Question

20．矩陣M=$[\begin{array}{l}{3}&{6}\\{5}&{2}\end{array}]$有屬于特征值λ₁=8的一個特征向量e₁=$[\begin{array}{l}{6}\\{5}\end{array}]$，及屬于特征值λ₂=-3的一個特征向量e₂=$[\begin{array}{l}{1}\\{-1}\end{array}]$．
（1）對于向量$\overrightarrow{α}=[\begin{array}{l}{3}\\{8}\end{array}]$，記做$\overrightarrow{α}={e}_{1}-3{e}_{2}$，利用這一表達式計算${M}^{3}\overrightarrow{α}$，${M}^{50}\overrightarrow{α}$；
（2）對于向量$\overrightarrow{β}=[\begin{array}{l}{8}\\{3}\end{array}]$，計算 ${M}^{5}\overrightarrow{β}$，${M}^{100}\overrightarrow{β}$．

Answer 1

分析 本題（1）可以將向量$\overrightarrow{α}=[\begin{array}{l}{3}\\{8}\end{array}]$轉(zhuǎn)化為向量e₁=$[\begin{array}{l}{6}\\{5}\end{array}]$、向量e₂=$[\begin{array}{l}{1}\\{-1}\end{array}]$的線性組合，再利用特征值的計算規(guī)律，求出算${M}^{3}\overrightarrow{α}$，${M}^{50}\overrightarrow{α}$；（2）類似（1）的計算，計算 ${M}^{5}\overrightarrow{β}$，${M}^{100}\overrightarrow{β}$，得到本題結(jié)論．

解答 解：（1）∵矩陣M=$[\begin{array}{l}{3}&{6}\\{5}&{2}\end{array}]$有屬于特征值λ₁=8的一個特征向量e₁=$[\begin{array}{l}{6}\\{5}\end{array}]$，及屬于特征值λ₂=-3的一個特征向量e₂=$[\begin{array}{l}{1}\\{-1}\end{array}]$，
∴$[\begin{array}{l}{3}&{6}\\{5}&{2}\end{array}]$×$[\begin{array}{l}{6}\\{5}\end{array}]$=8×$[\begin{array}{l}{6}\\{5}\end{array}]$，$[\begin{array}{l}{3}&{6}\\{5}&{2}\end{array}]$×$[\begin{array}{l}{1}\\{-1}\end{array}]$=-3×$[\begin{array}{l}{1}\\{-1}\end{array}]$．
∵$[\begin{array}{l}{3}\\{8}\end{array}]$=$[\begin{array}{l}{6}\\{5}\end{array}]$-3×$[\begin{array}{l}{1}\\{-1}\end{array}]$，
∴${M}^{3}\overrightarrow{α}$=M³（$[\begin{array}{l}{6}\\{5}\end{array}]$-3×$[\begin{array}{l}{1}\\{-1}\end{array}]$）=M³×$[\begin{array}{l}{6}\\{5}\end{array}]$-3×M³×$[\begin{array}{l}{1}\\{-1}\end{array}]$=8³×$[\begin{array}{l}{6}\\{5}\end{array}]$-3×（-3）³×$[\begin{array}{l}{1}\\{-1}\end{array}]$=$[\begin{array}{l}{6×{8}^{3}+{3}^{4}}\\{5×{8}^{3}-{3}^{4}}\end{array}]$=$[\begin{array}{l}{3153}\\{2479}\end{array}]$．
${M}^{50}\overrightarrow{α}$=M⁵⁰（$[\begin{array}{l}{6}\\{5}\end{array}]$-3×$[\begin{array}{l}{1}\\{-1}\end{array}]$）=M⁵⁰×$[\begin{array}{l}{6}\\{5}\end{array}]$-3×M⁵⁰×$[\begin{array}{l}{1}\\{-1}\end{array}]$=8⁵⁰×$[\begin{array}{l}{6}\\{5}\end{array}]$-3×（-3）⁵⁰×$[\begin{array}{l}{1}\\{-1}\end{array}]$=$[\begin{array}{l}{6×{8}^{50}+（-3）^{51}}\\{5×{8}^{50}-（-3）^{51}}\end{array}]$．
（2）∵$[\begin{array}{l}{8}\\{3}\end{array}]$=$[\begin{array}{l}{6}\\{5}\end{array}]$+2×$[\begin{array}{l}{1}\\{-1}\end{array}]$，

∴${M}^{5}\overrightarrow{β}$=M⁵（$[\begin{array}{l}{6}\\{5}\end{array}]$+2×$[\begin{array}{l}{1}\\{-1}\end{array}]$）=M⁵×$[\begin{array}{l}{6}\\{5}\end{array}]$+2×M⁵×$[\begin{array}{l}{1}\\{-1}\end{array}]$=8⁵×$[\begin{array}{l}{6}\\{5}\end{array}]$+2×（-3）⁵×$[\begin{array}{l}{1}\\{-1}\end{array}]$=$[\begin{array}{l}{196122}\\{164326}\end{array}]$．

${M}^{100}\overrightarrow{β}$=M¹⁰⁰（$[\begin{array}{l}{6}\\{5}\end{array}]$+2×$[\begin{array}{l}{1}\\{-1}\end{array}]$）=M¹⁰⁰×$[\begin{array}{l}{6}\\{5}\end{array}]$+2×M¹⁰⁰×$[\begin{array}{l}{1}\\{-1}\end{array}]$=8¹⁰⁰×$[\begin{array}{l}{6}\\{5}\end{array}]$+2×（-3）¹⁰⁰×$[\begin{array}{l}{1}\\{-1}\end{array}]$=$[\begin{array}{l}{6×{8}^{100}+2×{3}^{100}}\\{5×{8}^{100}-2×{3}^{100}}\end{array}]$．

點評 本題考查的是利用矩陣的特征值和特征向量，進行矩陣與向量的積的運算，本題難度不大，屬于基礎(chǔ)題．