题目

题目：设A为三阶矩阵.α11,α2,α3,为三维线性无关的列向量.又 (O)_(1)=(alpha )_(1)+4(alpha )_(2),-|||-(O)_(2)=(a)_(1)+(a)_(2) , (O)_(3)=3(alpha )_(3)设A为三阶矩阵.α11,α2,α3,为三维线性无关的列向量.又 (O)_(1)=(alpha )_(1)+4(alpha )_(2),-|||-(O)_(2)=(a)_(1)+(a)_(2) , (O)_(3)=3(alpha )_(3)

题目：

$\begin{aligned}&\text{设}A\text{为三阶矩阵}.\alpha_1,\alpha_2,\alpha_3,\text{为三维线性无关的列向量}.\text{又}A\alpha_1=\alpha_1+4\alpha_2,\\&A\alpha_2=\alpha_1+\alpha_2,\quad A\alpha_3=3\alpha_3.\end{aligned}$

$(1)\text\\{证明:矩阵}A\text{ 可相似对角化:} \$2)\text{设}P=(\alpha_1,\alpha_2,\alpha_3)=\begin{pmatrix}1&-1&0\\0&1&0\\2&1&1\end{pmatrix},\text{求} A.$

题目解答

答案

解答：

（1）证明： $\begin{aligned} &\text{令}P{=}(\alpha_{1},\alpha_{2},\alpha_{3}), \\ &{\text{由}}(A\alpha_{1},A\alpha_{2},A\alpha_{1}){=}(\alpha_{1}+4\alpha_{2},\alpha_{1}+\alpha_{2},3\alpha_{3}){\text{得}} \\ &\boldsymbol{AP}=\boldsymbol{P}\left[\begin{matrix}{1}&{1}&{0}\\{4}&{1}&{0}\\{0}&{0}&{3}\\\end{matrix}\right],\boldsymbol{从而}\boldsymbol{P}^{-1}\boldsymbol{AP}=\left[\begin{matrix}{1}&{1}&{0}\\{0}&{1}&{1}\\{0}&{0}&{1}\\\end{matrix}\right]=\boldsymbol{B}, \end{aligned}$