Question

3．已知矩阵A=$[\begin{array}{l}{1}&{2}\\{0}&{-2}\end{array}]$，矩阵B的逆矩阵B^-1=$[\begin{array}{l}{1}&{-\frac{1}{2}}\\{0}&{2}\end{array}]$，求矩阵AB．

Answer 1

分析 依题意，利用矩阵变换求得B=（B^-1）^-1=$[\begin{array}{cc}\frac{2}{2}&\frac{\frac{1}{2}}{2}\\ \frac{0}{2}&\frac{1}{2}\end{array}\right.]$=$[\begin{array}{l}{1}&{\frac{1}{4}}\\{0}&{\frac{1}{2}}\end{array}]$，再利用矩阵乘法的性质可求得答案．

解答 解：∵B^-1=$[\begin{array}{l}{1}&{-\frac{1}{2}}\\{0}&{2}\end{array}]$，
∴B=（B^-1）^-1=$[\begin{array}{cc}\frac{2}{2}&\frac{\frac{1}{2}}{2}\\ \frac{0}{2}&\frac{1}{2}\end{array}\right.]$=$[\begin{array}{l}{1}&{\frac{1}{4}}\\{0}&{\frac{1}{2}}\end{array}]$，又A=$[\begin{array}{l}{1}&{2}\\{0}&{-2}\end{array}]$，
∴AB=$[\begin{array}{cc}1&2\\ 0&-2\end{array}]$ $[\begin{array}{l}{1}&{\frac{1}{4}}\\{0}&{\frac{1}{2}}\end{array}]$=$[\begin{array}{cc}1&\frac{5}{4}\\ 0&-1\end{array}]$．

点评 本题考查逆变换与逆矩阵，考查矩阵乘法的性质，属于中档题．