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CIE 9231 2023 November Paper 12 Q3

A Level / CIE / FP1

CIE 9231 2023 Nov Paper 12 Paper · Question 3

题目

Problem

The matrix M\mathbf{M} is given by M=(k001)(1011)\mathbf{M}=\begin{pmatrix}k&0\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}, where kk is a constant and k≠0k\ne0 and k≠1k\ne1.

(a) The matrix M\mathbf{M} represents a sequence of two geometrical transformations. State the type of each transformation, and make clear the order in which they are applied.

(2)

The unit square in the xx–yy plane is transformed by M\mathbf{M} onto parallelogram OPQROPQR.

(b) Find, in terms of kk, the area of parallelogram OPQROPQR and the matrix which transforms OPQROPQR onto the unit square.

(3)

(c) Show that the line through the origin with gradient 1k−1\dfrac{1}{k-1} is invariant under the transformation in the xx–yy plane represented by M\mathbf{M}.

(3)
题目中文翻译

矩阵 M\mathbf{M} 为 M=(k001)(1011)\mathbf{M}=\begin{pmatrix}k&0\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix},其中 kk 为常数,且 k≠0k\ne0、k≠1k\ne1。

(a) 矩阵 M\mathbf{M} 表示依次进行的两个几何变换。写出每个变换的类型,并明确指出它们的施行顺序。

(2)

xx–yy 平面内的单位正方形经 M\mathbf{M} 变换后成为平行四边形 OPQROPQR。

(b) 用 kk 表示平行四边形 OPQROPQR 的面积,以及将 OPQROPQR 变换为单位正方形的矩阵。

(3)

(c) 证明,在 xx–yy 平面内,经过原点且斜率为 1k−1\dfrac{1}{k-1} 的直线在 M\mathbf{M} 所表示的变换下保持不变。

(3)
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

右侧矩阵先作用,先剪切,后拉伸。题目只要求类型与次序,可在描述中顺便明确两矩阵的几何含义。

答题过程

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The transformation is a shear followed by a stretch.

The shear fixes the yy-axis and maps (1,0)(1,0) to (1,1)(1,1). The subsequent stretch is parallel to the xx-axis with scale factor kk.

(b)

解法一

思路

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面积倍率取行列式绝对值,不能把负k当作负面积。逆矩阵按两行两列的标准公式计算。

答题过程

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The combined matrix is

M=(k011).\mathbf M=\begin{pmatrix}k&0\\1&1\end{pmatrix}.

Thus det⁡M=k\det\mathbf M=k. Since the original square has area 11, the parallelogram has area ∣k∣\boxed{|k|}.

As k≠0k\ne0, a matrix taking the parallelogram back onto the unit square is

M−1=1k(10−1k).\boxed{\mathbf M^{-1}=\frac1k\begin{pmatrix}1&0\\-1&k\end{pmatrix}}.

(c)

解法一

思路

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直接取所给直线上的一般点,证明其像等于该点向量乘k。k非零保证整条直线映回自身。

答题过程

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A general point on the given line has position vector (x,x/(k−1))(x,x/(k-1)). Its image is

M(xx/(k−1))= (kxx+x/(k−1))= (kxkx/(k−1))= k(xx/(k−1)).\begin{align*} \mathbf M\begin{pmatrix}x\\x/(k-1)\end{pmatrix} =&\,\begin{pmatrix}kx\\x+x/(k-1)\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}kx\\kx/(k-1)\end{pmatrix}\\[4mm] =&\,k\begin{pmatrix}x\\x/(k-1)\end{pmatrix}. \end{align*}

Thus the image satisfies Y=X/(k−1)Y=X/(k-1). Since k≠0k\ne0, the image covers the same line, proving it is invariant.