Question

2．已知矩阵$A=[{\begin{array}{l}{-1}&2\\ 1&x\end{array}}]，B=[{\begin{array}{l}1&1\\ 2&{-1}\end{array}}]$，向量$\overrightarrow{a}$=$[{\begin{array}{l}2\\ y\end{array}}]$，若A$\overrightarrow{a}$=B$\overrightarrow{a}$，求实数x，y的值．

Answer 1

分析 利用矩阵与矩阵相乘的运算法则和矩阵相等的性质直接求解．

解答 解：∵矩阵$A=[{\begin{array}{l}{-1}&2\\ 1&x\end{array}}]，B=[{\begin{array}{l}1&1\\ 2&{-1}\end{array}}]$，向量$\overrightarrow{a}$=$[{\begin{array}{l}2\\ y\end{array}}]$，
∴A$\overrightarrow{a}$=$[\begin{array}{l}{-1}&{2}\\{1}&{x}\end{array}][\begin{array}{l}{2}\\{y}\end{array}]$=$[\begin{array}{l}{-2+2y}\\{2+xy}\end{array}]$，
B$\overrightarrow{a}$=$[\begin{array}{l}{1}&{1}\\{2}&{-1}\end{array}][\begin{array}{l}{2}\\{y}\end{array}]$=$[\begin{array}{l}{2+y}\\{4-y}\end{array}]$，
∵A$\overrightarrow{a}$=B$\overrightarrow{a}$，∴$[\begin{array}{l}{-2+2y}\\{2+xy}\end{array}]$=$[\begin{array}{l}{2+y}\\{4-y}\end{array}]$，
∴$\left\{\begin{array}{l}{-2+2y=2+y}\\{2+xy=4-y}\end{array}\right.$，解得x=-$\frac{1}{2}$，y=4．

点评 本题考查实数值的求法，是基础题，解题时要认真审题，注意矩阵与矩阵相乘的运算法则和矩阵相等的性质的合理运用．