题目
Problem
7. The matrix A \mathbf{A} A
A = ( 4 − 5 − 3 2 ) \mathbf{A} = \begin{pmatrix} 4 & -5 \\ -3 & 2 \end{pmatrix} A = ( 4 − 3 − 5 2 ) The transformation represented by A \mathbf{A} A T T T T ′ T' T ′
Given that the area of triangle T T T 23 23 23 2 ^2 2
(a) determine the area of triangle T ′ T' T ′
(2)
The point P P P ( 3 p + 2 , 2 p − 1 ) (3p + 2, 2p - 1) ( 3 p + 2 , 2 p − 1 ) p p p A \mathbf{A} A P P P P ′ P' P ′ ( 17 , − 18 ) (17, -18) ( 17 , − 18 )
(b) Determine the value of p p p
(2)
Given that
B = ( 0 1 − 1 0 ) \mathbf{B} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} B = ( 0 − 1 1 0 ) (c) describe fully the single geometrical transformation represented by matrix B \mathbf{B} B
(2)
The transformation represented by matrix A \mathbf{A} A C \mathbf{C} C B \mathbf{B} B
(d) Determine C \mathbf{C} C
(3)
题目中文翻译
矩阵 A \mathbf{A} A
A = ( 4 − 5 − 3 2 ) \mathbf{A} = \begin{pmatrix} 4 & -5 \\ -3 & 2 \end{pmatrix} A = ( 4 − 3 − 5 2 ) A \mathbf{A} A T T T T ′ T' T ′
已知三角形 T T T 23 23 23 2 ^2 2
(a) 确定三角形 T ′ T' T ′
(2)
点 P P P ( 3 p + 2 , 2 p − 1 ) (3p + 2, 2p - 1) ( 3 p + 2 , 2 p − 1 ) p p p A \mathbf{A} A P P P ( 17 , − 18 ) (17, -18) ( 17 , − 18 ) P ′ P' P ′
(b) 确定 p p p
(2)
已知
B = ( 0 1 − 1 0 ) \mathbf{B} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} B = ( 0 − 1 1 0 ) (c) 完整描述矩阵 B \mathbf{B} B
(2)
矩阵 A \mathbf{A} A C \mathbf{C} C B \mathbf{B} B
(d) 确定矩阵 C \mathbf{C} C
(3)
解答
(a)
解法一
思路
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二维线性变换会把面积乘以变换矩阵行列式的绝对值。先求 det A \det\mathbf{A} det A ∣ det A ∣ |\det\mathbf{A}| ∣ det A ∣
答题过程
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The area scale factor is
∣ det A ∣ = ∣ 4 ( 2 ) − ( − 5 ) ( − 3 ) ∣ = ∣ 8 − 15 ∣ = 7. \begin{align*}
&\,|\det\mathbf{A}|\\
=&\,\big|4(2)-(-5)(-3)\big|\\
=&\,|8-15|\\
=&\,7.
\end{align*} = = = ∣ det A ∣ 4 ( 2 ) − ( − 5 ) ( − 3 ) ∣8 − 15∣ 7. Therefore,
Area ( T ′ ) = 7 ( 23 ) = 161 cm 2 . \begin{align*}
\operatorname{Area}(T')=&\,7(23)\\
=&\,\boxed{161\text{ cm}^2}.
\end{align*} Area ( T ′ ) = = 7 ( 23 ) 161 cm 2 .
(b)
解法一
思路
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把点 P P P A \mathbf{A} A P ′ = ( 17 , − 18 ) P'=(17,-18) P ′ = ( 17 , − 18 ) p p p
答题过程
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Since A \mathbf{A} A P P P P ′ P' P ′
( 4 − 5 − 3 2 ) ( 3 p + 2 2 p − 1 ) = ( 17 − 18 ) . \begin{align*}
&\,\begin{pmatrix}
4&-5\\
-3&2
\end{pmatrix}
\begin{pmatrix}
3p+2\\
2p-1
\end{pmatrix}\\
=&\,\begin{pmatrix}
17\\
-18
\end{pmatrix}.
\end{align*} = ( 4 − 3 − 5 2 ) ( 3 p + 2 2 p − 1 ) ( 17 − 18 ) . Using the first coordinate,
4 ( 3 p + 2 ) − 5 ( 2 p − 1 ) = 17. \begin{align*}
&\,4(3p+2)-5(2p-1)\\
=&\,17.
\end{align*} = 4 ( 3 p + 2 ) − 5 ( 2 p − 1 ) 17. Expanding and collecting terms,
12 p + 8 − 10 p + 5 = 17 , 2 p + 13 = 17 , 2 p = 4 , p = 2. \begin{align*}
12p+8-10p+5=&\,17,\\
2p+13=&\,17,\\
2p=&\,4,\\
p=&\,2.
\end{align*} 12 p + 8 − 10 p + 5 = 2 p + 13 = 2 p = p = 17 , 17 , 4 , 2. Indeed, the second coordinate gives
− 3 ( 3 p + 2 ) + 2 ( 2 p − 1 ) = − 3 ( 8 ) + 2 ( 3 ) = − 24 + 6 = − 18 \begin{align*}
&\,-3(3p+2)+2(2p-1)\\
=&\,-3(8)+2(3)\\
=&\,-24+6\\
=&\,-18
\end{align*} = = = − 3 ( 3 p + 2 ) + 2 ( 2 p − 1 ) − 3 ( 8 ) + 2 ( 3 ) − 24 + 6 − 18 when p = 2 p=2 p = 2
p = 2 . \boxed{p=2}. p = 2 .
解法二
思路
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用逆矩阵把已知的像点 P ′ P' P ′ P P P P P P 3 p + 2 3p+2 3 p + 2 2 p − 1 2p-1 2 p − 1 p p p
答题过程
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The determinant of A \mathbf A A
det A = 4 ( 2 ) − ( − 5 ) ( − 3 ) = 8 − 15 = − 7. \begin{align*}
&\,\det\mathbf A\\
=&\,4(2)-(-5)(-3)\\
=&\,8-15\\
=&\,-7.
\end{align*} = = = det A 4 ( 2 ) − ( − 5 ) ( − 3 ) 8 − 15 − 7. Hence A \mathbf A A
A − 1 = − 1 7 ( 2 5 3 4 ) . \mathbf A^{-1}=-\frac17
\begin{pmatrix}
2&5\\
3&4
\end{pmatrix}. A − 1 = − 7 1 ( 2 3 5 4 ) . Applying A − 1 \mathbf A^{-1} A − 1 P ′ P' P ′
( 3 p + 2 2 p − 1 ) = − 1 7 ( 2 5 3 4 ) ( 17 − 18 ) = − 1 7 ( 34 − 90 51 − 72 ) = − 1 7 ( − 56 − 21 ) = ( 8 3 ) . \begin{align*}
&\,\begin{pmatrix}
3p+2\\
2p-1
\end{pmatrix}\\
=&\,-\frac17
\begin{pmatrix}
2&5\\
3&4
\end{pmatrix}
\begin{pmatrix}
17\\
-18
\end{pmatrix}\\
=&\,-\frac17
\begin{pmatrix}
34-90\\
51-72
\end{pmatrix}\\
=&\,-\frac17
\begin{pmatrix}
-56\\
-21
\end{pmatrix}\\
=&\,\begin{pmatrix}
8\\
3
\end{pmatrix}.
\end{align*} = = = = ( 3 p + 2 2 p − 1 ) − 7 1 ( 2 3 5 4 ) ( 17 − 18 ) − 7 1 ( 34 − 90 51 − 72 ) − 7 1 ( − 56 − 21 ) ( 8 3 ) . Therefore,
3 p + 2 = 8 , 3 p = 6 , p = 2. \begin{align*}
3p+2=&\,8,\\
3p=&\,6,\\
p=&\,2.
\end{align*} 3 p + 2 = 3 p = p = 8 , 6 , 2. The other coordinate is consistent, since 2 ( 2 ) − 1 = 3 2(2)-1=3 2 ( 2 ) − 1 = 3
p = 2 . \boxed{p=2}. p = 2 .
(c)
解法一
思路
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观察标准旋转矩阵。矩阵 B \mathbf{B} B ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 0 , − 1 ) (0,-1) ( 0 , − 1 ) 90 ∘ 90^\circ 9 0 ∘
答题过程
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The matrix
B = ( 0 1 − 1 0 ) \mathbf{B}=
\begin{pmatrix}
0&1\\
-1&0
\end{pmatrix} B = ( 0 − 1 1 0 ) represents a rotation through 90 ∘ 90^\circ 9 0 ∘
(d)
解法一
思路
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先进行 A \mathbf{A} A C \mathbf{C} C C A \mathbf{CA} CA C A = B \mathbf{CA}=\mathbf{B} CA = B A − 1 \mathbf{A}^{-1} A − 1 C = B A − 1 \mathbf{C}=\mathbf{BA}^{-1} C = BA − 1
答题过程
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Since the transformation represented by A \mathbf{A} A C \mathbf{C} C
C A = B . \mathbf{CA}=\mathbf{B}. CA = B . Now
A − 1 = 1 4 ( 2 ) − ( − 5 ) ( − 3 ) × ( 2 5 3 4 ) = − 1 7 ( 2 5 3 4 ) . \begin{align*}
&\,\mathbf{A}^{-1}\\
=&\,\frac{1}{4(2)-(-5)(-3)}\\
&\,\hspace{2pt}\times
\begin{pmatrix}
2&5\\
3&4
\end{pmatrix}\\
=&\,-\frac17
\begin{pmatrix}
2&5\\
3&4
\end{pmatrix}.
\end{align*} = = A − 1 4 ( 2 ) − ( − 5 ) ( − 3 ) 1 × ( 2 3 5 4 ) − 7 1 ( 2 3 5 4 ) . Therefore,
C = B A − 1 = − 1 7 ( 0 1 − 1 0 ) ( 2 5 3 4 ) = − 1 7 ( 3 4 − 2 − 5 ) . \begin{align*}
&\,\mathbf{C}\\
=&\,\mathbf{BA}^{-1}\\
=&\,-\frac17
\begin{pmatrix}
0&1\\
-1&0
\end{pmatrix}
\begin{pmatrix}
2&5\\
3&4
\end{pmatrix}\\
=&\,-\frac17
\begin{pmatrix}
3&4\\
-2&-5
\end{pmatrix}.
\end{align*} = = = C BA − 1 − 7 1 ( 0 − 1 1 0 ) ( 2 3 5 4 ) − 7 1 ( 3 − 2 4 − 5 ) . Hence
C = ( − 3 7 − 4 7 2 7 5 7 ) . \boxed{
\mathbf{C}=
\begin{pmatrix}
-\dfrac37&-\dfrac47\\
\dfrac27&\dfrac57
\end{pmatrix}
}. C = − 7 3 7 2 − 7 4 7 5 .
解法二
思路
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设 C \mathbf C C A \mathbf A A C \mathbf C C C A = B \mathbf{CA}=\mathbf B CA = B B \mathbf B B C \mathbf C C
答题过程
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Let
C = ( a b c d ) . \mathbf C=
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}. C = ( a c b d ) . Since C A = B \mathbf{CA}=\mathbf B CA = B
( a b c d ) ( 4 − 5 − 3 2 ) = ( 4 a − 3 b − 5 a + 2 b 4 c − 3 d − 5 c + 2 d ) = ( 0 1 − 1 0 ) . \begin{align*}
&\,\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}
\begin{pmatrix}
4&-5\\
-3&2
\end{pmatrix}\\
=&\,\begin{pmatrix}
4a-3b&-5a+2b\\
4c-3d&-5c+2d
\end{pmatrix}\\
=&\,\begin{pmatrix}
0&1\\
-1&0
\end{pmatrix}.
\end{align*} = = ( a c b d ) ( 4 − 3 − 5 2 ) ( 4 a − 3 b 4 c − 3 d − 5 a + 2 b − 5 c + 2 d ) ( 0 − 1 1 0 ) . Equating the entries in the first row,
4 a − 3 b = 0 , − 5 a + 2 b = 1. \begin{align*}
4a-3b=&\,0,\\
-5a+2b=&\,1.
\end{align*} 4 a − 3 b = − 5 a + 2 b = 0 , 1. Multiplying the first equation by 2 2 2 3 3 3
8 a − 6 b = 0 , − 15 a + 6 b = 3. \begin{align*}
8a-6b=&\,0,\\
-15a+6b=&\,3.
\end{align*} 8 a − 6 b = − 15 a + 6 b = 0 , 3. Adding these equations,
− 7 a = 3 , a = − 3 7 . \begin{align*}
-7a=&\,3,\\
a=&\,-\frac37.
\end{align*} − 7 a = a = 3 , − 7 3 . Substituting into 4 a − 3 b = 0 4a-3b=0 4 a − 3 b = 0
− 12 7 − 3 b = 0 , − 3 b = 12 7 , b = − 4 7 . \begin{align*}
-\frac{12}{7}-3b=&\,0,\\
-3b=&\,\frac{12}{7},\\
b=&\,-\frac47.
\end{align*} − 7 12 − 3 b = − 3 b = b = 0 , 7 12 , − 7 4 . Equating the entries in the second row,
4 c − 3 d = − 1 , − 5 c + 2 d = 0. \begin{align*}
4c-3d=&\,-1,\\
-5c+2d=&\,0.
\end{align*} 4 c − 3 d = − 5 c + 2 d = − 1 , 0. Multiplying the first equation by 2 2 2 3 3 3
8 c − 6 d = − 2 , − 15 c + 6 d = 0. \begin{align*}
8c-6d=&\,-2,\\
-15c+6d=&\,0.
\end{align*} 8 c − 6 d = − 15 c + 6 d = − 2 , 0. Adding these equations,
− 7 c = − 2 , c = 2 7 . \begin{align*}
-7c=&\,-2,\\
c=&\,\frac27.
\end{align*} − 7 c = c = − 2 , 7 2 . Substituting into − 5 c + 2 d = 0 -5c+2d=0 − 5 c + 2 d = 0
− 10 7 + 2 d = 0 , 2 d = 10 7 , d = 5 7 . \begin{align*}
-\frac{10}{7}+2d=&\,0,\\
2d=&\,\frac{10}{7},\\
d=&\,\frac57.
\end{align*} − 7 10 + 2 d = 2 d = d = 0 , 7 10 , 7 5 . Hence
C = ( − 3 7 − 4 7 2 7 5 7 ) . \boxed{
\mathbf C=
\begin{pmatrix}
-\dfrac37&-\dfrac47\\
\dfrac27&\dfrac57
\end{pmatrix}
}. C = − 7 3 7 2 − 7 4 7 5 . 