题目
Problem
The matrix A A A
A = ( − 1 2 12 0 1 0 0 0 3 ) . A=\begin{pmatrix}
-1&2&12\\
0&1&0\\
0&0&3
\end{pmatrix}. A = − 1 0 0 2 1 0 12 0 3 . Use the characteristic equation of A A A
A 4 = p A 2 + q I , A^4=pA^2+qI, A 4 = p A 2 + q I , where p p p q q q
[6]
题目中文翻译
给定矩阵 A A A
A = ( − 1 2 12 0 1 0 0 0 3 ) A=\begin{pmatrix}
-1&2&12\\
0&1&0\\
0&0&3
\end{pmatrix} A = − 1 0 0 2 1 0 12 0 3 利用 A A A
A 4 = p A 2 + q I A^4=pA^2+qI A 4 = p A 2 + q I 其中 p p p q q q
(6)
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.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
解法一
思路
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由上三角矩阵的三个特征值写出特征多项式。用 Cayley–Hamilton 定理先表示 A 3 A^3 A 3 A A A A 3 A^3 A 3
答题过程
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The characteristic equation is
( λ + 1 ) ( λ − 1 ) ( λ − 3 ) = λ 3 − 3 λ 2 − λ + 3 = 0. (\lambda+1)(\lambda-1)(\lambda-3)=\lambda^3-3\lambda^2-\lambda+3=0. ( λ + 1 ) ( λ − 1 ) ( λ − 3 ) = λ 3 − 3 λ 2 − λ + 3 = 0. By the Cayley–Hamilton theorem, A 3 = 3 A 2 + A − 3 I A^3=3A^2+A-3I A 3 = 3 A 2 + A − 3 I A A A A 3 A^3 A 3
A 4 = 3 A 3 + A 2 − 3 A = 3 ( 3 A 2 + A − 3 I ) + A 2 − 3 A = 9 A 2 + 3 A − 9 I + A 2 − 3 A = 10 A 2 − 9 I . \begin{align*}
A^4=&\,3A^3+A^2-3A\\
=&\,3(3A^2+A-3I)+A^2-3A\\
=&\,9A^2+3A-9I+A^2-3A\\
=&\,10A^2-9I.
\end{align*} A 4 = = = = 3 A 3 + A 2 − 3 A 3 ( 3 A 2 + A − 3 I ) + A 2 − 3 A 9 A 2 + 3 A − 9 I + A 2 − 3 A 10 A 2 − 9 I . Thus ( p , q ) = ( 10 , − 9 ) (p,q)=(10,-9) ( p , q ) = ( 10 , − 9 )
解法二
思路
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也可保留特征多项式的因式。补乘一个 A + 3 I A+3I A + 3 I A 2 A^2 A 2
答题过程
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By the Cayley–Hamilton theorem, ( A 2 − I ) ( A − 3 I ) = 0 (A^2-I)(A-3I)=0 ( A 2 − I ) ( A − 3 I ) = 0 A + 3 I A+3I A + 3 I
( A 2 − I ) ( A 2 − 9 I ) = A 4 − 10 A 2 + 9 I = 0. (A^2-I)(A^2-9I)=A^4-10A^2+9I=0. ( A 2 − I ) ( A 2 − 9 I ) = A 4 − 10 A 2 + 9 I = 0. Therefore A 4 = 10 A 2 − 9 I A^4=10A^2-9I A 4 = 10 A 2 − 9 I
