题目
Problem
The matrices A \mathbf{A} A B \mathbf{B} B C \mathbf{C} C
A = ( 1 2 3 2 1 3 3 2 5 ) , B = ( 0 − 2 − 1 3 0 0 ) and C = ( − 2 − 1 1 1 1 3 ) . \begin{align*}
\mathbf{A}=&\,\begin{pmatrix}1&2&3\\2&1&3\\3&2&5\end{pmatrix},\\
\mathbf{B}=&\,\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix}\quad\text{and}\quad\mathbf{C}=\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}.
\end{align*} A = B = 1 2 3 2 1 2 3 3 5 , 0 − 1 0 − 2 3 0 and C = ( − 2 1 − 1 1 1 3 ) . (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 )
(3)
(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)
Let M = ( 3 0 0 1 ) \mathbf{M}=\begin{pmatrix}3&0\\0&1\end{pmatrix} M = ( 3 0 0 1 )
(c) Give full details of the transformation represented by M \mathbf{M} M
(2)
(d) Find the matrix N \mathbf{N} N N M = C A B \mathbf{NM}=\mathbf{CAB} NM = CAB
(3)
题目中文翻译
矩阵 A \mathbf{A} A B \mathbf{B} B C \mathbf{C} C
A = ( 1 2 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&2&3\\2&1&3\\3&2&5\end{pmatrix},\\
\mathbf{B}=&\,\begin{pmatrix}0&-2\\-1&3\\0&0\end{pmatrix},\quad\mathbf{C}=\begin{pmatrix}-2&-1&1\\1&1&3\end{pmatrix}.
\end{align*} A = B = 1 2 3 2 1 2 3 3 5 , 0 − 1 0 − 2 3 0 , C = ( − 2 1 − 1 1 1 3 ) . (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 )
(3)
(b) 求矩阵 C A B \mathbf{CAB} CAB x x x y y y
(5)
设 M = ( 3 0 0 1 ) \mathbf{M}=\begin{pmatrix}3&0\\0&1\end{pmatrix} M = ( 3 0 0 1 )
(c) 完整描述矩阵 M \mathbf{M} M
(2)
(d) 求满足 N M = C A B \mathbf{NM}=\mathbf{CAB} NM = CAB N \mathbf{N} N
(3)
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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先计算AB，尺寸为3乘2，再左乘C得到2乘2矩阵，逐个列出最终元素的点积。
答题过程
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First,
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 . Hence
C A B = ( − 2 − 1 1 1 1 3 ) ( − 2 4 − 1 − 1 − 2 0 ) = ( 4 + 1 − 2 − 8 + 1 + 0 − 2 − 1 − 6 4 − 1 + 0 ) = ( 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+0\\
-2-1-6&4-1+0
\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 + 0 4 − 1 + 0 ) ( 3 − 9 − 7 3 ) .
(b)
解法一
思路
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原点直线y=mx上的点变换后仍需满足相同斜率，再检查竖直线。
答题过程
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A point ( x , m x ) (x,mx) ( x , m x )
( X , Y ) = ( ( 3 − 7 m ) x , ( − 9 + 3 m ) x ) . (X,Y)=((3-7m)x,(-9+3m)x). ( X , Y ) = (( 3 − 7 m ) x , ( − 9 + 3 m ) x ) . For the line to be invariant, Y = m X Y=mX Y = 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, since ( 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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横坐标乘3，纵坐标不变，是沿横轴方向的单向拉伸。
答题过程
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The transformation is a stretch parallel to the x x x 3 3 3 .
(d)
解法一
思路
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M在N的右边，因此应在原等式右乘M的逆；效果是第一列除以3。
答题过程
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Since N M = C A B \mathbf{NM}=\mathbf{CAB} NM = CAB
N = C A B M − 1 = ( 3 − 7 − 9 3 ) ( 1 / 3 0 0 1 ) = ( 1 − 7 − 3 3 ) . \begin{align*}
\mathbf N=&\,\mathbf{CAB}\,\mathbf M^{-1}\\[4mm]
=&\,\begin{pmatrix}3&-7\\-9&3\end{pmatrix}
\begin{pmatrix}1/3&0\\0&1\end{pmatrix}\\[4mm]
=&\,\boxed{\begin{pmatrix}1&-7\\-3&3\end{pmatrix}}.
\end{align*} N = = = CAB M − 1 ( 3 − 9 − 7 3 ) ( 1/3 0 0 1 ) ( 1 − 3 − 7 3 ) . 