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CIE 9231 2026 June Paper 13 Q2

A Level / CIE / FP1

CIE 9231 2026 June Paper 13 Paper · Question 2

题目

Problem

The matrix M\mathbf{M} is given by

M=(122−122122122)(1201).\mathbf{M}=\begin{pmatrix}\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\\\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2}\end{pmatrix}\begin{pmatrix}1&2\\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) Find the equations of the invariant lines, through the origin, of the transformation represented by M\mathbf{M}.

(5)

题目中文翻译

矩阵 M\mathbf{M} 为

M=(122−122122122)(1201).\mathbf{M}=\begin{pmatrix}\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\\\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2}\end{pmatrix}\begin{pmatrix}1&2\\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。

完整描述每个变换,并明确它们的施行顺序。

(4)

(b) 求矩阵 M\mathbf{M} 所表示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

展开

矩阵作用于列向量时,右边矩阵先作用。完整描述错切的固定轴与点的像,以及旋转的角度、方向和中心。

答题过程

展开

The transformations are applied in this order:

  1. A shear with the xx-axis fixed, mapping (0,1)(0,1) to (2,1)(2,1), represented by (1201)\begin{pmatrix}1&2\\0&1\end{pmatrix}.
  2. An anticlockwise rotation through π/4\pi/4 about the origin, represented by (cos⁡(π/4)−sin⁡(π/4)sin⁡(π/4)cos⁡(π/4))\begin{pmatrix}\cos(\pi/4)&-\sin(\pi/4)\\\sin(\pi/4)&\cos(\pi/4)\end{pmatrix}.

(b)

解法一

思路

展开

不变直线要求像点仍在同一直线上,不要求每点固定。设y=mx,比较像点斜率,另检查竖直线。

答题过程

展开

Multiplying the matrices,

M=22(1113).\mathbf M=\frac{\sqrt2}{2}\begin{pmatrix}1&1\\1&3\end{pmatrix}.

For a point on y=mxy=mx, its image (X,Y)(X,Y) is given by

X= 22(1+m)x,Y= 22(1+3m)x.\begin{align*} X=&\,\frac{\sqrt2}{2}(1+m)x,\\[4mm] Y=&\,\frac{\sqrt2}{2}(1+3m)x. \end{align*}

The line is invariant when Y=mXY=mX, so

1+3m= m(1+m),m2−2m−1= 0,m= 1±2.\begin{align*} 1+3m=&\,m(1+m), \\[4mm] m^2-2m-1=&\,0, \\[4mm] m=&\,1\pm\sqrt2. \end{align*}

The vertical line x=0x=0 is not invariant, since (0,1)(0,1) maps to (2/2,32/2)(\sqrt2/2,3\sqrt2/2), whose first coordinate is non-zero. Hence the two invariant lines through the origin are

y=(1+2)x,y=(1−2)x.\boxed{y=(1+\sqrt2)x},\qquad\boxed{y=(1-\sqrt2)x}.