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CIE 9231 2023 June Paper 13 Q4

A Level / CIE / FP1

CIE 9231 2023 June Paper 13 Paper · Question 4

题目

Problem

The matrix M\mathbf{M} is given by

M=(cos⁡2θ−sin⁡2θsin⁡2θcos⁡2θ)(1k01),\mathbf{M}=\begin{pmatrix}\cos2\theta&-\sin2\theta\\\sin2\theta&\cos2\theta\end{pmatrix}\begin{pmatrix}1&k\\0&1\end{pmatrix},

where 0<θ<π0<\theta<\pi and kk is a non-zero constant. The matrix M\mathbf{M} represents a sequence of two geometrical transformations, one of which is a shear.

(a) Describe fully the other transformation and state the order in which the transformations are applied.

(3)

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

(2)

(c) Find, in terms of kk, the value of tan⁡θ\tan\theta for which M−I\mathbf{M}-\mathbf{I} is singular.

(5)

(d) Given that k=23k=2\sqrt3 and θ=13π\theta=\frac13\pi, show that the invariant points of the transformation represented by M\mathbf{M} lie on the line 3y+3x=03y+\sqrt3x=0.

(4)
题目中文翻译

矩阵 M\mathbf{M} 为

M=(cos⁡2θ−sin⁡2θsin⁡2θcos⁡2θ)(1k01),\mathbf{M}=\begin{pmatrix}\cos2\theta&-\sin2\theta\\\sin2\theta&\cos2\theta\end{pmatrix}\begin{pmatrix}1&k\\0&1\end{pmatrix},

其中 0<θ<π0<\theta<\pi,kk 为非零常数。矩阵 M\mathbf{M} 表示依次进行的两个几何变换,其中一个是错切变换。

(a) 完整描述另一个变换,并写出两个变换的施行顺序。

(3)

(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) 求使 M−I\mathbf{M}-\mathbf{I} 为奇异矩阵的 tan⁡θ\tan\theta 的值,用 kk 表示。

(5)

(d) 已知 k=23k=2\sqrt3 且 θ=13π\theta=\frac13\pi,证明 M\mathbf{M} 所表示的变换的不变点位于直线 3y+3x=03y+\sqrt3x=0 上。

(4)
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,另一矩阵是转角为2θ的标准旋转矩阵。

答题过程

展开

The other transformation is a rotation anticlockwise through angle 2θ2\theta about the origin.

The shear is applied first, followed by this rotation.

(b)

解法一

思路

展开

复合矩阵取逆时倒转次序,剪切参数变号、旋转角变号。

答题过程

展开 M−1=(1−k01)(cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θ).\boxed{\mathbf M^{-1} =\begin{pmatrix}1&-k\\0&1\end{pmatrix} \begin{pmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{pmatrix}}.

Neither factor is I\mathbf I, since k≠0k\ne0 and 0<θ<π0<\theta<\pi.

(c)

解法一

思路

展开

先算M减单位矩阵的行列式,再用倍角式把方程降为sinθ与cosθ。除以sinθ前用题目角范围确认非零,再排除cosθ为零。

答题过程

展开

Write c=cos⁡2θc=\cos2\theta and s=sin⁡2θs=\sin2\theta. Then

M−I=(c−1kc−ssks+c−1).\mathbf M-\mathbf I=\begin{pmatrix}c-1&kc-s\\s&ks+c-1\end{pmatrix}.

For this matrix to be singular,

0= (c−1)(ks+c−1)−s(kc−s)= −ks+(c−1)2+s2= 2−2c−ks.\begin{align*} 0=&\,(c-1)(ks+c-1)-s(kc-s)\\[4mm] =&\,-ks+(c-1)^2+s^2\\[4mm] =&\,2-2c-ks. \end{align*}

Using the double-angle identities,

4sin⁡2θ=2ksin⁡θcos⁡θ.4\sin^2\theta=2k\sin\theta\cos\theta.

Since 0<θ<π0<\theta<\pi, sin⁡θ>0\sin\theta>0, so 2sin⁡θ=kcos⁡θ2\sin\theta=k\cos\theta. Here cos⁡θ\cos\theta cannot be zero, as that would force sin⁡θ=0\sin\theta=0. Thus

tan⁡θ=k2.\boxed{\tan\theta=\frac k2}.

(d)

解法一

思路

展开

代入参数先算M,再用固定点的像等于自身得到线性方程。两个条件其实是同一条直线。

答题过程

展开

For k=23k=2\sqrt3 and θ=π/3\theta=\pi/3,

M= (−1/2−3/23/2−1/2)(12301)= (−1/2−33/23/25/2).\begin{align*} \mathbf M =&\,\begin{pmatrix}-1/2&-\sqrt3/2\\\sqrt3/2&-1/2\end{pmatrix} \begin{pmatrix}1&2\sqrt3\\0&1\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}-1/2&-3\sqrt3/2\\\sqrt3/2&5/2\end{pmatrix}. \end{align*}

An invariant point satisfies M(x,y)T=(x,y)T\mathbf M(x,y)^{\mathsf T}=(x,y)^{\mathsf T}, giving

−x2−332y= x,32x+52y= y.\begin{align*} -\frac x2-\frac{3\sqrt3}{2}y=&\,x,\\[4mm] \frac{\sqrt3}{2}x+\frac52y=&\,y. \end{align*}

These simplify to x+3y=0x+\sqrt3y=0 and 3x+3y=0\sqrt3x+3y=0, which are equivalent. Therefore the invariant points lie on

3y+3x=0.\boxed{3y+\sqrt3x=0}.

Conversely every point on this line satisfies both equations, so the whole line consists of invariant points.