Question

9．计算：$[\begin{array}{l}{1}&{2}\\{2}&{1}\end{array}]$$[\begin{array}{l}{-1}&{2}\\{3}&{-4}\end{array}]$=$[\begin{array}{l}{5}&{-6}\\{1}&{0}\end{array}]$．

Answer 1

分析 通过矩阵乘法的性质计算即得结论．

解答 解：$[\begin{array}{l}{1}&{2}\\{2}&{1}\end{array}]$$[\begin{array}{l}{-1}&{2}\\{3}&{-4}\end{array}]$=$[\begin{array}{l}{1×（-1）+2×3}&{1×2+2×（-4）}\\{2×（-1）+1×3}&{2×2+1×（-4）}\end{array}]$=$[\begin{array}{l}{5}&{-6}\\{1}&{0}\end{array}]$，
故答案为：$[\begin{array}{l}{5}&{-6}\\{1}&{0}\end{array}]$．

点评 本题考查矩阵乘法的运算，注意解题方法的积累，属于基础题．