题目

Problem

The matrices $\mathbf{A}$ , $\mathbf{P}$ and $\mathbf{D}$ are given by

\begin{align*} \mathbf{A}=&\,\begin{pmatrix}8&-5\\7&-4\end{pmatrix},\qquad \mathbf{P}=\begin{pmatrix}5&3\\7&3\end{pmatrix},\\ \mathbf{D}=&\,\begin{pmatrix}1&0\\0&3\end{pmatrix}. \end{align*}

Prove by mathematical induction that, for all positive integers $n$ , $\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}$ .

(7)

题目中文翻译

矩阵 $\mathbf{A}$ 、 $\mathbf{P}$ 和 $\mathbf{D}$ 为

\begin{align*} \mathbf{A}=&\,\begin{pmatrix}8&-5\\7&-4\end{pmatrix},\qquad \mathbf{P}=\begin{pmatrix}5&3\\7&3\end{pmatrix},\\ \mathbf{D}=&\,\begin{pmatrix}1&0\\0&3\end{pmatrix}. \end{align*}

用数学归纳法证明：对于所有正整数 $n$ ， $\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}$ 。

(7)

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

展开

先直接算出P的逆矩阵并验证n=1，再假设n=k成立，利用中间的P逆与P相消推出下一步。

答题过程

展开

Since $\det\mathbf P=15-21=-6\ne0$ ,

\begin{align*} \mathbf P^{-1}=&\,-\frac16\begin{pmatrix}3&-3\\-7&5\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}-1/2&1/2\\7/6&-5/6\end{pmatrix}. \end{align*}

For the base case $n=1$ ,

\begin{align*} \mathbf P\mathbf D\mathbf P^{-1} =&\,\begin{pmatrix}5&9\\7&9\end{pmatrix} \begin{pmatrix}-1/2&1/2\\7/6&-5/6\end{pmatrix} \\[4mm] =&\,\begin{pmatrix} -5/2+63/6&5/2-45/6\\ -7/2+63/6&7/2-45/6 \end{pmatrix} \\[4mm] =&\,\begin{pmatrix}8&-5\\7&-4\end{pmatrix}=\mathbf A. \end{align*}

Thus the result is true for $n=1$ .

Assume for some positive integer $k$ that $\mathbf A^k=\mathbf P\mathbf D^k\mathbf P^{-1}$ . Then

\begin{align*} \mathbf A^{k+1} =&\,\mathbf A^k\mathbf A \\[4mm] =&\,(\mathbf P\mathbf D^k\mathbf P^{-1}) (\mathbf P\mathbf D\mathbf P^{-1}) \\[4mm] =&\,\mathbf P\mathbf D^k(\mathbf P^{-1}\mathbf P) \mathbf D\mathbf P^{-1} \\[4mm] =&\,\mathbf P\mathbf D^{k+1}\mathbf P^{-1}. \end{align*}

Therefore the result holds for $k+1$ . By mathematical induction,

\boxed{\mathbf A^n=\mathbf P\mathbf D^n\mathbf P^{-1}}

for every positive integer $n$ .

CIE 9231 2026 June Paper 13 Q3

题目

解答

解法一

思路

答题过程