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CIE 9231 2022 June Paper 12 Q7

A Level / CIE / FP1

CIE 9231 2022 June Paper 12 Paper · Question 7

题目

Problem

The matrix A\mathbf A is given by

A=(1234k6789).\mathbf A=\begin{pmatrix}1&2&3\\4&k&6\\7&8&9\end{pmatrix}.

(a) Find the set of values of kk for which A\mathbf A is non-singular.

(3)

(b) Given that A\mathbf A is non-singular, find, in terms of kk, the entries in the top row of A−1\mathbf A^{-1}.

(4)

(c) Given that B=(100010)\mathbf B=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}, give an example of a matrix C\mathbf C such that BAC=(21k4)\mathbf B\mathbf A\mathbf C=\begin{pmatrix}2&1\\k&4\end{pmatrix}.

(4)

(d) Find the set of values of kk for which the transformation in the xx–yy plane represented by (21k4)\begin{pmatrix}2&1\\k&4\end{pmatrix} has two distinct invariant lines through the origin.

(6)
题目中文翻译

矩阵 A\mathbf A 为

A=(1234k6789).\mathbf A=\begin{pmatrix}1&2&3\\4&k&6\\7&8&9\end{pmatrix}.

(a) 求使 A\mathbf A 为非奇异矩阵的 kk 的取值集合。

(3)

(b) 已知 A\mathbf A 为非奇异矩阵,用 kk 表示 A−1\mathbf A^{-1} 第一行的各元素。

(4)

(c) 已知 B=(100010)\mathbf B=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix},给出一个矩阵 C\mathbf C 的例子,使得 BAC=(21k4)\mathbf B\mathbf A\mathbf C=\begin{pmatrix}2&1\\k&4\end{pmatrix}。

(4)

(d) 求使矩阵 (21k4)\begin{pmatrix}2&1\\k&4\end{pmatrix} 所表示的 xx–yy 平面内的变换具有两条经过原点的不同不变直线的 kk 的取值集合。

(6)
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

展开

按第一行展开行列式,并把“非奇异”转成行列式不为零。

答题过程

展开

Expanding along the first row,

det⁡A= (9k−48)−2(36−42)+3(32−7k)= 9k−48+12+96−21k= 60−12k.\begin{align*} \det\mathbf A=&\,(9k-48)-2(36-42)+3(32-7k)\\[4mm] =&\,9k-48+12+96-21k\\[4mm] =&\,60-12k. \end{align*}

Thus A\mathbf A is non-singular when 60−12k≠060-12k\ne0, giving k≠5\boxed{k\ne5}.

(b)

解法一

思路

展开

逆矩阵第一行来自余因子矩阵的第一列;注意转置与第二个余因子的负号。

答题过程

展开

Since A−1=(det⁡A)−1(cof⁡A)T\mathbf A^{-1}=(\det\mathbf A)^{-1}(\operatorname{cof}\mathbf A)^{\mathsf T}, its top row uses the cofactors C11,C21,C31C_{11},C_{21},C_{31}:

C11= 9k−48,C21= −(18−24)=6,C31= 12−3k.\begin{align*} C_{11}=&\,9k-48,\\[4mm] C_{21}=&\,-(18-24)=6,\\[4mm] C_{31}=&\,12-3k. \end{align*}

Hence the required row is

160−12k(9k−48612−3k),k≠5.\boxed{\frac1{60-12k}\begin{pmatrix}9k-48&6&12-3k\end{pmatrix}},\qquad k\ne5.

(c)

解法一

思路

展开

左乘 BB 选出前两行;右边只需把所得矩阵的前两列交换,并舍去第三列。直接写出实现这个操作的矩阵并验算。

答题过程

展开

First,

BA=(1234k6).\mathbf B\mathbf A=\begin{pmatrix}1&2&3\\4&k&6\end{pmatrix}.

Choose

C=(011000).\boxed{\mathbf C=\begin{pmatrix}0&1\\1&0\\0&0\end{pmatrix}}.

Then

BAC= (1234k6)(011000)= (21k4),\begin{align*} \mathbf B\mathbf A\mathbf C =&\,\begin{pmatrix}1&2&3\\4&k&6\end{pmatrix} \begin{pmatrix}0&1\\1&0\\0&0\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}2&1\\k&4\end{pmatrix}, \end{align*}

as required.

(d)

解法一

思路

展开

先求不变方向的两个实斜率,再排除其中一条直线被压成原点的情形;两个不同的实斜率本身还不够。

答题过程

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A nonvertical line through the origin has equation y=mxy=mx. Under the transformation,

(x′y′)= (21k4)(xmx)= ((2+m)x(k+4m)x).\begin{align*} \begin{pmatrix}x'\\y'\end{pmatrix} =&\,\begin{pmatrix}2&1\\k&4\end{pmatrix}\begin{pmatrix}x\\mx\end{pmatrix}\\[4mm] =&\,\begin{pmatrix}(2+m)x\\(k+4m)x\end{pmatrix}. \end{align*}

For its image to be the same line, k+4m=m(2+m)k+4m=m(2+m), with 2+m≠02+m\ne0. Thus

m2−2m−k=0.m^2-2m-k=0.

Two distinct real slopes require

(−2)2−4(1)(−k)>0⟹k>−1.(-2)^2-4(1)(-k)>0\quad\Longrightarrow\quad k>-1.

Their values are m=1±1+km=1\pm\sqrt{1+k}, and the scale factors along these directions are

2+m=3±1+k.2+m=3\pm\sqrt{1+k}.

For k>−1k>-1, one of these vanishes precisely when k=8k=8. At k=8k=8, the direction m=−2m=-2 is mapped entirely to the origin, so it does not give a line mapped onto itself. This value must be excluded.

The vertical line is not invariant: (0,y)(0,y) maps to (y,4y)(y,4y), which is not vertical. Therefore no additional invariant line was omitted, and the required set is

k>−1,k≠8.\boxed{k>-1,\qquad k\ne8}.