Question

13．已知矩陣A=$[\begin{array}{l}{a}&\\{c}&4zup0h5\end{array}]$，若矩陣A屬于特征值λ₁=3的一個(gè)特征向量為$\overrightarrow{α}$₁=$[\begin{array}{l}{1}\\{1}\end{array}]$，屬于特征值λ₂=1的一個(gè)特征向量$\overrightarrow{α}$₂=
$[\begin{array}{l}{1}\\{-1}\end{array}]$．
（1）求矩陣A；
（2）若向量$\overrightarrow{β}$=$[\begin{array}{l}{4}\\{2}\end{array}]$，求A²⁰¹⁷β．

Answer 1

分析 （1）由$[\begin{array}{l}{a}&\\{c}&d0ulfkp\end{array}][\begin{array}{l}{1}\\{1}\end{array}]=3[\begin{array}{l}{1}\\{1}\end{array}]$及$[\begin{array}{l}{a}&\\{c}&fjv0qlg\end{array}]$$[\begin{array}{l}{1}\\{-1}\end{array}]$=$[\begin{array}{l}{1}\\{-1}\end{array}]$，列方程組求出a，b，c，d，由此能求出矩陣A．
（2）設(shè)$\overrightarrow{β}$=$m\overrightarrow{{α}_{1}}+n\overrightarrow{{α}_{2}}$，則$[\begin{array}{l}{4}\\{2}\end{array}]=m[\begin{array}{l}{1}\\{1}\end{array}]+n[\begin{array}{l}{1}\\{-1}\end{array}]$，列方程組求出$\overrightarrow{β}=3\overrightarrow{{α}_{1}}+\overrightarrow{{α}_{2}}$，由此能求出A²⁰¹⁷β．

解答 解：（1）由$[\begin{array}{l}{a}&\\{c}&cwirmgb\end{array}][\begin{array}{l}{1}\\{1}\end{array}]=3[\begin{array}{l}{1}\\{1}\end{array}]$及$[\begin{array}{l}{a}&\\{c}&n2ssuwm\end{array}]$$[\begin{array}{l}{1}\\{-1}\end{array}]$=$[\begin{array}{l}{1}\\{-1}\end{array}]$，
得$\left\{\begin{array}{l}{a+b=3}\\{a-b=1}\\{c+d=3}\\{c-d=-1}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=2}\\{b=1}\\{c=1}\\{d=2}\end{array}\right.$，
∴A=$[\begin{array}{l}{2}&{1}\\{1}&{2}\end{array}]$．
（2）設(shè)$\overrightarrow{β}$=$m\overrightarrow{{α}_{1}}+n\overrightarrow{{α}_{2}}$，則$[\begin{array}{l}{4}\\{2}\end{array}]=m[\begin{array}{l}{1}\\{1}\end{array}]+n[\begin{array}{l}{1}\\{-1}\end{array}]$，
∴$\left\{\begin{array}{l}{4=m+n}\\{2=m-n}\end{array}\right.$，解得$\left\{\begin{array}{l}{m=3}\\{n=1}\end{array}\right.$，
∴$\overrightarrow{β}=3\overrightarrow{{α}_{1}}+\overrightarrow{{α}_{2}}$，
∴A²⁰¹⁷β=$3{{λ}_{1}}^{2017}\overrightarrow{{α}_{1}}+{{λ}_{2}}^{2017}\overrightarrow{{α}_{2}}$=$[\begin{array}{l}{{3}^{2018}+1}\\{{3}^{2018}-1}\end{array}]$．

點(diǎn)評(píng) 本題考查矩陣的求法，考查矩陣、特征向量、特征值等基礎(chǔ)知識(shí)，考查推理論能力、運(yùn)算求解能力，考查化歸與轉(zhuǎn)化思想、函數(shù)與方程思想，是中檔題．