l}{1}&{0}&{2}&{-1}\\{0}&{1}&{-4}&{3}\end{array}]$，
∴A^-1=$[\begin{array}{l}{2}&{-1}\\{-4}&{3}\end{array}]$；
（2）∵AX=A^-1，∴X=A^-1A^-1=$[\begin{array}{l}{2}&{-1}\\{-4}&{3}\end{array}]$$[\begin{array}{l}{2}&{-1}\\{-4}&{3}\end{array}]$=$[\begin{array}{l}{8}&{-5}\\{-20}&{13}\end{array}]$，
即$X=[{\begin{array}{l}8&{-5}\\{-20}&{13}\end{array}}]$．

点评 本题考查矩阵乘法，注意解题方法的积累，属于基础题．