Question

10．已知矩阵A=$（\begin{array}{l}{1}&{2}\\{-1}&{4}\end{array}）$．
（1）求A的逆矩阵A^-1；
（2）求矩阵A的特征值λ₁、λ₂和对应的一个特征向量$\overrightarrow{{α}_{1}}$、$\overrightarrow{α_2}$．

Answer 1

分析 （1）通过变换可得$[\begin{array}{l}{1}&{2}&{1}&{0}\\{-1}&{4}&{0}&{1}\end{array}]$→$[\begin{array}{l}{1}&{0}&{\frac{2}{3}}&{-\frac{1}{3}}\\{0}&{1}&{\frac{1}{6}}&{\frac{1}{6}}\end{array}]$，即得结论；
（2）令其特征多项式f（λ）为0，即可求得特征值，再分别在f（λ）=0代入特征值即可求得分别对应的特征向量．

解答 解：（1）∵$[\begin{array}{l}{1}&{2}&{1}&{0}\\{-1}&{4}&{0}&{1}\end{array}]$→$[\begin{array}{l}{1}&{2}&{1}&{0}\\{0}&{6}&{1}&{1}\end{array}]$→$[\begin{array}{l}{1}&{2}&{1}&{0}\\{0}&{1}&{\frac{1}{6}}&{\frac{1}{6}}\end{array}]$→$[\begin{array}{l}{1}&{0}&{\frac{2}{3}}&{-\frac{1}{3}}\\{0}&{1}&{\frac{1}{6}}&{\frac{1}{6}}\end{array}]$，
∴A^-1=$[\begin{array}{l}{\frac{2}{3}}&{-\frac{1}{3}}\\{\frac{1}{6}}&{\frac{1}{6}}\end{array}]$；
（2）∵A=$（\begin{array}{l}{1}&{2}\\{-1}&{4}\end{array}）$，
∴f（λ）=$|\begin{array}{l}{λ-1}&{-2}\\{1}&{λ-4}\end{array}|$=λ²-5λ+6=（λ-2）（λ-3）=0，
解得λ₁=2，λ₂=3，
设$\overrightarrow{{α}_{1}}$=$[\begin{array}{l}{{x}_{1}}\\{{y}_{1}}\end{array}]$，$\overrightarrow{{α}_{2}}$=$[\begin{array}{l}{{x}_{2}}\\{{y}_{2}}\end{array}]$，
∵$[\begin{array}{l}{1}&{2}\\{-1}&{4}\end{array}]$$[\begin{array}{l}{{x}_{1}}\\{{y}_{1}}\end{array}]$=2$[\begin{array}{l}{{x}_{1}}\\{{y}_{1}}\end{array}]$，∴$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=1}\end{array}\right.$，即$\overrightarrow{{α}_{1}}$=$[\begin{array}{l}{2}\\{1}\end{array}]$，
∵∵$[\begin{array}{l}{1}&{2}\\{-1}&{4}\end{array}]$$[\begin{array}{l}{{x}_{2}}\\{{y}_{2}}\end{array}]$=3$[\begin{array}{l}{{x}_{2}}\\{{y}_{2}}\end{array}]$，∴$\left\{\begin{array}{l}{{x}_{2}=1}\\{{y}_{2}=1}\end{array}\right.$，即$\overrightarrow{{α}_{2}}$=$[\begin{array}{l}{1}\\{1}\end{array}]$．

点评 本题考查矩阵特征值的求法及求其对应的特征向量，属于基础题．