Question

12．矩阵A=$[\begin{array}{l}{a}&{k}\\{0}&{1}\end{array}]$（k≠0）的一个特征向量为$\overrightarrow{a}$=$[\begin{array}{l}k\\-1\end{array}]$，A的逆矩阵A^-1对应的变换将点（3，1）变为点（1，1）．则a+k=3．

Answer 1

分析 由$[\begin{array}{l}{a}&{k}\\{0}&{1}\end{array}]$$[\begin{array}{l}k\\-1\end{array}]$=λ$[\begin{array}{l}k\\-1\end{array}]$，即$\left\{\begin{array}{l}{ak-k=kλ}\\{-1=-λ}\end{array}\right.$，由k≠0，解得：$\left\{\begin{array}{l}{λ=1}\\{a=2}\end{array}\right.$，根据矩阵的运算性质可知：A$[\begin{array}{l}{1}\\{1}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，$[\begin{array}{l}{2}&{k}\\{0}&{1}\end{array}]$$[\begin{array}{l}{1}\\{1}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，即可求得k的值，求得a+k的值．

解答 解：设特征向量为$\overrightarrow{a}$=$[\begin{array}{l}k\\-1\end{array}]$，对应的特征值为λ，λ，
则$[\begin{array}{l}{a}&{k}\\{0}&{1}\end{array}]$$[\begin{array}{l}k\\-1\end{array}]$=λ$[\begin{array}{l}k\\-1\end{array}]$，即$\left\{\begin{array}{l}{ak-k=kλ}\\{-1=-λ}\end{array}\right.$，
由k≠0，解得：$\left\{\begin{array}{l}{λ=1}\\{a=2}\end{array}\right.$，
由A^-1$[\begin{array}{l}{3}\\{1}\end{array}]$=$[\begin{array}{l}{1}\\{1}\end{array}]$，即A$[\begin{array}{l}{1}\\{1}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，
$[\begin{array}{l}{2}&{k}\\{0}&{1}\end{array}]$$[\begin{array}{l}{1}\\{1}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，即$\left\{\begin{array}{l}{2+k=3}\\{1=1}\end{array}\right.$，解得：k=1，
∴a+k=2+1=3，
故答案为：3．

点评 本题考查矩阵的运算，考查矩阵的特征值与特征向量的计算，考查计算能力，属于中档题．