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FP1 2023 Jan Q7

A Level / Edexcel / FP1

IAL 2023 Jan Paper · Question 7

题目

Problem

(i)

P=(0−1−10)\mathbf{P} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}

The matrix P\mathbf{P} represents a geometrical transformation UU

(a) Describe UU fully as a single geometrical transformation.

(2)

The transformation VV, represented by the 2×22 \times 2 matrix Q\mathbf{Q}, is a rotation through 240∘240^\circ anticlockwise about the origin followed by an enlargement about (0,0)(0, 0) with scale factor 66

(b) Determine the matrix Q\mathbf{Q}, giving each entry in exact numerical form.

(2)

Given that UU followed by VV is the transformation TT, which is represented by the matrix R\mathbf{R}

(c) determine the matrix R\mathbf{R}

(2)

(ii) The transformation WW is represented by the matrix

(−223232)\begin{pmatrix} -2 & 2\sqrt{3} \\ 2\sqrt{3} & 2 \end{pmatrix}

Show that there is a real number λ\lambda for which WW maps the point (λ,1)(\lambda, 1) onto the point (4λ,4)(4\lambda, 4), giving the exact value of λ\lambda

(5)
题目中文翻译

(i)

P=(0−1−10)\mathbf{P} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}

矩阵 P\mathbf{P} 表示几何变换 UU。

(a) 完整描述 UU 为单次几何变换。

(2)

变换 VV 由 2×22 \times 2 矩阵 Q\mathbf{Q} 表示，是绕原点逆时针旋转 240∘240^\circ 后接以 (0,0)(0, 0) 为中心、比例因子为 66 的位似变换。

(b) 确定矩阵 Q\mathbf{Q}，每个元素以精确数值形式表示。

(2)

已知 UU 后接 VV 是变换 TT，由矩阵 R\mathbf{R} 表示，

(c) 确定矩阵 R\mathbf{R}。

(2)

(ii) 变换 WW 由矩阵

(−223232)\begin{pmatrix} -2 & 2\sqrt{3} \\ 2\sqrt{3} & 2 \end{pmatrix}

表示。

证明：存在实数 λ\lambda，使得 WW 将点 (λ,1)(\lambda, 1) 映射到点 (4λ,4)(4\lambda, 4)，并给出 λ\lambda 的精确值。

(5)

解答

(i)(a)

解法一

思路

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观察矩阵对一般点 (x,y)(x,y) 的作用：它把点变为 (−y,−x)(-y,-x)，这正是关于直线 y=−xy=-x 对称的坐标变化。

答题过程

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Since

 (0−1−10)(xy)= (−y−x),\begin{align*} &\,\begin{pmatrix}0&-1\\-1&0\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}\\ =&\,\begin{pmatrix}-y\\-x\end{pmatrix}, \end{align*}

UU is a reflection in the line y=−xy=-x.

(i)(b)

解法一

思路

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先写出逆时针旋转 240∘240^\circ 的标准矩阵，再用位似比例因子 6 乘矩阵中的每个元素。

答题过程

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The rotation matrix is

 (cos⁡240∘−sin⁡240∘sin⁡240∘cos⁡240∘)= (−1232−32−12).\begin{align*} &\,\begin{pmatrix} \cos240^\circ&-\sin240^\circ\\ \sin240^\circ&\cos240^\circ \end{pmatrix}\\ =&\,\begin{pmatrix} -\dfrac12&\dfrac{\sqrt3}{2}\\ -\dfrac{\sqrt3}{2}&-\dfrac12 \end{pmatrix}. \end{align*}

Applying the enlargement with scale factor 6 gives

Q=(−333−33−3).\boxed{ \mathbf Q= \begin{pmatrix} -3&3\sqrt3\\ -3\sqrt3&-3 \end{pmatrix}}.

(i)(c)

解法一

思路

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先施行 UU、再施行 VV，因此列向量先左乘 P\mathbf P，再左乘 Q\mathbf Q；合成矩阵的正确次序是 QP\mathbf Q\mathbf P。

答题过程

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Since UU is followed by VV,

R=QP.\mathbf R=\mathbf Q\mathbf P.

Thus

 R= (−333−33−3) ×(0−1−10).\begin{align*} &\,\mathbf R\\ =&\,\begin{pmatrix} -3&3\sqrt3\\ -3\sqrt3&-3 \end{pmatrix}\\ &\,\hspace{2pt}\times\begin{pmatrix} 0&-1\\ -1&0 \end{pmatrix}. \end{align*}

The four entries are

 r11= (−3)(0)+(33)(−1)= −33,\begin{align*} &\,r_{11}\\ =&\,(-3)(0)+(3\sqrt3)(-1)\\ =&\,-3\sqrt3, \end{align*}  r12= (−3)(−1)+(33)(0)= 3,\begin{align*} &\,r_{12}\\ =&\,(-3)(-1)+(3\sqrt3)(0)\\ =&\,3, \end{align*}  r21= (−33)(0)+(−3)(−1)= 3,\begin{align*} &\,r_{21}\\ =&\,(-3\sqrt3)(0)+(-3)(-1)\\ =&\,3, \end{align*}

and

 r22= (−33)(−1)+(−3)(0)= 33.\begin{align*} &\,r_{22}\\ =&\,(-3\sqrt3)(-1)+(-3)(0)\\ =&\,3\sqrt3. \end{align*}

Therefore,

R=(−333333).\mathbf R= \boxed{\begin{pmatrix} -3\sqrt3&3\\ 3&3\sqrt3 \end{pmatrix}}.

(ii)

解法一

思路

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直接把点 (λ,1)(\lambda,1) 写成列向量并施行矩阵 WW。分别比较两个坐标，两个方程都应给出同一个 λ\lambda；这也完成“确实存在”的验证。

答题过程

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The required mapping gives

 (−223232)(λ1)= ((−2)λ+(23)(1)(23)λ+2(1))= (4λ4).\begin{align*} &\,\begin{pmatrix} -2&2\sqrt3\\ 2\sqrt3&2 \end{pmatrix} \begin{pmatrix}\lambda\\1\end{pmatrix}\\ =&\,\begin{pmatrix} (-2)\lambda+(2\sqrt3)(1)\\ (2\sqrt3)\lambda+2(1) \end{pmatrix}\\ =&\,\begin{pmatrix}4\lambda\\4\end{pmatrix}. \end{align*}

Hence

−2λ+23= 4λ,23λ+2= 4.\begin{align*} -2\lambda+2\sqrt3=&\,4\lambda,\\ 2\sqrt3\lambda+2=&\,4. \end{align*}

The first equation gives

6λ=23,6\lambda=2\sqrt3,

so

λ=33.\lambda=\frac{\sqrt3}{3}.

The second equation gives

23λ=2,2\sqrt3\lambda=2,

so

λ=13=33.\lambda=\frac1{\sqrt3}=\frac{\sqrt3}{3}.

Both coordinates therefore give the same real value. Hence

λ=33.\boxed{\lambda=\frac{\sqrt3}{3}}.

解法二

思路

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对映射矩阵求逆，再把像点 (4λ,4)(4\lambda,4) 逆映射回原来的点 (λ,1)(\lambda,1)。比较所得列向量的两个分量，同样得到一致的实数值。

答题过程

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Let

W=(−223232).\mathbf W= \begin{pmatrix} -2&2\sqrt3\\ 2\sqrt3&2 \end{pmatrix}.

Since

 det⁡W= (−2)(2)−(23)2= −4−12= −16,\begin{align*} &\,\det\mathbf W\\ =&\,(-2)(2)-(2\sqrt3)^2\\ =&\,-4-12\\ =&\,-16, \end{align*}  W−1= −116 ×(2−23−23−2).\begin{align*} &\,\mathbf W^{-1}\\ =&\,-\frac1{16}\\ &\,\hspace{2pt}\times\begin{pmatrix} 2&-2\sqrt3\\ -2\sqrt3&-2 \end{pmatrix}. \end{align*}

Thus

 (λ1)= W−1(4λ4)= −116 ×(8λ−83−83λ−8).\begin{align*} &\,\begin{pmatrix}\lambda\\1\end{pmatrix}\\ =&\,\mathbf W^{-1} \begin{pmatrix}4\lambda\\4\end{pmatrix}\\ =&\,-\frac1{16}\\ &\,\hspace{2pt}\times\begin{pmatrix} 8\lambda-8\sqrt3\\ -8\sqrt3\lambda-8 \end{pmatrix}. \end{align*}

The two entries before multiplying by −1/16-1/16 are

 2(4λ)+(−23)(4)= 8λ−83,\begin{align*} &\,2(4\lambda)+(-2\sqrt3)(4)\\ =&\,8\lambda-8\sqrt3, \end{align*}

and

 (−23)(4λ)+(−2)(4)= −83λ−8.\begin{align*} &\,(-2\sqrt3)(4\lambda)+(-2)(4)\\ =&\,-8\sqrt3\lambda-8. \end{align*}

Equating the first components,

λ= −116(8λ−83)= −λ+32.\begin{align*} \lambda=&\,-\frac1{16}(8\lambda-8\sqrt3)\\ =&\,\frac{-\lambda+\sqrt3}{2}. \end{align*}

Multiplying by 22 and rearranging,

2λ= −λ+3,3λ= 3,λ= 33.\begin{align*} 2\lambda=&\,-\lambda+\sqrt3,\\ 3\lambda=&\,\sqrt3,\\ \lambda=&\,\frac{\sqrt3}{3}. \end{align*}

Equating the second components,

1= −116(−83λ−8)= 3λ+12.\begin{align*} 1=&\,-\frac1{16}(-8\sqrt3\lambda-8)\\ =&\,\frac{\sqrt3\lambda+1}{2}. \end{align*}

Multiplying by 22 and rearranging,

2= 3λ+1,3λ= 1,λ= 33.\begin{align*} 2=&\,\sqrt3\lambda+1,\\ \sqrt3\lambda=&\,1,\\ \lambda=&\,\frac{\sqrt3}{3}. \end{align*}

Therefore,

λ=33.\boxed{\lambda=\frac{\sqrt3}{3}}.