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IAL 2020 May FP1 Q6

A Level / Edexcel / FP1

IAL 2020 May Paper · Question 6

题目

Problem

6. (i) A=(1003)A = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}

(a) Describe fully the single transformation represented by the matrix AA.

(2)

The matrix BB represents a rotation of 45°45° clockwise about the origin.

(b) Write down the matrix BB, giving each element of the matrix in exact form.

(1)

The transformation represented by matrix AA followed by the transformation represented by matrix BB is represented by the matrix CC.

(c) Determine CC.

(2)

(ii) The trapezium TT has vertices at the points (2,0)(-2, 0), (2,k)(-2, k), (5,8)(5, 8) and (5,0)(5, 0), where kk is a positive constant. Trapezium TT is transformed onto the trapezium TT' by the matrix

(5123)\begin{pmatrix} 5 & 1 \\ -2 & 3 \end{pmatrix}

Given that the area of trapezium TT' is 510510 square units, calculate the exact value of kk.

(5)
题目中文翻译
  1. (i) A=(1003)A = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}

(a) 完全描述矩阵 AA 所表示的单个变换。

矩阵 BB 表示绕原点顺时针旋转 45°45°

(b) 写出矩阵 BB,矩阵的每个元素用精确形式表示。

先进行矩阵 AA 表示的变换,再进行矩阵 BB 表示的变换,所得复合变换由矩阵 CC 表示。

(c) 求 CC

(ii) 梯形 TT 的顶点为 (2,0)(-2, 0)(2,k)(-2, k)(5,8)(5, 8)(5,0)(5, 0),其中 kk 是正常数。梯形 TT 在矩阵

(5123)\begin{pmatrix} 5 & 1 \\ -2 & 3 \end{pmatrix}

表示的变换下变成梯形 TT'

已知梯形 TT' 的面积为 510510 平方单位,求 kk 的精确值。

解答

(i)(a)

解法一

思路

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矩阵 AA 保持 xx 坐标不变,并把 yy 坐标乘以 33,因此它表示平行于 yy 轴的拉伸。

答题过程

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The matrix AA represents a stretch parallel to the yy-axis with scale factor 33.

(i)(b)

解法一

思路

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顺时针旋转 45°45° 等价于逆时针旋转 45°-45°,将该角代入标准旋转矩阵。

答题过程

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The matrix representing a clockwise rotation through 45°45° about the origin is

B=(22222222).\boxed{ B= \begin{pmatrix} \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} \\ -\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} \end{pmatrix} }.

(i)(c)

解法一

思路

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变换顺序是先 AABB,所以复合矩阵是 BABA,不能写成 ABAB

答题过程

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Since the transformation represented by AA is followed by the transformation represented by BB,

C=BA.C=BA.

Right-multiplication by AA leaves the first column of BB unchanged and multiplies its second column by 33. Hence

C=(2232222322).C= \boxed{ \begin{pmatrix} \dfrac{\sqrt{2}}{2} & \dfrac{3\sqrt{2}}{2} \\ -\dfrac{\sqrt{2}}{2} & \dfrac{3\sqrt{2}}{2} \end{pmatrix} }.

(ii)

解法一

思路

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先用梯形面积公式求 TT 的面积。矩阵变换的面积比例因子是行列式的绝对值,因此用 TT 的面积乘以该比例因子,并令结果等于 510510

答题过程

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The parallel sides of TT have lengths kk and 88, and the perpendicular distance between them is

5(2)=7.5-(-2)=7.

Therefore,

Area of T=12(k+8)(7).\text{Area of }T=\frac{1}{2}(k+8)(7).

The determinant of the transformation matrix is

5123=5(3)1(2)=17.\begin{align*} \begin{vmatrix} 5 & 1 \\ -2 & 3 \end{vmatrix} =&\,5(3)-1(-2) \\ =&\,17. \end{align*}

Hence the area scale factor is 1717. Since the area of TT' is 510510,

17[12(k+8)(7)]=510.17\left[\frac{1}{2}(k+8)(7)\right]=510.

Thus

k+8=607,k+8=\frac{60}{7},

so

k=47.\boxed{k=\frac{4}{7}}.

解法二

思路

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把梯形 TT 的四个顶点分别乘以变换矩阵,得到 TT' 的四个顶点。保持原顶点的环绕顺序,使用鞋带公式表示 TT' 的面积,再令其等于 510510

答题过程

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Let the transformed vertices, in the given order, be V1V_1, V2V_2, V3V_3 and V4V_4. Then

(2,0)(10,4),(2,k)(10+k,4+3k),(5,8)(33,14),(5,0)(25,10).\begin{align*} (-2,0)\mapsto&\,(-10,4), \\ (-2,k)\mapsto&\,(-10+k,4+3k), \\ (5,8)\mapsto&\,(33,14), \\ (5,0)\mapsto&\,(25,-10). \end{align*}

For the downward products in the shoelace formula,

S1=(10)(4+3k)+14(10+k)+33(10)+25(4)=41016k.\begin{align*} S_1 =&\,(-10)(4+3k)+14(-10+k) \\ +&\,\hspace{2pt}33(-10)+25(4) \\ =&\,-410-16k. \end{align*}

For the upward products,

S2=4(10+k)+33(4+3k)+14(25)+(10)(10)=542+103k.\begin{align*} S_2 =&\,4(-10+k)+33(4+3k) \\ +&\,\hspace{2pt}14(25)+(-10)(-10) \\ =&\,542+103k. \end{align*}

Therefore, using the shoelace formula and k>0k>0,

Area of T=12S1S2=12952119k=12(952+119k).\begin{align*} \text{Area of }T' =&\,\frac{1}{2}|S_1-S_2| \\ =&\,\frac{1}{2}|-952-119k| \\ =&\,\frac{1}{2}(952+119k). \end{align*}

Therefore,

12(952+119k)=510,\frac{1}{2}(952+119k)=510,

which gives

k=47.\boxed{k=\frac{4}{7}}.