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CIE 9231 2023 June Paper 12 Q4

A Level / CIE / FP1

CIE 9231 2023 June Paper 12 Paper · Question 4

题目

Problem

The matrix M\mathbf{M} is given by M=(ab2c2a)\mathbf{M}=\begin{pmatrix}a&b^2\\c^2&a\end{pmatrix}, where aa, bb, cc are real constants and b≠0b\ne0.

(a) Show that M\mathbf{M} does not represent a rotation about the origin.

(2)

(b) Find the equations of the invariant lines, through the origin, of the transformation in the xx–yy plane represented by M\mathbf{M}.

(5)

It is given that M\mathbf{M} represents the sequence of two transformations in the xx–yy plane given by an enlargement, centre the origin, scale factor 55 followed by a shear, xx-axis fixed, with (0,1)(0,1) mapped to (5,1)(5,1).

(c) Find M\mathbf{M}.

(3)

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

Given that the area of triangle DEFDEF is 12 cm212\text{ cm}^2, find the area of triangle PQRPQR.

(2)
题目中文翻译

矩阵 M\mathbf{M} 为 M=(ab2c2a)\mathbf{M}=\begin{pmatrix}a&b^2\\c^2&a\end{pmatrix},其中 aa、bb、cc 为实常数,且 b≠0b\ne0。

(a) 证明 M\mathbf{M} 不表示绕原点的旋转变换。

(2)

(b) 求 M\mathbf{M} 所表示的 xx–yy 平面变换的所有经过原点的不变直线方程。

(5)

已知 M\mathbf{M} 表示在 xx–yy 平面内依次进行的两个变换:先作以原点为中心、比例因子为 55 的位似变换,再作保持 xx 轴不动、将 (0,1)(0,1) 映射至 (5,1)(5,1) 的错切变换。

(c) 求 M\mathbf{M}。

(3)

(d) xx–yy 平面内的三角形 DEFDEF 经 M\mathbf{M} 变换后成为三角形 PQRPQR。

已知三角形 DEFDEF 的面积为 12 cm212\text{ cm}^2,求三角形 PQRPQR 的面积。

(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

旋转矩阵的两个非对角元素互为相反数,而这里两个平方都非负,且b平方严格正,不能互为相反数。

答题过程

展开

A rotation about the origin has matrix

(cos⁡θ−sin⁡θsin⁡θcos⁡θ).\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}.

If M\mathbf M represented such a rotation, we would require b2=−c2b^2=-c^2, or b2+c2=0b^2+c^2=0. Since b,cb,c are real, this forces b=c=0b=c=0, contradicting b≠0b\ne0.

Therefore M\mathbf M does not represent a rotation about the origin.

(c)

解法一

思路

展开

放大矩阵先作用,剪切矩阵放在左边。剪切的非对角元素为5,再乘放大因子得到25。

答题过程

展开

The enlargement has matrix 5I5\mathbf I, and the shear has matrix (1501)\begin{pmatrix}1&5\\0&1\end{pmatrix}. Thus

M= (1501)(5005)= (52505).\begin{align*} \mathbf M=&\,\begin{pmatrix}1&5\\0&1\end{pmatrix} \begin{pmatrix}5&0\\0&5\end{pmatrix}\\[4mm] =&\,\boxed{\begin{pmatrix}5&25\\0&5\end{pmatrix}}. \end{align*}

(d)

解法一

思路

展开

面积倍率是行列式绝对值25,不是剪切系数。

答题过程

展开 ∣det⁡M∣=25.|\det\mathbf M|=25.

Therefore

Area⁡(PQR)=25×12=300 cm2.\boxed{\operatorname{Area}(PQR)=25\times12=300\text{ cm}^2}.