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CIE 9231 2022 November Paper 12 Q3

A Level / CIE / FP1

CIE 9231 2022 Nov Paper 12 Paper · Question 3

题目

Problem

The matrix M\mathbf M is given by

M=(100k)(10k1),\mathbf M=\begin{pmatrix}1&0\\0&k\end{pmatrix}\begin{pmatrix}1&0\\k&1\end{pmatrix},

where kk is a constant and k≠0k\ne0 or 11.

(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)

(b) Write M−1\mathbf M^{-1} as the product of two matrices, neither of which is I\mathbf I.

(2)

(c) Show that the invariant points of the transformation represented by M\mathbf M lie on the line y=k21−kxy=\dfrac{k^2}{1-k}x.

(4)

(d) The triangle ABCABC in the xx–yy plane is transformed by M\mathbf M onto triangle DEFDEF.

Find the value of kk for which the area of triangle DEFDEF is equal to the area of triangle ABCABC.

(2)
题目中文翻译

矩阵 M\mathbf M 为

M=(100k)(10k1),\mathbf M=\begin{pmatrix}1&0\\0&k\end{pmatrix}\begin{pmatrix}1&0\\k&1\end{pmatrix},

其中 kk 为常数,且 kk 不等于 00 或 11。

(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,并明确说明它们的作用顺序。

(2)

(b) 将 M−1\mathbf M^{-1} 写成两个矩阵的乘积,且这两个矩阵均不是 I\mathbf I。

(2)

(c) 证明 M\mathbf M 所表示的变换的不变点位于直线 y=k21−kxy=\dfrac{k^2}{1-k}x 上。

(4)

(d) xx–yy 平面内的三角形 ABCABC 经 M\mathbf M 变换后得到三角形 DEFDEF。

求使三角形 DEFDEF 与三角形 ABCABC 的面积相等的 kk 的值。

(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

右侧剪切先作用,再沿纵轴拉伸。

答题过程

展开

The sequence is a shear followed by a stretch. The stretch is parallel to the yy-axis with scale factor kk.

(b)

解法一

思路

展开

将复合变换的次序倒转,各因子取逆。

答题过程

展开 M−1=(10−k1)(1001/k).\boxed{\mathbf M^{-1} =\begin{pmatrix}1&0\\-k&1\end{pmatrix} \begin{pmatrix}1&0\\0&1/k\end{pmatrix}}.

Neither factor is I\mathbf I, since k≠0,1k\ne0,1.

(c)

解法一

思路

展开

不变点要求坐标不变。横坐标自动满足,纵坐标给一条直线,利用k不等于1可解出斜率。

答题过程

展开

The combined matrix is

M=(10k2k).\mathbf M=\begin{pmatrix}1&0\\k^2&k\end{pmatrix}.

Thus (x,y)(x,y) maps to (x,k2x+ky)(x,k^2x+ky). For an invariant point,

k2x+ky=y⟹(1−k)y=k2x.k^2x+ky=y\quad\Longrightarrow\quad (1-k)y=k^2x.

Since k≠1k\ne1, the invariant points lie on

y=k21−kx.\boxed{y=\frac{k^2}{1-k}x}.

Conversely every point on this line is fixed, so it is the entire line of invariant points.

(d)

解法一

思路

展开

相等面积要求行列式绝对值为1,再排除题目已经禁用的k=1。

答题过程

展开

Since det⁡M=k\det\mathbf M=k, the area scale factor is ∣k∣|k|. Equal areas require ∣k∣=1|k|=1, so k=±1k=\pm1.

The value k=1k=1 is excluded by the question. Therefore k=−1\boxed{k=-1}.