题目
Problem
5.
A = ( 1 2 − 3 2 3 2 1 2 ) A=\begin{pmatrix}
\frac12 & -\frac{\sqrt3}{2} \\
\frac{\sqrt3}{2} & \frac12
\end{pmatrix} A = ( 2 1 2 3 − 2 3 2 1 ) (a) Describe fully the single geometrical transformation represented by the matrix A A A
(3)
(b) Hence write down the matrix A 6 A^6 A 6
(1)
The transformation represented by the matrix C C C B B B A A A
Given that
B = ( 2 3 − 7 − 4 5 3 ) B=\begin{pmatrix}
2\sqrt3 & -7 \\
-4 & 5\sqrt3
\end{pmatrix} B = ( 2 3 − 4 − 7 5 3 ) (c) find the matrix C C C
(4)
题目中文翻译
5.
A = ( 1 2 − 3 2 3 2 1 2 ) A=\begin{pmatrix}
\frac12 & -\frac{\sqrt3}{2} \\
\frac{\sqrt3}{2} & \frac12
\end{pmatrix} A = ( 2 1 2 3 − 2 3 2 1 ) (a) 完整描述矩阵 A A A
(3)
(b) 由此写出矩阵 A 6 A^6 A 6
(1)
矩阵 C C C B B B A A A
已知
B = ( 2 3 − 7 − 4 5 3 ) B=\begin{pmatrix}
2\sqrt3 & -7 \\
-4 & 5\sqrt3
\end{pmatrix} B = ( 2 3 − 4 − 7 5 3 ) (c) 求矩阵 C C C
(4)
解答
(a)
解法一
思路
展开
把 A A A θ \theta θ cos θ \cos\theta cos θ sin θ \sin\theta sin θ
答题过程
展开
The standard matrix for an anticlockwise rotation through an angle θ \theta θ
( cos θ − sin θ sin θ cos θ ) . \begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}. ( cos θ sin θ − sin θ cos θ ) . For the given matrix,
cos θ = 1 2 , sin θ = 3 2 , \begin{align*}
\cos\theta=&\,\frac12,\\
\sin\theta=&\,\frac{\sqrt3}{2},
\end{align*} cos θ = sin θ = 2 1 , 2 3 , so θ = 60 ∘ \theta=60^\circ θ = 6 0 ∘ A A A
a rotation through 60 ∘ anticlockwise about the origin . \boxed{\begin{gathered}
\text{a rotation through }60^\circ\\
\text{anticlockwise}\\
\text{about the origin}.
\end{gathered}} a rotation through 6 0 ∘ anticlockwise about the origin .
(b)
解法一
思路
展开
承接 (a)，连续应用六次 A A A 6 × 60 ∘ = 360 ∘ 6\times60^\circ=360^\circ 6 × 6 0 ∘ = 36 0 ∘
答题过程
展开
From part (a), A 6 A^6 A 6
6 × 60 ∘ = 360 ∘ . 6\times60^\circ=360^\circ. 6 × 6 0 ∘ = 36 0 ∘ . Hence
A 6 = ( 1 0 0 1 ) . \boxed{A^6=\begin{pmatrix}1&0\\0&1\end{pmatrix}}. A 6 = ( 1 0 0 1 ) .
(c)
解法一
思路
展开
先做 C C C B B B B C BC B C B C = A BC=A B C = A B − 1 B^{-1} B − 1 C = B − 1 A C=B^{-1}A C = B − 1 A
答题过程
展开
Since the transformation represented by C C C B B B
B C = A . BC=A. B C = A . Now
det B = ( 2 3 ) ( 5 3 ) − ( − 7 ) ( − 4 ) = 30 − 28 = 2. \begin{align*}
&\,\det B\\
=&\,(2\sqrt3)(5\sqrt3)-(-7)(-4) \\
=&\,30-28=2.
\end{align*} = = det B ( 2 3 ) ( 5 3 ) − ( − 7 ) ( − 4 ) 30 − 28 = 2. Therefore,
B − 1 = 1 2 ( 5 3 7 4 2 3 ) . B^{-1}=\frac12
\begin{pmatrix}
5\sqrt3 & 7 \\
4 & 2\sqrt3
\end{pmatrix}. B − 1 = 2 1 ( 5 3 4 7 2 3 ) . Hence
C = B − 1 A = 1 2 × ( 5 3 2 + 7 3 2 − 15 2 + 7 2 2 + 3 − 2 3 + 3 ) = 1 2 ( 6 3 − 4 5 − 3 ) . \begin{align*}
&\,C\\
=&\,B^{-1}A \\
=&\,\tfrac12\\
&\,\times
\begin{pmatrix}
\begin{gathered}\tfrac{5\sqrt3}{2}\\+\tfrac{7\sqrt3}{2}\end{gathered}
& -\tfrac{15}{2}+\tfrac72 \\
2+3 & \begin{gathered}-2\sqrt3\\+\sqrt3\end{gathered}
\end{pmatrix} \\
=&\,\tfrac12
\begin{pmatrix}
6\sqrt3 & -4 \\
5 & -\sqrt3
\end{pmatrix}.
\end{align*} = = = C B − 1 A 2 1 × 2 5 3 + 2 7 3 2 + 3 − 2 15 + 2 7 − 2 3 + 3 2 1 ( 6 3 5 − 4 − 3 ) . Thus
C = ( 3 3 − 2 5 2 − 3 2 ) . \boxed{C=\begin{pmatrix}
3\sqrt3 & -2 \\
\frac52 & -\frac{\sqrt3}{2}
\end{pmatrix}}. C = ( 3 3 2 5 − 2 − 2 3 ) .
解法二
思路
展开
也可以用联立方程法。把 C C C B C = A BC=A B C = A
答题过程
展开
Let
C = ( a b c d ) . C=\begin{pmatrix}a&b\\c&d\end{pmatrix}. C = ( a c b d ) . Since B C = A BC=A B C = A
( 2 3 − 7 − 4 5 3 ) ( a b c d ) = ( 1 2 − 3 2 3 2 1 2 ) . \begin{align*}
&\,\begin{pmatrix}2\sqrt3&-7\\-4&5\sqrt3\end{pmatrix}
\begin{pmatrix}a&b\\c&d\end{pmatrix}\\
=&\,\begin{pmatrix}\frac12&-\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac12\end{pmatrix}.
\end{align*} = ( 2 3 − 4 − 7 5 3 ) ( a c b d ) ( 2 1 2 3 − 2 3 2 1 ) . Equating corresponding entries gives
2 3 a − 7 c = 1 2 , − 4 a + 5 3 c = 3 2 , \begin{align*}
2\sqrt3a-7c=&\,\frac12, \\
-4a+5\sqrt3c=&\,\frac{\sqrt3}{2},
\end{align*} 2 3 a − 7 c = − 4 a + 5 3 c = 2 1 , 2 3 , and
2 3 b − 7 d = − 3 2 , − 4 b + 5 3 d = 1 2 . \begin{align*}
2\sqrt3b-7d=&\,-\frac{\sqrt3}{2}, \\
-4b+5\sqrt3d=&\,\frac12.
\end{align*} 2 3 b − 7 d = − 4 b + 5 3 d = − 2 3 , 2 1 . For the first pair, multiply the first equation by 2 2 2 3 \sqrt3 3
4 3 a − 14 c = 1 , − 4 3 a + 15 c = 3 2 . \begin{align*}
4\sqrt3a-14c=&\,1, \\
-4\sqrt3a+15c=&\,\frac32.
\end{align*} 4 3 a − 14 c = − 4 3 a + 15 c = 1 , 2 3 . Adding gives c = 5 2 c=\frac52 c = 2 5
2 3 a − 35 2 = 1 2 , 2\sqrt3a-\frac{35}{2}=\frac12, 2 3 a − 2 35 = 2 1 , so
a = 3 3 , c = 5 2 . a=3\sqrt3,
\qquad
c=\frac52. a = 3 3 , c = 2 5 . For the second pair, multiply the first equation by 2 2 2 3 \sqrt3 3
4 3 b − 14 d = − 3 , − 4 3 b + 15 d = 3 2 . \begin{align*}
4\sqrt3b-14d=&\,-\sqrt3, \\
-4\sqrt3b+15d=&\,\frac{\sqrt3}{2}.
\end{align*} 4 3 b − 14 d = − 4 3 b + 15 d = − 3 , 2 3 . Adding gives d = − 3 2 d=-\frac{\sqrt3}{2} d = − 2 3 b = − 2 b=-2 b = − 2
b = − 2 , d = − 3 2 . b=-2,
\qquad
d=-\frac{\sqrt3}{2}. b = − 2 , d = − 2 3 . Therefore,
C = ( 3 3 − 2 5 2 − 3 2 ) . \boxed{C=\begin{pmatrix}
3\sqrt3 & -2 \\
\frac52 & -\frac{\sqrt3}{2}
\end{pmatrix}}. C = ( 3 3 2 5 − 2 − 2 3 ) . 