Question

4．已知矩阵A=$[\begin{array}{l}{1}&{2}\\{-1}&{4}\end{array}]$，向量$\overrightarrow{a}$=$[\begin{array}{l}{5}\\{3}\end{array}]$，计算A⁵$\overrightarrow{a}$．

Answer 1

分析 令f（λ）=$|\begin{array}{l}{λ-1}&{-2}\\{1}&{λ-4}\end{array}|$=λ²-5λ+6=0，解得λ=2或3．分别对应的一个特征向量为$[\begin{array}{l}{2}\\{1}\end{array}]$；$[\begin{array}{l}{1}\\{1}\end{array}]$．设$[\begin{array}{l}{5}\\{3}\end{array}]$=m$[\begin{array}{l}{2}\\{1}\end{array}]$++n$[\begin{array}{l}{1}\\{1}\end{array}]$．解得m，n，即可得出．

解答 解：∵f（λ）=$|\begin{array}{l}{λ-1}&{-2}\\{1}&{λ-4}\end{array}|$=λ²-5λ+6，由f（λ）=0，解得λ=2或3．
当λ=2时，对应的一个特征向量为α₁=$[\begin{array}{l}{2}\\{1}\end{array}]$；当λ=3时，对应的一个特征向量为α₂=$[\begin{array}{l}{1}\\{1}\end{array}]$．
设$[\begin{array}{l}{5}\\{3}\end{array}]$=m$[\begin{array}{l}{2}\\{1}\end{array}]$++n$[\begin{array}{l}{1}\\{1}\end{array}]$．解得$\left\{\begin{array}{l}{m=2}\\{n=1}\end{array}\right.$．
∴A⁵$\overrightarrow{a}$=2×2⁵$[\begin{array}{l}{2}\\{1}\end{array}]$+1×3⁵$[\begin{array}{l}{1}\\{1}\end{array}]$=$[\begin{array}{l}{371}\\{307}\end{array}]$．

点评 本题考查了矩阵与变换、特征向量，考查了推理能力与计算能力，属于中档题．