Question

4．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 先求出A=$[\begin{array}{l}{1}&{0}&{0}\\{0}&{\frac{1}{2}}&{\frac{3}{2}}\\{0}&{1}&{\frac{5}{2}}\end{array}]$，再求出A的转置矩阵，由此能求出|A|，A^-1，（A^*）^-1．

解答 解：∵A=$\frac{1}{2}$$[\begin{array}{l}{2}&{0}&{0}\\{0}&{1}&{3}\\{0}&{2}&{5}\end{array}]$=$[\begin{array}{l}{1}&{0}&{0}\\{0}&{\frac{1}{2}}&{\frac{3}{2}}\\{0}&{1}&{\frac{5}{2}}\end{array}]$，
∴|A|=$|\begin{array}{l}{1}&{0}&{0}\\{0}&{\frac{1}{2}}&{\frac{3}{2}}\\{0}&{1}&{\frac{5}{2}}\end{array}|$=1×$\frac{1}{2}×\frac{5}{2}$-1×$1×\frac{3}{2}$=-$\frac{1}{4}$；
A₁₁=$|\begin{array}{l}{\frac{1}{2}}&{\frac{3}{2}}\\{1}&{\frac{5}{2}}\end{array}|$=-$\frac{1}{4}$，A₁₂=-$|\begin{array}{l}{0}&{\frac{3}{2}}\\{0}&{\frac{5}{2}}\end{array}|$=0，A₁₃=$|\begin{array}{l}{0}&{\frac{1}{2}}\\{0}&{1}\end{array}|$=0，
A₂₁=-$|\begin{array}{l}{0}&{0}\\{1}&{\frac{5}{2}}\end{array}|$=0，A₂₂=$|\begin{array}{l}{1}&{0}\\{0}&{\frac{5}{2}}\end{array}|$=$\frac{5}{2}$，A₂₃=-$|\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}|$=-1，
${A}_{31}^{\;}$=$|\begin{array}{l}{0}&{0}\\{\frac{1}{2}}&{\frac{3}{2}}\end{array}|$=0，${A}_{32}^{\;}$=-$|\begin{array}{l}{1}&{0}\\{0}&{\frac{3}{2}}\end{array}|$=-$\frac{3}{2}$，${A}_{33}^{\;}$=$|\begin{array}{l}{1}&{0}\\{0}&{\frac{1}{2}}\end{array}|$=$\frac{1}{2}$，
∴A^*=$[\begin{array}{l}{-\frac{1}{4}}&{0}&{0}\\{0}&{\frac{5}{2}}&{-\frac{3}{2}}\\{0}&{-1}&{\frac{1}{2}}\end{array}]$，
∴${A}^{-1}=\frac{1}{|A|}•{A}^{*}$=-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^*|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 本题考查矩阵的行列式、逆矩阵、转置矩阵的求法，是中档题，解题时要熟练掌握基本概念．