Question

2．已知矩阵A=$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$，B=$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$，求矩阵A^-1B．

Answer 1

分析 设矩阵A-1=$[\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}]$，通过AA^-1为单位矩阵可得A^-1，进而可得结论．

解答 解：设矩阵A的逆矩阵为$[\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}]$，
则$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$$[\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}]$=$[\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}]$，即$[\begin{array}{l}{-a}&{-b}\\{2c}&{2d}\end{array}]$=$[\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}]$，
故a=-1，b=0，c=0，d=$\frac{1}{2}$，
从而A^-1=$[\begin{array}{l}{-1}&{0}\\{0}&{\frac{1}{2}}\end{array}]$，
∴A^-1B=$[\begin{array}{l}{-1}&{0}\\{0}&{\frac{1}{2}}\end{array}]$$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$．

点评 本题考查逆矩阵、矩阵的乘法，考查运算求解能力，属于基础题．