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CIE 9231 2025 November Paper 11 Q2

A Level / CIE / FP1

CIE 9231 2025 Nov Paper 11 Paper · Question 2

题目

Problem

The matrices A\mathbf{A} and B\mathbf{B} are given by

A=(13201)andB=(10321).\mathbf{A}=\begin{pmatrix}1&\frac32\\0&1\end{pmatrix}\quad\text{and}\quad \mathbf{B}=\begin{pmatrix}1&0\\\frac32&1\end{pmatrix}.

(a) Give full details of the geometrical transformation in the xx-yy plane represented by A\mathbf{A}.

(2)

(b) Give full details of the geometrical transformation in the xx-yy plane represented by B\mathbf{B}.

(2)

The triangle DEFDEF in the xx-yy plane is transformed by AB\mathbf{AB} onto triangle PQRPQR.

(c) Show that the triangles DEFDEF and PQRPQR have the same area.

(2)

(d) Find the equations of the invariant lines, through the origin, of the transformation in the xx-yy plane represented by AB\mathbf{AB}.

(5)

题目中文翻译

矩阵 A\mathbf{A} 和 B\mathbf{B} 为

A=(13201)和B=(10321).\mathbf{A}=\begin{pmatrix}1&\frac32\\0&1\end{pmatrix}\quad\text{和}\quad \mathbf{B}=\begin{pmatrix}1&0\\\frac32&1\end{pmatrix}.

(a) 完整描述 A\mathbf{A} 所表示的 xx-yy 平面上的几何变换。

(2)

(b) 完整描述 B\mathbf{B} 所表示的 xx-yy 平面上的几何变换。

(2)

xx-yy 平面上的三角形 DEFDEF 经 AB\mathbf{AB} 变换后得到三角形 PQRPQR。

(c) 证明三角形 DEFDEF 和 PQRPQR 的面积相同。

(2)

(d) 求 AB\mathbf{AB} 所表示的 xx-yy 平面上的变换中,经过原点的不变直线的方程。

(5)

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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矩阵使x增加原y的1.5倍、y不变,属于固定横轴的错切。

答题过程

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The transformation is a shear with the xx-axis fixed, mapping (0,1)(0,1) to (3/2,1)(3/2,1). In coordinates it is

(x,y)⟼(x+32y,y).(x,y)\longmapsto(x+\tfrac32y,y).

(b)

解法一

思路

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矩阵使y增加原x的1.5倍、x不变,属于固定纵轴的错切。

答题过程

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The transformation is a shear with the yy-axis fixed, mapping (1,0)(1,0) to (1,3/2)(1,3/2). In coordinates it is

(x,y)⟼(x,y+32x).(x,y)\longmapsto(x,y+\tfrac32x).

(c)

解法一

思路

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面积倍数是复合变换行列式的绝对值,可以用两个行列式相乘计算。

答题过程

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The area scale factor is

∣det⁡(AB)∣=∣det⁡Adet⁡B∣=∣1⋅1∣=1.|\det(\mathbf A\mathbf B)| =|\det\mathbf A\det\mathbf B|=|1\cdot1|=1.

Hence triangle PQRPQR has the same area as triangle DEFDEF.

(d)

解法一

思路

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先求复合矩阵,再设过原点的不变直线为y=mx,要求像点仍满足同一斜率关系。

答题过程

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Multiplying,

AB=(13/43/23/21).\mathbf A\mathbf B=\begin{pmatrix}13/4&3/2\\3/2&1\end{pmatrix}.

For a point (x,mx)(x,mx) on an invariant line, its image (X,Y)(X,Y) satisfies

X= 134x+32mx,Y= 32x+mx,Y= mX.\begin{align*} X=&\,\frac{13}{4}x+\frac32mx,\\[4mm] Y=&\,\frac32x+mx,\\[4mm] Y=&\,mX. \end{align*}

Thus

32+m= 134m+32m2,2m2+3m−2= 0,(2m−1)(m+2)= 0.\begin{align*} \frac32+m=&\,\frac{13}{4}m+\frac32m^2, \\[4mm] 2m^2+3m-2=&\,0, \\[4mm] (2m-1)(m+2)=&\,0. \end{align*}

The vertical line x=0x=0 is not invariant, since the image of (0,1)(0,1) has first coordinate 3/23/2. Therefore the invariant lines are

y=12x,y=−2x.\boxed{y=\frac12x,\qquad y=-2x}.