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CIE 9231 2023 June Paper 12 Q1

A Level / CIE / FP1

CIE 9231 2023 June Paper 12 Paper · Question 1

题目

Problem

Let A=(3011)\mathbf{A}=\begin{pmatrix}3&0\\1&1\end{pmatrix}.

(a) Prove by mathematical induction that, for all positive integers nn,

2An=(2×3n03n−12).2\mathbf{A}^n=\begin{pmatrix}2\times3^n&0\\3^n-1&2\end{pmatrix}.
(5)

(b) Find, in terms of nn, the inverse of An\mathbf{A}^n.

(2)
题目中文翻译

设 A=(3011)\mathbf{A}=\begin{pmatrix}3&0\\1&1\end{pmatrix}。

(a) 用数学归纳法证明,对于所有正整数 nn,

2An=(2×3n03n−12).2\mathbf{A}^n=\begin{pmatrix}2\times3^n&0\\3^n-1&2\end{pmatrix}.
(5)

(b) 求 An\mathbf{A}^n 的逆矩阵,用 nn 表示。

(2)
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

先验n=1;归纳时将 2Ak2A^k 右乘 AA,左下元素合并后正好是下一次幂的目标值。

答题过程

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For n=1n=1,

2A= (6022)= (2×31031−12),\begin{align*} 2\mathbf A=&\,\begin{pmatrix}6&0\\2&2\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}2\times3^1&0\\3^1-1&2\end{pmatrix}, \end{align*}

so the statement holds.

Assume it holds for some positive integer kk, so

2Ak=(2×3k03k−12).2\mathbf A^k=\begin{pmatrix}2\times3^k&0\\3^k-1&2\end{pmatrix}.

Then

2Ak+1= (2Ak)A= (2×3k03k−12)(3011)= (2×3k+103(3k−1)+22)= (2×3k+103k+1−12).\begin{align*} 2\mathbf A^{k+1} =&\,(2\mathbf A^k)\mathbf A\\[4mm] =&\,\begin{pmatrix}2\times3^k&0\\3^k-1&2\end{pmatrix} \begin{pmatrix}3&0\\1&1\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}2\times3^{k+1}&0\\3(3^k-1)+2&2\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}2\times3^{k+1}&0\\3^{k+1}-1&2\end{pmatrix}. \end{align*}

This is the required expression for n=k+1n=k+1. Since it holds for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1, the result follows by induction for all positive integers nn.

(b)

解法一

思路

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先把上一问除以2得到A的n次幂,再用两行两列矩阵的逆公式。注意常数2在取逆前后的不同位置。

答题过程

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By (a),

An=(3n0(3n−1)/21).\mathbf A^n=\begin{pmatrix}3^n&0\\(3^n-1)/2&1\end{pmatrix}.

Its determinant is

det⁡(An)=3n≠0.\det(\mathbf A^n)=3^n\ne0.

Therefore

(An)−1=3−n(10(1−3n)/23n).\boxed{(\mathbf A^n)^{-1} =3^{-n}\begin{pmatrix}1&0\\(1-3^n)/2&3^n\end{pmatrix}}.