Question

6．设矩阵A满足：A$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$，求矩阵A的逆矩阵A^-1．

Answer 1

分析 设B=$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$，求得B^*，则B^-1=$\frac{1}{丨B丨}$×B^*，由矩阵的乘法，A=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$×B^-1，即可求得矩阵A，则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$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$，设B=$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$，则丨B丨=6，B^*=$[\begin{array}{l}{6}&{-2}\\{0}&{1}\end{array}]$，
则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=$\frac{1}{丨A丨}$×$[\begin{array}{l}{\frac{1}{2}}&{0}\\{0}&{-1}\end{array}]$=$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$，
矩阵A的逆矩阵A^-1=$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$．

点评 本题考查矩阵与逆矩阵，考查逆矩阵的求法，矩阵的乘法，考查计算能力，属于中档题．