Question

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

Answer 1

分析 先求出AB=$[\begin{array}{l}{1}&{2}\\{0}&{2}\end{array}]$，由此利用矩阵的变换能求出AB的逆矩阵（AB）^-1．

解答 解：∵矩阵$A=[{\begin{array}{l}1&0\\ 0&2\end{array}}]，B=[{\begin{array}{l}1&{\frac{1}{2}}\\ 0&1\end{array}}]$，
∴AB=$[\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}][\begin{array}{l}{1}&{\frac{1}{2}}\\{0}&{1}\end{array}]$=$[\begin{array}{l}{1}&{2}\\{0}&{2}\end{array}]$，
∵$[\begin{array}{l}{1}&{2}&{\;}&{1}&{0}\\{0}&{2}&{\;}&{0}&{1}\end{array}]$→$[\begin{array}{l}{1}&{2}&{\;}&{1}&{0}\\{0}&{1}&{\;}&{0}&{\frac{1}{2}}\end{array}]$→$[\begin{array}{l}{1}&{0}&{\;}&{1}&{-1}\\{0}&{1}&{\;}&{0}&{\frac{1}{2}}\end{array}]$
∴AB的逆矩阵（AB）^-1=$[\begin{array}{l}{1}&{-1}\\{0}&{\frac{1}{2}}\end{array}]$．
故答案为：$[\begin{array}{l}{1}&{-1}\\{0}&{\frac{1}{2}}\end{array}]$．

点评 本题考查矩阵乘积的逆矩阵的求法，考查矩阵的乘积、逆矩阵等基础知识，考查推理论证能力、运算求解能力，考查化归与转化思想、函数与方程思想，是基础题．