题目
Problem
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 ′
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
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
矩阵 A \mathbf{A} A C \mathbf{C} C B \mathbf{B} B
(d) 确定矩阵 C \mathbf{C} C
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}|
=&\,\left|4(2)-(-5)(-3)\right|\\
=&\,|8-15|\\
=&\,7.
\end{align*} ∣ det A ∣ = = = ∣ 4 ( 2 ) − ( − 5 ) ( − 3 ) ∣ ∣8 − 15∣ 7. Therefore,
Area ( T ′ ) = 7 ( 23 ) = 161 cm 2 . \operatorname{Area}(T')=7(23)=\boxed{161\text{ cm}^2}. 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{pmatrix}
4&-5\\
-3&2
\end{pmatrix}
\begin{pmatrix}
3p+2\\
2p-1
\end{pmatrix}
=
\begin{pmatrix}
17\\
-18
\end{pmatrix}. ( 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 , 2 p + 13 = 17 , p = 2. \begin{align*}
4(3p+2)-5(2p-1)=&\,17,\\
2p+13=&\,17,\\
p=&\,2.
\end{align*} 4 ( 3 p + 2 ) − 5 ( 2 p − 1 ) = 2 p + 13 = p = 17 , 17 , 2. Indeed, the second coordinate gives
− 3 ( 3 p + 2 ) + 2 ( 2 p − 1 ) = − 18 -3(3p+2)+2(2p-1)=-18 − 3 ( 3 p + 2 ) + 2 ( 2 p − 1 ) = − 18 when p = 2 p=2 p = 2
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)}
\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 . 