题目
Problem
The matrix A \mathbf{A} A
A = ( 1 k − 1 3 1 k 3 1 − 1 ) , \mathbf{A}=\begin{pmatrix}1&k&-1\\3&1&k\\3&1&-1\end{pmatrix}, A = 1 3 3 k 1 1 − 1 k − 1 , where k k k
(a) Find the values of k k k A \mathbf{A} A
(3)
The matrices B \mathbf{B} B C \mathbf{C} C
B = ( 1 0 0 − 1 1 0 ) and C = ( 0 1 0 1 0 1 ) . \mathbf{B}=\begin{pmatrix}1&0\\0&-1\\1&0\end{pmatrix}\quad\text{and}\quad
\mathbf{C}=\begin{pmatrix}0&1&0\\1&0&1\end{pmatrix}. B = 1 0 1 0 − 1 0 and C = ( 0 1 1 0 0 1 ) . (b) Show that C A B = ( k + 3 − 1 2 − k − 1 ) \mathbf{CAB}=\begin{pmatrix}k+3&-1\\2&-k-1\end{pmatrix} CAB = ( k + 3 2 − 1 − k − 1 )
(3)
(c) It is given that k = − 2 k=-2 k = − 2
Show that the transformation in the x x x y y y C A B \mathbf{CAB} CAB
(5)
题目中文翻译
矩阵 A \mathbf{A} A
A = ( 1 k − 1 3 1 k 3 1 − 1 ) , \mathbf{A}=\begin{pmatrix}1&k&-1\\3&1&k\\3&1&-1\end{pmatrix}, A = 1 3 3 k 1 1 − 1 k − 1 , 其中 k k k
(a) 求使 A \mathbf{A} A k k k
(3)
矩阵 B \mathbf{B} B C \mathbf{C} C
B = ( 1 0 0 − 1 1 0 ) 和 C = ( 0 1 0 1 0 1 ) . \mathbf{B}=\begin{pmatrix}1&0\\0&-1\\1&0\end{pmatrix}\quad\text{和}\quad
\mathbf{C}=\begin{pmatrix}0&1&0\\1&0&1\end{pmatrix}. B = 1 0 1 0 − 1 0 和 C = ( 0 1 1 0 0 1 ) . (b) 证明 C A B = ( k + 3 − 1 2 − k − 1 ) \mathbf{CAB}=\begin{pmatrix}k+3&-1\\2&-k-1\end{pmatrix} CAB = ( k + 3 2 − 1 − k − 1 )
(3)
(c) 已知 k = − 2 k=-2 k = − 2
证明 C A B \mathbf{CAB} CAB x x x y y y
(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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矩阵奇异当且仅当行列式为零。沿第一行展开，并保留代数符号。
答题过程
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Expanding along the first row,
det A = 1 ( − 1 − k ) − k ( − 3 − 3 k ) − 1 ( 3 − 3 ) = 3 k 2 + 2 k − 1 = ( 3 k − 1 ) ( k + 1 ) . \begin{align*}
\det\mathbf A
=&\,1(-1-k)-k(-3-3k)-1(3-3) \\[4mm]
=&\,3k^2+2k-1 \\[4mm]
=&\,(3k-1)(k+1).
\end{align*} det A = = = 1 ( − 1 − k ) − k ( − 3 − 3 k ) − 1 ( 3 − 3 ) 3 k 2 + 2 k − 1 ( 3 k − 1 ) ( k + 1 ) . Thus A \mathbf A A
k = 1 3 or k = − 1 . \boxed{k=\frac13\text{ or }k=-1}. k = 3 1 or k = − 1 .
(b)
解法一
思路
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先算AB，再左乘C；展示中间矩阵以自然推出给定结果。
答题过程
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First,
A B = ( 1 − 1 − k 3 + k − 1 3 − 1 − 1 ) = ( 0 − k k + 3 − 1 2 − 1 ) . \begin{align*}
\mathbf A\mathbf B
=&\,\begin{pmatrix}
1-1&-k\\
3+k&-1\\
3-1&-1
\end{pmatrix}\\[4mm]
=&\,\begin{pmatrix}0&-k\\k+3&-1\\2&-1\end{pmatrix}.
\end{align*} AB = = 1 − 1 3 + k 3 − 1 − k − 1 − 1 0 k + 3 2 − k − 1 − 1 . Hence
C A B = ( 0 1 0 1 0 1 ) ( 0 − k k + 3 − 1 2 − 1 ) = ( k + 3 − 1 0 + 2 − k − 1 ) = ( k + 3 − 1 2 − k − 1 ) , \begin{align*}
\mathbf C\mathbf A\mathbf B
=&\,\begin{pmatrix}0&1&0\\1&0&1\end{pmatrix}
\begin{pmatrix}0&-k\\k+3&-1\\2&-1\end{pmatrix} \\[4mm]
=&\,\begin{pmatrix}k+3&-1\\0+2&-k-1\end{pmatrix} \\[4mm]
=&\,\boxed{\begin{pmatrix}k+3&-1\\2&-k-1\end{pmatrix}},
\end{align*} CAB = = = ( 0 1 1 0 0 1 ) 0 k + 3 2 − k − 1 − 1 ( k + 3 0 + 2 − 1 − k − 1 ) ( k + 3 2 − 1 − k − 1 ) , as required.
(c)
解法一
思路
展开
本题说没有不变直线，不能只考虑过原点的直线。对一般y=mx+c比较像点，再单独排除竖直线。
答题过程
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When k = − 2 k=-2 k = − 2
T = ( 1 − 1 2 1 ) . \mathbf T=\begin{pmatrix}1&-1\\2&1\end{pmatrix}. T = ( 1 2 − 1 1 ) . Thus the image coordinates are
( X , Y ) = ( x − y , 2 x + y ) . (X,Y)=(x-y,2x+y). ( X , Y ) = ( x − y , 2 x + y ) . Suppose a non-vertical line y = m x + c y=mx+c y = m x + c Y = m X + c Y=mX+c Y = m X + c
2 x + m x + c = m ( x − m x − c ) + c , 2x+mx+c=m(x-mx-c)+c, 2 x + m x + c = m ( x − m x − c ) + c , so
( m 2 + 2 ) x + m c = 0 (m^2+2)x+mc=0 ( m 2 + 2 ) x + m c = 0 for every real x x x m 2 + 2 = 0 m^2+2=0 m 2 + 2 = 0
A vertical line x = c x=c x = c ( c , y ) (c,y) ( c , y ) X = c − y X=c-y X = c − y y y y
Therefore the transformation has no invariant lines \boxed{\text{no invariant lines}} no invariant lines
