Question

12．矩陣A=$[\begin{array}{l}{a}&{k}\\{0}&{1}\end{array}]$（k≠0）的一個特征向量為$\overrightarrow{a}$=$[\begin{array}{l}k\\-1\end{array}]$，A的逆矩陣A^-1對應(yīng)的變換將點（3，1）變?yōu)辄c（1，1）．則a+k=3．

Answer 1

分析 由$[\begin{array}{l}{a}&{k}\\{0}&{1}\end{array}]$$[\begin{array}{l}k\\-1\end{array}]$=λ$[\begin{array}{l}k\\-1\end{array}]$，即$\left\{\begin{array}{l}{ak-k=kλ}\\{-1=-λ}\end{array}\right.$，由k≠0，解得：$\left\{\begin{array}{l}{λ=1}\\{a=2}\end{array}\right.$，根據(jù)矩陣的運算性質(zhì)可知：A$[\begin{array}{l}{1}\\{1}\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，$[\begin{array}{l}{2}&{k}\\{0}&{1}\end{array}]$$[\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：設(shè)特征向量為$\overrightarrow{a}$=$[\begin{array}{l}k\\-1\end{array}]$，對應(yīng)的特征值為λ，λ，
則$[\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，
故答案為：3．

點評 本題考查矩陣的運算，考查矩陣的特征值與特征向量的計算，考查計算能力，屬于中檔題．