题目
Problem
Let k k k A \mathbf{A} A B \mathbf{B} B C \mathbf{C} C
A = ( 1 k 3 2 1 3 3 2 5 ) , B = ( 0 − 2 − 1 3 0 0 ) , C = ( − 2 − 1 1 1 1 3 ) . \begin{align*}
\mathbf{A}=&\,\begin{pmatrix}1&k&3\\2&1&3\\3&2&5\end{pmatrix},\\
\mathbf{B}=&\,\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix},\qquad
\mathbf{C}=\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}.
\end{align*} A = B = 1 2 3 k 1 2 3 3 5 , 0 − 1 0 − 2 3 0 , C = ( − 2 1 − 1 1 1 3 ) . It is given that A \mathbf{A} A
(a) Show that C A B = ( 3 − 7 − 9 3 ) \mathbf{CAB}=\begin{pmatrix}3&-7\\-9&3\end{pmatrix} CAB = ( 3 − 9 − 7 3 )
(5)
(b) Find the equations of the invariant lines, through the origin, of the transformation in the x x x y y y C A B \mathbf{CAB} CAB
(5)
(c) The matrices D \mathbf{D} D E \mathbf{E} E F \mathbf{F} F x x x y y y
D \mathbf{D} D E \mathbf{E} E x x x F \mathbf{F} F y = x y=x y = x
Given that C A B = D − 9 E F \mathbf{CAB}=\mathbf{D}-9\mathbf{EF} CAB = D − 9 EF D \mathbf{D} D E \mathbf{E} E F \mathbf{F} F
(5)
题目中文翻译
设 k k k A \mathbf{A} A B \mathbf{B} B C \mathbf{C} C
A = ( 1 k 3 2 1 3 3 2 5 ) , B = ( 0 − 2 − 1 3 0 0 ) , C = ( − 2 − 1 1 1 1 3 ) . \begin{align*}
\mathbf{A}=&\,\begin{pmatrix}1&k&3\\2&1&3\\3&2&5\end{pmatrix},\\
\mathbf{B}=&\,\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix},\qquad
\mathbf{C}=\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}.
\end{align*} A = B = 1 2 3 k 1 2 3 3 5 , 0 − 1 0 − 2 3 0 , C = ( − 2 1 − 1 1 1 3 ) . 已知 A \mathbf{A} A
(a) 证明 C A B = ( 3 − 7 − 9 3 ) \mathbf{CAB}=\begin{pmatrix}3&-7\\-9&3\end{pmatrix} CAB = ( 3 − 9 − 7 3 )
(5)
(b) 求矩阵 C A B \mathbf{CAB} CAB x x x y y y
(5)
(c) 矩阵 D \mathbf{D} D E \mathbf{E} E F \mathbf{F} F x x x y y y
D \mathbf{D} D E \mathbf{E} E x x x F \mathbf{F} F y = x y=x y = x
已知 C A B = D − 9 E F \mathbf{CAB}=\mathbf{D}-9\mathbf{EF} CAB = D − 9 EF D \mathbf{D} D E \mathbf{E} E F \mathbf{F} F
(5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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先由A奇异求k，再按矩阵尺寸算AB和CAB，不能在未知k时直接套最终矩阵。
答题过程
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Since A \mathbf A A
0 = det A = ( 5 − 6 ) − k ( 10 − 9 ) + 3 ( 4 − 3 ) = 2 − k . \begin{align*}
0=\det\mathbf A
=&\,(5-6)-k(10-9)+3(4-3)\\[4mm]
=&\,2-k.
\end{align*} 0 = det A = = ( 5 − 6 ) − k ( 10 − 9 ) + 3 ( 4 − 3 ) 2 − k . Thus k = 2 k=2 k = 2
A B = ( 1 2 3 2 1 3 3 2 5 ) ( 0 − 2 − 1 3 0 0 ) = ( − 2 4 − 1 − 1 − 2 0 ) . \begin{align*}
\mathbf{AB}
=&\,\begin{pmatrix}1&2&3\\2&1&3\\3&2&5\end{pmatrix}
\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix}\\[4mm]
=&\,\begin{pmatrix}-2&4\\-1&-1\\-2&0\end{pmatrix}.
\end{align*} AB = = 1 2 3 2 1 2 3 3 5 0 − 1 0 − 2 3 0 − 2 − 1 − 2 4 − 1 0 . Then
C A B = ( − 2 − 1 1 1 1 3 ) ( − 2 4 − 1 − 1 − 2 0 ) = ( 4 + 1 − 2 − 8 + 1 − 2 − 1 − 6 4 − 1 ) = ( 3 − 7 − 9 3 ) , \begin{align*}
\mathbf{CAB}
=&\,\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}
\begin{pmatrix}-2&4\\-1&-1\\-2&0\end{pmatrix}\\[4mm]
=&\,\begin{pmatrix}4+1-2&-8+1\\-2-1-6&4-1\end{pmatrix}\\[4mm]
=&\,\boxed{\begin{pmatrix}3&-7\\-9&3\end{pmatrix}},
\end{align*} CAB = = = ( − 2 1 − 1 1 1 3 ) − 2 − 1 − 2 4 − 1 0 ( 4 + 1 − 2 − 2 − 1 − 6 − 8 + 1 4 − 1 ) ( 3 − 9 − 7 3 ) , as required.
(b)
解法一
思路
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原点直线上的点映射后仍满足y=mx，得出斜率方程，并另查竖直线。
答题过程
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The image of ( x , m x ) (x,mx) ( x , m x ) ( ( 3 − 7 m ) x , ( − 9 + 3 m ) x ) ((3-7m)x,(-9+3m)x) (( 3 − 7 m ) x , ( − 9 + 3 m ) x )
− 9 + 3 m = m ( 3 − 7 m ) ⟹ 7 m 2 = 9. -9+3m=m(3-7m)\quad\Longrightarrow\quad 7m^2=9. − 9 + 3 m = m ( 3 − 7 m ) ⟹ 7 m 2 = 9. Thus m = ± 3 / 7 m=\pm3/\sqrt7 m = ± 3/ 7 3 − 7 m = 0 3-7m=0 3 − 7 m = 0
The vertical line is not invariant because ( 0 , y ) (0,y) ( 0 , y ) ( − 7 y , 3 y ) (-7y,3y) ( − 7 y , 3 y )
y = 3 7 x and y = − 3 7 x . \boxed{y=\frac3{\sqrt7}x\quad\text{and}\quad y=-\frac3{\sqrt7}x}. y = 7 3 x and y = − 7 3 x .
(c)
解法一
思路
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按每种几何变换的标准形式设矩阵，先乘EF，再逐个元素与已知矩阵比较。
答题过程
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Let the enlargement scale factor be d d d e e e
D = ( d 0 0 d ) , E = ( e 0 0 1 ) , F = ( 0 1 1 0 ) . \begin{align*}
\mathbf D=&\,\begin{pmatrix}d&0\\0&d\end{pmatrix},\\[4mm]
\mathbf E=&\,\begin{pmatrix}e&0\\0&1\end{pmatrix},\\[4mm]
\mathbf F=&\,\begin{pmatrix}0&1\\1&0\end{pmatrix}.
\end{align*} D = E = F = ( d 0 0 d ) , ( e 0 0 1 ) , ( 0 1 1 0 ) . Since E F = ( 0 e 1 0 ) \mathbf{EF}=\begin{pmatrix}0&e\\1&0\end{pmatrix} EF = ( 0 1 e 0 )
( 3 − 7 − 9 3 ) = D − 9 E F = ( d − 9 e − 9 d ) . \begin{align*}
\begin{pmatrix}3&-7\\-9&3\end{pmatrix}
=&\,\mathbf D-9\mathbf{EF}\\[4mm]
=&\,\begin{pmatrix}d&-9e\\-9&d\end{pmatrix}.
\end{align*} ( 3 − 9 − 7 3 ) = = D − 9 EF ( d − 9 − 9 e d ) . Hence d = 3 d=3 d = 3 e = 7 / 9 e=7/9 e = 7/9
D = ( 3 0 0 3 ) , E = ( 7 / 9 0 0 1 ) , F = ( 0 1 1 0 ) . \begin{align*}
\mathbf D=&\,\boxed{\begin{pmatrix}3&0\\0&3\end{pmatrix}},\\[4mm]
\mathbf E=&\,\boxed{\begin{pmatrix}7/9&0\\0&1\end{pmatrix}},\\[4mm]
\mathbf F=&\,\boxed{\begin{pmatrix}0&1\\1&0\end{pmatrix}}.
\end{align*} D = E = F = ( 3 0 0 3 ) , ( 7/9 0 0 1 ) , ( 0 1 1 0 ) . 