(b) Use the characteristic equation of P to find P−1.
[4]
The 3×3 matrix A has distinct non-zero eigenvalues a, 21, 2 with corresponding eigenvectors
100,−120,11−1
respectively.
(c) Find A−1 in terms of a.
[5]
题目中文翻译
给定矩阵 P
P=100−12011−1
(a) 写出 P 的特征值。
(1)
(b) 利用 P 的特征方程求 P−1。
(4)
三阶矩阵 A 有互不相同且非零的特征值 a、21、2,其对应的特征向量依次为
100,−120,11−1
(c) 求用 a 表示的 A−1。
(5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
By the Cayley–Hamilton theorem, P3−2P2−P+2I=0. Multiplying by P−1 gives 2P−1=I+2P−P2. For example, (P2)12=1(−1)+(−1)2=−3 and (P2)13=1(1)+(−1)1+1(−1)=−1. Thus
P2=100−340−111,
we obtain
P−1=100212102321−1.
(c)
解法一
思路
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题给三个特征向量恰好是 P 的列。逆矩阵的特征值取倒数,按相同顺序放入 D,再计算 PDP−1。
答题过程
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The columns of P are the stated eigenvectors of A. They are also eigenvectors of A−1, with eigenvalues 1/a,2,1/2 respectively. Therefore