CalcGospel 國際數學圖譜
返回

FP1 2025 Jan Q7

A Level / Edexcel / FP1

IAL 2025 Jan Paper · Question 7

题目

Problem

(i)

A=(−1232−32−12)\mathbf{A} = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}

(a) Describe fully the single transformation represented by the matrix A\mathbf{A}.

(2)

The matrix B\mathbf{B} represents a stretch scale factor 2 parallel to the xx-axis.

(b) Write down the matrix B\mathbf{B}.

(1)

The transformation represented by matrix B\mathbf{B} followed by the transformation represented by matrix A\mathbf{A} is the transformation represented by the matrix C\mathbf{C}.

(c) Determine C\mathbf{C}.

(2)

(ii)

M=(k−2−12k)\mathbf{M} = \begin{pmatrix} k & -2 \\ -1 & 2k \end{pmatrix}

where kk is a constant.

Given that the transformation represented by matrix M\mathbf{M} maps the point (k,k)(k, k) onto the point (35,91)(35, 91)

(a) determine the value of kk.

(4)

A quadrilateral QQ is transformed to another quadrilateral Q′Q' by the matrix M\mathbf{M}.

Given that Q′Q' has area 336

(b) use the value of kk found in part (ii)(a) to determine the area of QQ.

(2)
题目中文翻译

(i)

A=(−1232−32−12)\mathbf{A} = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}

(a) 完整描述矩阵 A\mathbf A 所代表的单一几何变换。

(2)

矩阵 B\mathbf B 表示沿 xx 轴方向、比例因子为 2 的拉伸变换。

(b) 写出矩阵 B\mathbf B。

(1)

先进行矩阵 B\mathbf B 所代表的变换，再进行矩阵 A\mathbf A 所代表的变换，其合成变换由矩阵 C\mathbf C 表示。

(c) 求 C\mathbf C。

(2)

(ii)

M=(k−2−12k)\mathbf{M} = \begin{pmatrix} k & -2 \\ -1 & 2k \end{pmatrix}

其中 kk 为常数。

已知矩阵 M\mathbf M 所代表的变换将点 (k,k)(k,k) 映射到点 (35,91)(35,91)。

(a) 求 kk 的值。

(4)

四边形 QQ 经矩阵 M\mathbf M 变换为另一个四边形 Q′Q'。

已知 Q′Q' 的面积为 336，

(b) 利用在 (ii)(a) 中求得的 kk 的值，求 QQ 的面积。

(2)

解答

(i)(a)

解法一

思路

展开

把矩阵与标准旋转矩阵比较。左上角给出 cos⁡θ=−12\cos\theta=-\frac12，左下角给出 sin⁡θ=−32\sin\theta=-\frac{\sqrt3}{2}，因此角度位于第三象限。完整描述还必须写出旋转中心。

答题过程

展开

The standard matrix for an anticlockwise rotation through θ\theta about the origin is

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

Here,

cos⁡θ=−12,sin⁡θ=−32.\begin{gathered} \cos\theta=-\frac12,\\ \sin\theta=-\frac{\sqrt3}{2}. \end{gathered}

Thus θ=240∘\theta=240^\circ. Therefore, A\mathbf A represents a rotation through 240∘240^\circ anticlockwise about the origin.

(i)(b)

解法一

思路

展开

沿 xx 轴方向、比例因子为 2 的拉伸把横坐标乘以 2，而纵坐标保持不变。

答题过程

展开

The stretch maps

(x,y)⟼(2x,y).(x,y)\longmapsto(2x,y).

Hence

B=(2001).\boxed{ \mathbf B= \begin{pmatrix} 2&0\\ 0&1 \end{pmatrix}}.

(i)(c)

解法一

思路

展开

先进行 B\mathbf B、再进行 A\mathbf A，所以对列向量而言复合矩阵的顺序是 AB\mathbf A\mathbf B。矩阵乘法不能随意交换次序。

答题过程

展开

Since the transformation represented by B\mathbf B is followed by that represented by A\mathbf A,

C= AB= (−1232−32−12)(2001)= (−132−3−12).\begin{align*} \mathbf C =&\,\mathbf A\mathbf B\\ =&\,\begin{pmatrix} -\frac12&\frac{\sqrt3}{2}\\ -\frac{\sqrt3}{2}&-\frac12 \end{pmatrix} \begin{pmatrix} 2&0\\ 0&1 \end{pmatrix}\\ =&\,\boxed{ \begin{pmatrix} -1&\frac{\sqrt3}{2}\\ -\sqrt3&-\frac12 \end{pmatrix}}. \end{align*}

Writing C=(cij)\mathbf C=(c_{ij}), the entries are obtained as follows:

c11= (−12)(2)+32(0)= −1,c12= (−12)(0)+32(1)= 32,c21= (−32)(2)−12(0)= −3,c22= (−32)(0)−12(1)= −12.\begin{align*} c_{11}=&\,\big(-\tfrac12\big)(2)+\tfrac{\sqrt3}{2}(0)\\ =&\,-1,\\ c_{12}=&\,\big(-\tfrac12\big)(0)+\tfrac{\sqrt3}{2}(1)\\ =&\,\frac{\sqrt3}{2},\\ c_{21}=&\,\big(-\tfrac{\sqrt3}{2}\big)(2)-\tfrac12(0)\\ =&\,-\sqrt3,\\ c_{22}=&\,\big(-\tfrac{\sqrt3}{2}\big)(0)-\tfrac12(1)\\ =&\,-\frac12. \end{align*}

(ii)(a)

解法一

思路

展开

把点 (k,k)(k,k) 写成列向量并左乘 M\mathbf M，再分别比较像点的两个坐标。两个所得方程必须同时成立；将第二个方程减去第一个方程的两倍，会直接得到关于 kk 的一次方程。

答题过程

展开

The given mapping gives

(k−2−12k)(kk)=(3591).\begin{pmatrix} k&-2\\ -1&2k \end{pmatrix} \begin{pmatrix}k\\k\end{pmatrix} = \begin{pmatrix}35\\91\end{pmatrix}.

Therefore,

k2−2k=35k^2-2k=35

and

2k2−k=91.2k^2-k=91.

Subtracting twice the first equation from the second,

 (2k2−k)−2(k2−2k)= 91−70 3k=21.\begin{align*} &\,(2k^2-k)-2(k^2-2k)\\ =&\,91-70\\ &\,3k=21. \end{align*}

Hence

k=7.\boxed{k=7}.

Checking in both original equations,

72−2(7)= 49−14= 35,2(7)2−7= 98−7= 91.\begin{align*} 7^2-2(7)=&\,49-14\\ =&\,35,\\ 2(7)^2-7=&\,98-7\\ =&\,91. \end{align*}

解法二

思路

展开

也可以用逆矩阵把像点还原。先排除会使矩阵奇异的两个参数值，再用逆矩阵表示原像点的坐标。因为原像点的横、纵坐标都等于 kk，两个对应分子必须相等，由此得到一次方程。

答题过程

展开

The determinant is

det⁡M=2k2−2.\det\mathbf M=2k^2-2.

If k=1k=1, the point (1,1)(1,1) maps to (−1,1)(-1,1), not (35,91)(35,91). If k=−1k=-1, the point (−1,−1)(-1,-1) maps to (3,3)(3,3), not (35,91)(35,91). Hence k≠±1k\neq\pm1, and M\mathbf M is invertible.

Its inverse is

M−1\mathbf M^{-1} =12k2−2(2k21k).=\frac1{2k^2-2} \begin{pmatrix}2k&2\\1&k\end{pmatrix}.

Applying this to the image point,

 (kk)= M−1(3591)= 12k2−2(2k21k) ×(3591)= 12k2−2(70k+18235+91k).\begin{align*} &\,\begin{pmatrix}k\\k\end{pmatrix}\\ =&\,\mathbf M^{-1}\begin{pmatrix}35\\91\end{pmatrix}\\ =&\,\frac1{2k^2-2}\begin{pmatrix}2k&2\\1&k\end{pmatrix}\\ &\,\hspace{2pt}\times\begin{pmatrix}35\\91\end{pmatrix}\\ =&\,\frac1{2k^2-2}\begin{pmatrix}70k+182\\35+91k\end{pmatrix}. \end{align*}

The two numerator entries come from

 (2k)(35)+2(91)= 70k+182, 1(35)+k(91)= 35+91k.\begin{align*} &\,(2k)(35)+2(91)\\ =&\,70k+182,\\ &\,1(35)+k(91)\\ =&\,35+91k. \end{align*}

The two coordinates of the original point are equal, and the common denominator is non-zero. Therefore,

70k+182= 35+91k147= 21kk= 7.\begin{align*} 70k+182=&\,35+91k\\ 147=&\,21k\\ k=&\,7. \end{align*}

This value satisfies both original mapping equations, since

72−2(7)=35,2(7)2−7=91.\begin{gathered} 7^2-2(7)=35,\\ 2(7)^2-7=91. \end{gathered}

Hence k=7\boxed{k=7}.

(ii)(b)

解法一

思路

展开

矩阵变换把面积乘以行列式的绝对值。把 k=7k=7 代入 M\mathbf M 求出面积比例因子，再由 Q′Q' 的面积反求 QQ 的面积。

答题过程

展开

When k=7k=7,

M=(7−2−114).\mathbf M= \begin{pmatrix} 7&-2\\ -1&14 \end{pmatrix}.

The area scale factor is

 ∣det⁡M∣= ∣7(14)−(−2)(−1)∣= 96.\begin{align*} &\,\big|\det\mathbf M\big|\\ =&\,\big|7(14)-(-2)(-1)\big|\\ =&\,96. \end{align*}

Therefore,

336=96×Area⁡(Q),336=96\times\operatorname{Area}(Q),

so

Area⁡(Q)=33696=72.\boxed{\operatorname{Area}(Q)=\frac{336}{96}=\frac72}.