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

分析 由矩陣的運(yùn)算性質(zhì)|A|=$(\frac{1}{2})^{3}$×2×(1×5-2×3),求得|A|,再求出A的轉(zhuǎn)置矩陣,由A-1=$\frac{1}{丨A丨}$×A*,求出,A-1,根據(jù)矩陣A*丨I進(jìn)行初等行變換即可求得(A*-1

解答 解:A=$\frac{1}{2}$$(\begin{array}{l}{2}&{0}&{0}\\{0}&{1}&{3}\\{0}&{2}&{5}\end{array})$,設(shè)B=$(\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}]$.

點(diǎn)評 本題考查逆變換及逆矩陣,考查求|A|,轉(zhuǎn)置矩陣及逆矩陣的方法,考查計(jì)算能力,屬于中檔題.

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