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CIE 9231 2023 November Paper 22 Q6

A Level / CIE / FP2

CIE 9231 2023 Nov Paper 22 Paper · Question 6

题目

Problem

The matrix PP is given by

P=(1−1102100−1).P=\begin{pmatrix} 1&-1&1\\ 0&2&1\\ 0&0&-1 \end{pmatrix}.

(a) State the eigenvalues of PP.

[1]

(b) Use the characteristic equation of PP to find P−1P^{-1}.

[4]

The 3×33\times3 matrix AA has distinct non-zero eigenvalues aa, 12\frac{1}{2}, 22 with corresponding eigenvectors

(100),(−120),(11−1)\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad \begin{pmatrix}-1\\2\\0\end{pmatrix},\qquad \begin{pmatrix}1\\1\\-1\end{pmatrix}

respectively.

(c) Find A−1A^{-1} in terms of aa.

[5]
题目中文翻译

给定矩阵 PP

P=(1−1102100−1)P=\begin{pmatrix} 1&-1&1\\ 0&2&1\\ 0&0&-1 \end{pmatrix}

(a) 写出 PP 的特征值。

(1)

(b) 利用 PP 的特征方程求 P−1P^{-1}。

(4)

三阶矩阵 AA 有互不相同且非零的特征值 aa、12\frac{1}{2}、22,其对应的特征向量依次为

(100),(−120),(11−1)\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad \begin{pmatrix}-1\\2\\0\end{pmatrix},\qquad \begin{pmatrix}1\\1\\-1\end{pmatrix}

(c) 求用 aa 表示的 A−1A^{-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.

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

(a)

解法一

思路

展开

上三角矩阵的特征值由主对角线直接读出。

答题过程

展开

The eigenvalues of PP are 1,2,−1\boxed{1,2,-1}.

(b)

解法一

思路

展开

用特征值写出特征方程。由 Cayley–Hamilton 定理再乘 P−1P^{-1},将逆矩阵表示成 P2,P,IP^2,P,I 的组合。

答题过程

展开

The characteristic polynomial is

(λ−1)(λ−2)(λ+1)=λ3−2λ2−λ+2.(\lambda-1)(\lambda-2)(\lambda+1)=\lambda^3-2\lambda^2-\lambda+2.

By the Cayley–Hamilton theorem, P3−2P2−P+2I=0P^3-2P^2-P+2I=0. Multiplying by P−1P^{-1} gives 2P−1=I+2P−P22P^{-1}=I+2P-P^2. For example, (P2)12=1(−1)+(−1)2=−3(P^2)_{12}=1(-1)+(-1)2=-3 and (P2)13=1(1)+(−1)1+1(−1)=−1(P^2)_{13}=1(1)+(-1)1+1(-1)=-1. Thus

P2=(1−3−1041001),P^2=\begin{pmatrix}1&-3&-1\\0&4&1\\0&0&1\end{pmatrix},

we obtain

P−1=(112320121200−1).P^{-1}=\begin{pmatrix}1&\frac12&\frac32\\0&\frac12&\frac12\\0&0&-1\end{pmatrix}.

(c)

解法一

思路

展开

题给三个特征向量恰好是 PP 的列。逆矩阵的特征值取倒数,按相同顺序放入 DD,再计算 PDP−1PDP^{-1}。

答题过程

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The columns of PP are the stated eigenvectors of AA. They are also eigenvectors of A−1A^{-1}, with eigenvalues 1/a,2,1/21/a,2,1/2 respectively. Therefore

A−1= Pdiag⁡(1/a,2,1/2)P−1= (1/a−212041200−12) ×(112320121200−1)= (1a1−2a2a3−3a2a02320012).\begin{align*} A^{-1}=&\,P\operatorname{diag}(1/a,2,1/2)P^{-1}\\ =&\,\begin{pmatrix}1/a&-2&\frac12\\0&4&\frac12\\0&0&-\frac12\end{pmatrix}\\ &\,\hspace{2pt}\times\begin{pmatrix}1&\frac12&\frac32\\0&\frac12&\frac12\\0&0&-1\end{pmatrix}\\ =&\,\begin{pmatrix} \frac1a&\frac{1-2a}{2a}&\frac{3-3a}{2a}\\ 0&2&\frac32\\0&0&\frac12 \end{pmatrix}. \end{align*}