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CIE 9231 2026 June Paper 11 Q3

A Level / CIE / FP1

CIE 9231 2026 June Paper 11 Paper · Question 3

题目

Problem

The matrix M\mathbf{M} represents the sequence of two transformations in the xx-yy plane given by a stretch parallel to the xx-axis, scale factor kk (k≠0,1k\ne0,1), followed by a shear, xx-axis fixed, with (0,1)(0,1) mapped to (2,1)(2,1).

(a) Show that M=(k201)\mathbf{M}=\begin{pmatrix}k&2\\0&1\end{pmatrix}.

(4)

(b) The transformation represented by M\mathbf{M} has a line of invariant points.

Find, in terms of kk, the equation of this line.

(3)

The square SS in the xx-yy plane is transformed by M\mathbf{M} onto the parallelogram PP.

(c) Given that the area of SS is half of the area of PP, find the possible values of kk.

(2)

题目中文翻译

矩阵 M\mathbf{M} 表示在 xx-yy 平面上依次进行的两个变换:先进行平行于 xx 轴、伸缩因子为 kk(k≠0,1k\ne0,1)的伸缩变换,再进行以 xx 轴为不动轴、将 (0,1)(0,1) 映射到 (2,1)(2,1) 的错切变换。

(a) 证明 M=(k201)\mathbf{M}=\begin{pmatrix}k&2\\0&1\end{pmatrix}。

(4)

(b) 矩阵 M\mathbf{M} 所表示的变换有一条由不动点组成的直线。

用 kk 表示这条直线的方程。

(3)

xx-yy 平面上的正方形 SS 经 M\mathbf{M} 变换后得到平行四边形 PP。

(c) 已知 SS 的面积为 PP 的面积的一半,求 kk 的可能值。

(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

先伸缩再错切,采用列向量时,后做的变换矩阵要乘在左边。

答题过程

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The stretch and shear have matrices

A= (k001),B= (1201),\begin{align*} \mathbf A=&\,\begin{pmatrix}k&0\\0&1\end{pmatrix},\\[4mm] \mathbf B=&\,\begin{pmatrix}1&2\\0&1\end{pmatrix}, \end{align*}

respectively. The shear fixes (1,0)(1,0) and maps (0,1)(0,1) to (2,1)(2,1).

Since the stretch is applied first,

M= BA= (1k+2(0)1(0)+2(1)0k+1(0)0(0)+1(1))= (k201),\begin{align*} \mathbf M=&\,\mathbf B\mathbf A \\[4mm] =&\,\begin{pmatrix} 1k+2(0)&1(0)+2(1)\\ 0k+1(0)&0(0)+1(1) \end{pmatrix} \\[4mm] =&\,\boxed{\begin{pmatrix}k&2\\0&1\end{pmatrix}}, \end{align*}

as required.

(b)

解法一

思路

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不变点要求变换后的两个坐标都等于原坐标,而不仅是直线整体不变。

答题过程

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For a fixed point (x,y)(x,y),

M(xy)=(kx+2yy)=(xy).\mathbf M\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}kx+2y\\y\end{pmatrix} =\begin{pmatrix}x\\y\end{pmatrix}.

The second coordinate is automatically unchanged, and the first requires kx+2y=xkx+2y=x. Hence the line of invariant points is

y=1−k2x.\boxed{y=\frac{1-k}{2}x}.

(c)

解法一

思路

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面积倍数是行列式的绝对值,负行列式仍可能给出正确面积倍数。

答题过程

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The area scale factor is ∣det⁡M∣=∣k∣|\det\mathbf M|=|k|. Since the area of PP is twice that of SS,

∣k∣=2.|k|=2.

Thus k=2 or k=−2\boxed{k=2\text{ or }k=-2}; both satisfy k≠0,1k\ne0,1.