Question

16．設矩陣M=$[\begin{array}{l}{1}&{2}\\{x}&{y}\end{array}]$，N=$[\begin{array}{l}{2}&{4}\\{-1}&{-1}\end{array}]$，若MN=$[\begin{array}{l}{0}&{2}\\{5}&{13}\end{array}]$，求矩陣M的逆矩陣M^-1．

Answer 1

分析 先求出MN=$[\begin{array}{l}{0}&{2}\\{2x-y}&{4x-y}\end{array}]$，由MN=$[\begin{array}{l}{0}&{2}\\{5}&{13}\end{array}]$，列出方程組求出M=$[\begin{array}{l}{1}&{2}\\{4}&{3}\end{array}]$，由此能求出矩陣M的逆矩陣M^-1．

解答 解：∵M=$[\begin{array}{l}{1}&{2}\\{x}&{y}\end{array}]$，N=$[\begin{array}{l}{2}&{4}\\{-1}&{-1}\end{array}]$，
∴MN=$[\begin{array}{l}{0}&{2}\\{2x-y}&{4x-y}\end{array}]$，
∵MN=$[\begin{array}{l}{0}&{2}\\{5}&{13}\end{array}]$，∴$\left\{\begin{array}{l}{2x-y=5}\\{4x-y=13}\end{array}\right.$，
解得x=4，y=3，
∴M=$[\begin{array}{l}{1}&{2}\\{4}&{3}\end{array}]$，
∵（A|I）=$[\begin{array}{l}{1}&{2}&{\;}&{1}&{0}\\{4}&{3}&{\;}&{0}&{1}\end{array}]$→$[\begin{array}{l}{1}&{2}&{\;}&{1}&{0}\\{0}&{-5}&{\;}&{-4}&{1}\end{array}]$→$[\begin{array}{l}{1}&{0}&{\;}&{-\frac{3}{5}}&{\frac{2}{5}}\\{0}&{1}&{\;}&{\frac{4}{5}}&{-\frac{1}{5}}\end{array}]$．
∴矩陣M的逆矩陣M^-1=$[\begin{array}{l}{-\frac{3}{5}}&{\frac{2}{5}}\\{\frac{4}{5}}&{-\frac{1}{5}}\end{array}]$．

點評 本題考查逆矩陣的求法，涉及到矩陣與矩陣相乘、矩陣變換等基礎知識，考查推理論證能力、運算求解能力，考查化歸與轉化思想、函數與方程思想，是基礎題．