Question

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

Answer 1

分析 依題意，利用矩陣變換求得B=（B^-1）^-1=$[\begin{array}{cc}\frac{2}{2}&\frac{\frac{1}{2}}{2}\\ \frac{0}{2}&\frac{1}{2}\end{array}\right.]$=$[\begin{array}{l}{1}&{\frac{1}{4}}\\{0}&{\frac{1}{2}}\end{array}]$，再利用矩陣乘法的性質(zhì)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=$[\begin{array}{cc}1&2\\ 0&-2\end{array}]$ $[\begin{array}{l}{1}&{\frac{1}{4}}\\{0}&{\frac{1}{2}}\end{array}]$=$[\begin{array}{cc}1&\frac{5}{4}\\ 0&-1\end{array}]$．

點(diǎn)評 本題考查逆變換與逆矩陣，考查矩陣乘法的性質(zhì)，屬于中檔題．