Question
Problem
\frac12 & -\frac{\sqrt3}{2} \\
\frac{\sqrt3}{2} & \frac12
\end{pmatrix}$$
(a) Describe fully the single geometrical transformation represented by the matrix $A$.
(3)
(b) Hence write down the matrix $A^6$.
(1)
The transformation represented by the matrix $C$ followed by the transformation represented by the matrix $B$ is equivalent to the transformation represented by the matrix $A$.
Given that
$$B=\begin{pmatrix}
2\sqrt3 & -7 \\
-4 & 5\sqrt3
\end{pmatrix}$$
(c) find the matrix $C$, giving your answer in simplest form.
(4)
中文翻译
\frac12 & -\frac{\sqrt3}{2} \\
\frac{\sqrt3}{2} & \frac12
\end{pmatrix}$$
(a) 完整描述矩阵 $A$ 所表示的单一几何变换。
(3)
(b) 由此写出矩阵 $A^6$。
(1)
矩阵 $C$ 所表示的变换后接矩阵 $B$ 所表示的变换等价于矩阵 $A$ 所表示的变换。
已知
$$B=\begin{pmatrix}
2\sqrt3 & -7 \\
-4 & 5\sqrt3
\end{pmatrix}$$
(c) 求矩阵 $C$，将答案写成最简形式。
(4)
解答
(a)
解法一
思路
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把 A A A θ \theta θ cos θ \cos\theta cos θ sin θ \sin\theta sin θ
答题过程
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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 and sin θ = 3 2 , \cos\theta=\frac12
\qquad\text{and}\qquad
\sin\theta=\frac{\sqrt3}{2}, cos θ = 2 1 and sin θ = 2 3 , so θ = 60 ∘ \theta=60^\circ θ = 6 0 ∘ A A A
a rotation through 60 ∘ anticlockwise about the origin . \boxed{\text{a rotation through }60^\circ
\text{ anticlockwise about the origin}.} a rotation through 6 0 ∘ anticlockwise about the origin .
(b)
解法一
思路
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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
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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
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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
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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 \\
=&\,\frac12
\begin{pmatrix}
\frac{5\sqrt3}{2}+\frac{7\sqrt3}{2}
& -\frac{15}{2}+\frac72 \\
2+3 & -2\sqrt3+\sqrt3
\end{pmatrix} \\
=&\,\frac12
\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
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官方评分资料也给出联立方程法。把 C C C B C = A BC=A B C = A
答题过程
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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{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}. ( 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 ) . 