Question

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

Answer 1

分析 先求出AB=$[\begin{array}{l}{1}&{2}\\{0}&{2}\end{array}]$，由此利用矩陣的變換能求出AB的逆矩陣（AB）^-1．

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

點(diǎn)評(píng) 本題考查矩陣乘積的逆矩陣的求法，考查矩陣的乘積、逆矩陣等基礎(chǔ)知識(shí)，考查推理論證能力、運(yùn)算求解能力，考查化歸與轉(zhuǎn)化思想、函數(shù)與方程思想，是基礎(chǔ)題．