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CIE 9231 2024 June Paper 11 Q4

A Level / CIE / FP1

CIE 9231 2024 June Paper 11 Paper · Question 4

题目

Problem

The matrix M\mathbf{M} is given by

M=(12−12312312)(14001).\mathbf{M}=\begin{pmatrix}\frac12&-\frac12\sqrt3\\\frac12\sqrt3&\frac12\end{pmatrix}\begin{pmatrix}14&0\\0&1\end{pmatrix}.

(a) The matrix M\mathbf{M} represents a sequence of two geometrical transformations in the xx-yy plane.

Give full details of each transformation, and make clear the order in which they are applied.

(4)

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

(2)

(c) Find the equations of the invariant lines, through the origin, of the transformation represented by M\mathbf{M}.

(5)

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

Given that the area of triangle DEFDEF is 28 cm228\text{ cm}^2, find the area of triangle ABCABC.

(2)
题目中文翻译

矩阵 M\mathbf{M} 定义为

M=(12−12312312)(14001).\mathbf{M}=\begin{pmatrix}\frac12&-\frac12\sqrt3\\\frac12\sqrt3&\frac12\end{pmatrix}\begin{pmatrix}14&0\\0&1\end{pmatrix}.

(a) 矩阵 M\mathbf{M} 表示在 xx-yy 平面上依次进行的两个几何变换。

详细说明每个变换,并明确指出它们的实施顺序。

(4)

(b) 将 M−1\mathbf{M}^{-1} 写成两个矩阵的乘积,且这两个矩阵都不能是 I\mathbf{I}。

(2)

(c) 求矩阵 M\mathbf{M} 所表示的变换的所有经过原点的不变直线方程。

(5)

(d) xx-yy 平面上的三角形 ABCABC 经矩阵 M\mathbf{M} 变换后得到三角形 DEFDEF。

已知三角形 DEFDEF 的面积为 28 cm228\text{ cm}^2,求三角形 ABCABC 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

右边矩阵先作用,因此先横向拉伸,再绕原点逆时针转六十度。

答题过程

展开

First apply a one-way stretch parallel to the xx-axis with scale factor 1414.

Then apply a rotation through π3\frac\pi3 radians anticlockwise about the origin, since the left matrix is the standard rotation matrix with cos⁡(π/3)=12\cos(\pi/3)=\frac12 and sin⁡(π/3)=32\sin(\pi/3)=\frac{\sqrt3}{2}.

(b)

解法一

思路

展开

复合变换取逆时顺序倒转,各矩阵分别取逆。旋转逆矩阵把角度变号。

答题过程

展开

Writing M=RS\mathbf M=\mathbf R\mathbf S,

M−1= S−1R−1= (1/14001)(1/23/2−3/21/2).\begin{align*} \mathbf M^{-1}=&\,\mathbf S^{-1}\mathbf R^{-1}\\[4mm] =&\,\boxed{\begin{pmatrix}1/14&0\\0&1\end{pmatrix} \begin{pmatrix}1/2&\sqrt3/2\\-\sqrt3/2&1/2\end{pmatrix}}. \end{align*}

Both factors are different from I\mathbf I.

(c)

解法一

思路

展开

写出合成矩阵,把y=mx上的点映射后仍在该线上作为条件。最后检查竖直线没有遗漏。

答题过程

展开

Multiplying the matrices gives

M=(7−3/2731/2).\mathbf M=\begin{pmatrix}7&-\sqrt3/2\\7\sqrt3&1/2\end{pmatrix}.

The image of (x,mx)(x,mx) is

X= (7−32m)x,Y= (73+12m)x.\begin{align*} X=&\,\big(7-\tfrac{\sqrt3}{2}m\big)x,\\[4mm] Y=&\,\big(7\sqrt3+\tfrac12m\big)x. \end{align*}

For y=mxy=mx to be invariant, require Y=mXY=mX. Thus

73+m2= m(7−32m),3m2−13m+143= 0,(3m−7)(m−23)= 0.\begin{align*} 7\sqrt3+\frac m2=&\,m\big(7-\tfrac{\sqrt3}{2}m\big),\\[4mm] \sqrt3m^2-13m+14\sqrt3=&\,0,\\[4mm] (\sqrt3m-7)(m-2\sqrt3)=&\,0. \end{align*}

Therefore m=7/3m=7/\sqrt3 or m=23m=2\sqrt3. For these slopes the horizontal scale factors are 7/27/2 and 44, both nonzero, so both lines map onto themselves.

The vertical line is not invariant, since (0,y)(0,y) maps to (−3y/2,y/2)(-\sqrt3y/2,y/2). Hence the required equations are

y=23xandy=73x.\boxed{y=2\sqrt3x\quad\text{and}\quad y=\frac7{\sqrt3}x}.

(d)

解法一

思路

展开

旋转不改面积,拉伸使面积乘14,所以原三角形面积是像面积除以14。

答题过程

展开 ∣det⁡M∣=1×14=14.|\det\mathbf M|=1\times14=14.

Thus 28=14 Area⁡(ABC)28=14\,\operatorname{Area}(ABC), giving

Area⁡(ABC)=2 cm2.\boxed{\operatorname{Area}(ABC)=2\text{ cm}^2}.