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CIE 9231 2022 June Paper 13 Q5

A Level / CIE / FP1

CIE 9231 2022 June Paper 13 Paper · Question 5

题目

Problem

Let A=(1a01)\mathbf A=\begin{pmatrix}1&a\\0&1\end{pmatrix}, where aa is a positive constant.

(a) State the type of the geometrical transformation in the xx–yy plane represented by A\mathbf A.

(1)

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

An=(1na01).\mathbf A^n=\begin{pmatrix}1&na\\0&1\end{pmatrix}.
(5)

Let B=(bba−1a−1)\mathbf B=\begin{pmatrix}b&b\\a^{-1}&a^{-1}\end{pmatrix}, where bb is a positive constant.

(c) Find the equations of the invariant lines, through the origin, of the transformation in the xx–yy plane represented by AnB\mathbf A^n\mathbf B.

(6)
题目中文翻译

设 A=(1a01)\mathbf A=\begin{pmatrix}1&a\\0&1\end{pmatrix},其中 aa 是正常数。

(a) 写出 A\mathbf A 所表示的 xx–yy 平面内的几何变换的类型。

(1)

(b) 用数学归纳法证明,对于所有正整数 nn,

An=(1na01).\mathbf A^n=\begin{pmatrix}1&na\\0&1\end{pmatrix}.
(5)

设 B=(bba−1a−1)\mathbf B=\begin{pmatrix}b&b\\a^{-1}&a^{-1}\end{pmatrix},其中 bb 是正常数。

(c) 求 AnB\mathbf A^n\mathbf B 所表示的 xx–yy 平面内的变换的、经过原点的不变直线的方程。

(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

(a)

解法一

思路

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观察坐标变换:纵坐标不变,横坐标增加与纵坐标成比例的量。

答题过程

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The transformation is (x,y)↦(x+ay,y)(x,y)\mapsto(x+ay,y). It is a shear in the x-direction\boxed{\text{shear in the }x\text{-direction}}, with the xx-axis fixed and shear factor aa.

(b)

解法一

思路

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归纳法必须写出初始情形、明确的归纳假设、矩阵乘法及最终结论。

答题过程

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For n=1n=1,

A1=(1a01),\mathbf A^1=\begin{pmatrix}1&a\\0&1\end{pmatrix},

so the formula holds.

Assume that for some positive integer kk,

Ak=(1ka01).\mathbf A^k=\begin{pmatrix}1&ka\\0&1\end{pmatrix}.

Then

Ak+1= (1ka01)(1a01)= (1a+ka01)= (1(k+1)a01).\begin{align*} \mathbf A^{k+1} =&\,\begin{pmatrix}1&ka\\0&1\end{pmatrix}\begin{pmatrix}1&a\\0&1\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}1&a+ka\\0&1\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}1&(k+1)a\\0&1\end{pmatrix}. \end{align*}

Thus the formula holds for k+1k+1 whenever it holds for kk. By mathematical induction,

An=(1na01)\boxed{\mathbf A^n=\begin{pmatrix}1&na\\0&1\end{pmatrix}}

for every positive integer nn.