Question

6．設(shè)矩陣A滿足：A$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$，求矩陣A的逆矩陣A^-1．

Answer 1

分析 設(shè)B=$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$，求得B^*，則B^-1=$\frac{1}{丨B丨}$×B^*，由矩陣的乘法，A=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$×B^-1，即可求得矩陣A，則A^-1=$\frac{1}{丨A丨}$×$[\begin{array}{l}{\frac{1}{2}}&{0}\\{0}&{-1}\end{array}]$．，即可求得A^-1．

解答 解：A$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$，設(shè)B=$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$，則丨B丨=6，B^*=$[\begin{array}{l}{6}&{-2}\\{0}&{1}\end{array}]$，
則B^-1=$\frac{1}{丨B丨}$×B^*=$\frac{1}{6}$×$[\begin{array}{l}{6}&{-2}\\{0}&{1}\end{array}]$=$[\begin{array}{l}{1}&{-\frac{1}{3}}\\{0}&{\frac{1}{6}}\end{array}]$，
A=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$×B^-1=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$$[\begin{array}{l}{1}&{-\frac{1}{3}}\\{0}&{\frac{1}{6}}\end{array}]$=$[\begin{array}{l}{-1}&{0}\\{0}&{\frac{1}{2}}\end{array}]$，
A=$[\begin{array}{l}{-1}&{0}\\{0}&{\frac{1}{2}}\end{array}]$，丨A丨=-$\frac{1}{2}$，A^*=$[\begin{array}{l}{\frac{1}{2}}&{0}\\{0}&{-1}\end{array}]$
A^-1=$\frac{1}{丨A丨}$×$[\begin{array}{l}{\frac{1}{2}}&{0}\\{0}&{-1}\end{array}]$=$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$，
矩陣A的逆矩陣A^-1=$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$．

點(diǎn)評(píng) 本題考查矩陣與逆矩陣，考查逆矩陣的求法，矩陣的乘法，考查計(jì)算能力，屬于中檔題．