题目
Problem
(i)
P = ( 0 − 1 − 1 0 ) \mathbf{P} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} P = ( 0 − 1 − 1 0 )
The matrix P \mathbf{P} P U U U
(a) Describe U U U
(2)
The transformation V V V 2 × 2 2 \times 2 2 × 2 Q \mathbf{Q} Q 240 ° 240° 240° ( 0 , 0 ) (0, 0) ( 0 , 0 ) 6 6 6
(b) Determine the matrix Q \mathbf{Q} Q
(2)
Given that U U U V V V T T T R \mathbf{R} R
(c) determine the matrix R \mathbf{R} R
(2)
(ii) The transformation W W W
( − 2 2 3 2 3 2 ) \begin{pmatrix} -2 & 2\sqrt{3} \\ 2\sqrt{3} & 2 \end{pmatrix} ( − 2 2 3 2 3 2 )
Show that there is a real number λ \lambda λ W W W ( λ , 1 ) (\lambda, 1) ( λ , 1 ) ( 4 λ , 4 ) (4\lambda, 4) ( 4 λ , 4 ) λ \lambda λ
(5)
题目中文翻译
(i)
P = ( 0 − 1 − 1 0 ) \mathbf{P} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} P = ( 0 − 1 − 1 0 )
矩阵 P \mathbf{P} P U U U
(a) 完整描述 U U U
变换 V V V 2 × 2 2 \times 2 2 × 2 Q \mathbf{Q} Q 240 ° 240° 240° ( 0 , 0 ) (0, 0) ( 0 , 0 ) 6 6 6
(b) 确定矩阵 Q \mathbf{Q} Q
已知 U U U V V V T T T R \mathbf{R} R
(c) 确定矩阵 R \mathbf{R} R
(ii) 变换 W W W ( − 2 2 3 2 3 2 ) \begin{pmatrix} -2 & 2\sqrt{3} \\ 2\sqrt{3} & 2 \end{pmatrix} ( − 2 2 3 2 3 2 )
证明：存在实数 λ \lambda λ W W W ( λ , 1 ) (\lambda, 1) ( λ , 1 ) ( 4 λ , 4 ) (4\lambda, 4) ( 4 λ , 4 ) λ \lambda λ
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) ( x , y ) ( − y , − x ) (-y,-x) ( − y , − x ) y = − x y=-x y = − x
答题过程
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Since
( 0 − 1 − 1 0 ) ( x y ) = ( − y − x ) , \begin{pmatrix}0&-1\\-1&0\end{pmatrix}
\begin{pmatrix}x\\y\end{pmatrix}
=\begin{pmatrix}-y\\-x\end{pmatrix}, ( 0 − 1 − 1 0 ) ( x y ) = ( − y − x ) , U U U
reflection in the line y = − x . \boxed{\text{reflection in the line }y=-x}. reflection in the line y = − x .
(i)(b)
解法一
思路
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先写出逆时针旋转 240 ∘ 240^\circ 24 0 ∘
答题过程
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The rotation matrix is
( cos 240 ∘ − sin 240 ∘ sin 240 ∘ cos 240 ∘ ) = ( − 1 2 3 2 − 3 2 − 1 2 ) . \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*} = ( cos 24 0 ∘ sin 24 0 ∘ − sin 24 0 ∘ cos 24 0 ∘ ) − 2 1 − 2 3 2 3 − 2 1 . Applying the enlargement with scale factor 6 gives
Q = ( − 3 3 3 − 3 3 − 3 ) . \boxed{
\mathbf Q=
\begin{pmatrix}
-3&3\sqrt3\\
-3\sqrt3&-3
\end{pmatrix}}. Q = ( − 3 − 3 3 3 3 − 3 ) .
(i)(c)
解法一
思路
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先施行 U U U V V V P \mathbf P P Q \mathbf Q Q Q P \mathbf Q\mathbf P QP
答题过程
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Since U U U V V V
R = Q P = ( − 3 3 3 − 3 3 − 3 ) ( 0 − 1 − 1 0 ) = ( − 3 3 3 3 3 3 ) . \begin{align*}
\mathbf R
=&\,\mathbf Q\mathbf P\\
=&\,\begin{pmatrix}
-3&3\sqrt3\\
-3\sqrt3&-3
\end{pmatrix}
\begin{pmatrix}
0&-1\\
-1&0
\end{pmatrix}\\
=&\,\boxed{
\begin{pmatrix}
-3\sqrt3&3\\
3&3\sqrt3
\end{pmatrix}}.
\end{align*} R = = = QP ( − 3 − 3 3 3 3 − 3 ) ( 0 − 1 − 1 0 ) ( − 3 3 3 3 3 3 ) .
(ii)
解法一
思路
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直接把点 ( λ , 1 ) (\lambda,1) ( λ , 1 ) W W W λ \lambda λ
答题过程
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The required mapping gives
( − 2 2 3 2 3 2 ) ( λ 1 ) = ( 4 λ 4 ) . \begin{pmatrix}
-2&2\sqrt3\\
2\sqrt3&2
\end{pmatrix}
\begin{pmatrix}\lambda\\1\end{pmatrix}
=\begin{pmatrix}4\lambda\\4\end{pmatrix}. ( − 2 2 3 2 3 2 ) ( λ 1 ) = ( 4 λ 4 ) . Hence
− 2 λ + 2 3 = 4 λ , 2 3 λ + 2 = 4. \begin{align*}
-2\lambda+2\sqrt3=&\,4\lambda,\\
2\sqrt3\lambda+2=&\,4.
\end{align*} − 2 λ + 2 3 = 2 3 λ + 2 = 4 λ , 4. The first equation gives
6 λ = 2 3 , 6\lambda=2\sqrt3, 6 λ = 2 3 , so
λ = 3 3 . \lambda=\frac{\sqrt3}{3}. λ = 3 3 . The second equation gives
2 3 λ = 2 , 2\sqrt3\lambda=2, 2 3 λ = 2 , so
λ = 1 3 = 3 3 . \lambda=\frac1{\sqrt3}=\frac{\sqrt3}{3}. λ = 3 1 = 3 3 . Both coordinates therefore give the same real value. Hence
λ = 3 3 . \boxed{\lambda=\frac{\sqrt3}{3}}. λ = 3 3 .
解法二
思路
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官方替代路线是对映射矩阵求逆，再把像点 ( 4 λ , 4 ) (4\lambda,4) ( 4 λ , 4 ) ( λ , 1 ) (\lambda,1) ( λ , 1 )
答题过程
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Let
W = ( − 2 2 3 2 3 2 ) . \mathbf W=
\begin{pmatrix}
-2&2\sqrt3\\
2\sqrt3&2
\end{pmatrix}. W = ( − 2 2 3 2 3 2 ) . Since
det W = ( − 2 ) ( 2 ) − ( 2 3 ) 2 = − 16 , \det\mathbf W=(-2)(2)-(2\sqrt3)^2=-16, det W = ( − 2 ) ( 2 ) − ( 2 3 ) 2 = − 16 , W − 1 = − 1 16 ( 2 − 2 3 − 2 3 − 2 ) . \mathbf W^{-1}
=-\frac1{16}
\begin{pmatrix}
2&-2\sqrt3\\
-2\sqrt3&-2
\end{pmatrix}. W − 1 = − 16 1 ( 2 − 2 3 − 2 3 − 2 ) . Thus
( λ 1 ) = W − 1 ( 4 λ 4 ) = − 1 16 ( 8 λ − 8 3 − 8 3 λ − 8 ) . \begin{align*}
\begin{pmatrix}\lambda\\1\end{pmatrix}
=&\,\mathbf W^{-1}
\begin{pmatrix}4\lambda\\4\end{pmatrix}\\
=&\,-\frac1{16}
\begin{pmatrix}
8\lambda-8\sqrt3\\
-8\sqrt3\lambda-8
\end{pmatrix}.
\end{align*} ( λ 1 ) = = W − 1 ( 4 λ 4 ) − 16 1 ( 8 λ − 8 3 − 8 3 λ − 8 ) . Equating the first components,
λ = − λ + 3 2 , \lambda=\frac{-\lambda+\sqrt3}{2}, λ = 2 − λ + 3 , so
3 λ = 3 ⟹ λ = 3 3 . 3\lambda=\sqrt3
\quad\Longrightarrow\quad
\lambda=\frac{\sqrt3}{3}. 3 λ = 3 ⟹ λ = 3 3 . Equating the second components,
1 = 3 λ + 1 2 , 1=\frac{\sqrt3\lambda+1}{2}, 1 = 2 3 λ + 1 , which gives
3 λ = 1 ⟹ λ = 3 3 . \sqrt3\lambda=1
\quad\Longrightarrow\quad
\lambda=\frac{\sqrt3}{3}. 3 λ = 1 ⟹ λ = 3 3 . Therefore,
λ = 3 3 . \boxed{\lambda=\frac{\sqrt3}{3}}. λ = 3 3 . 