Question

2．设A=$\frac{1}{2}$$（\begin{array}{l}{2}&{0}&{0}\\{0}&{1}&{3}\\{0}&{2}&{5}\end{array}）$，求|A|，A^-1，（A^*）^-1．

Answer 1

分析 由矩阵的运算性质|A|=$（\frac{1}{2}）^{3}$×2×（1×5-2×3），求得|A|，再求出A的转置矩阵，由A^-1=$\frac{1}{丨A丨}$×A*，求出，A^-1，根据矩阵A*丨I进行初等行变换即可求得（A^*）^-1．

解答 解：A=$\frac{1}{2}$$（\begin{array}{l}{2}&{0}&{0}\\{0}&{1}&{3}\\{0}&{2}&{5}\end{array}）$，设B=$（\begin{array}{l}{2}&{0}&{0}\\{0}&{1}&{3}\\{0}&{2}&{5}\end{array}）$，
|A|=$（\frac{1}{2}）^{3}$×2×（1×5-2×3）=-$\frac{1}{4}$，
A₁₁=（$\frac{1}{2}$）²$|\begin{array}{l}{1}&{3}\\{2}&{5}\end{array}|$=-$\frac{1}{4}$，A₁₂=-（$\frac{1}{2}$）²$|\begin{array}{l}{0}&{3}\\{0}&{5}\end{array}|$=0，A₁₃=（$\frac{1}{2}$）²$|\begin{array}{l}{0}&{1}\\{0}&{2}\end{array}|$=0，
A₂₁=-（$\frac{1}{2}$）²$|\begin{array}{l}{0}&{0}\\{2}&{5}\end{array}|$=0，A₂₂=（$\frac{1}{2}$）²$|\begin{array}{l}{2}&{0}\\{0}&{5}\end{array}|$=$\frac{5}{2}$，A₂₃=-（$\frac{1}{2}$）²$|\begin{array}{l}{2}&{0}\\{0}&{2}\end{array}|$=-1，
A₃₁=（$\frac{1}{2}$）²$|\begin{array}{l}{0}&{0}\\{1}&{3}\end{array}|$=0，A₃₂=-（$\frac{1}{2}$）²$|\begin{array}{l}{2}&{0}\\{0}&{3}\end{array}|$=-$\frac{3}{2}$，A₃₃=（$\frac{1}{2}$）²$|\begin{array}{l}{2}&{0}\\{0}&{1}\end{array}|$=$\frac{1}{2}$，
∴A*=$[\begin{array}{l}{-\frac{1}{4}}&{0}&{0}\\{0}&{\frac{5}{2}}&{-\frac{3}{2}}\\{0}&{-1}&{\frac{1}{2}}\end{array}]$，
∴A^-1=$\frac{1}{丨A丨}$×A*=-4×$[\begin{array}{l}{-\frac{1}{4}}&{0}&{0}\\{0}&{\frac{5}{2}}&{-\frac{3}{2}}\\{0}&{-1}&{\frac{1}{2}}\end{array}]$=$[\begin{array}{l}{1}&{0}&{0}\\{0}&{-10}&{6}\\{0}&{4}&{-2}\end{array}]$，
∵（A*丨I）=$[\begin{array}{l}{\frac{1}{4}}&{0}&{0}&{1}&{0}&{0}\\{0}&{\frac{5}{2}}&{\frac{3}{2}}&{0}&{1}&{0}\\{0}&{-1}&{\frac{1}{2}}&{0}&{0}&{1}\end{array}]$→$[\begin{array}{l}{1}&{0}&{0}&{-4}&{0}&{0}\\{0}&{1}&{\frac{3}{5}}&{0}&{\frac{2}{5}}&{0}\\{0}&{0}&{\frac{1}{10}}&{0}&{\frac{2}{5}}&{1}\end{array}]$→$[\begin{array}{l}{1}&{0}&{0}&{-4}&{0}&{0}\\{0}&{1}&{0}&{0}&{-2}&{-6}\\{0}&{0}&{1}&{0}&{-4}&{-10}\end{array}]$，
∴（A^*）^-1=$[\begin{array}{l}{-4}&{0}&{0}\\{0}&{-2}&{-6}\\{0}&{-4}&{-10}\end{array}]$．

点评 本题考查逆变换及逆矩阵，考查求|A|，转置矩阵及逆矩阵的方法，考查计算能力，属于中档题．