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CIE 9231 2025 November Paper 12 Q4

A Level / CIE / FP1

CIE 9231 2025 Nov Paper 12 Paper · Question 4

题目

Problem

Let kk and mm be non-zero constants. The matrices A\mathbf{A}, B\mathbf{B} and C\mathbf{C} are given by

A= (01−1111),B= (k00m)andC= (2−11112).\begin{align*} \mathbf{A}=&\,\begin{pmatrix}0&1\\-1&1\\1&1\end{pmatrix},\\ \mathbf{B}=&\,\begin{pmatrix}k&0\\0&m\end{pmatrix}\quad\text{and}\\ \mathbf{C}=&\,\begin{pmatrix}2&-1&1\\1&1&2\end{pmatrix}. \end{align*}

(a) Give full details of the geometrical transformation in the xx-yy plane represented by the matrix B\mathbf{B} in each of the following cases.

(i) m=1m=1

(2)

(ii) m=km=k

(2)

(b) Show that the matrix ABC\mathbf{ABC} is singular.

(6)

题目中文翻译

设 kk 和 mm 为非零常数。矩阵 A\mathbf{A}、B\mathbf{B} 和 C\mathbf{C} 为

A= (01−1111),B= (k00m)和C= (2−11112).\begin{align*} \mathbf{A}=&\,\begin{pmatrix}0&1\\-1&1\\1&1\end{pmatrix},\\ \mathbf{B}=&\,\begin{pmatrix}k&0\\0&m\end{pmatrix}\quad\text{和}\\ \mathbf{C}=&\,\begin{pmatrix}2&-1&1\\1&1&2\end{pmatrix}. \end{align*}

(a) 对以下每种情况，完整描述矩阵 B\mathbf{B} 所表示的 xx-yy 平面上的几何变换。

(i) m=1m=1

(2)

(ii) m=km=k

(2)

(b) 证明矩阵 ABC\mathbf{ABC} 是奇异矩阵。

(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)(i)

解法一

思路

展开

m为1时y不变，x乘以k，是沿横轴方向的伸缩。

答题过程

展开

When m=1m=1, B=(k001)\mathbf B=\begin{pmatrix}k&0\\0&1\end{pmatrix}. It represents a stretch parallel to the xx-axis, with scale factor kk.

(a)(ii)

解法一

思路

展开

m等于k时两个坐标同时乘以k，是以原点为中心的放大变换。

答题过程

展开

When m=km=k, B=kI\mathbf B=k\mathbf I. It represents an enlargement with centre the origin and scale factor kk.

(b)

解法一

思路

展开

先展示两步矩阵乘法，再沿第一行展开行列式。把三个二阶子式分别化简，避免一次展开过长。

答题过程

展开

First,

AB=(0m−kmkm).\mathbf A\mathbf B=\begin{pmatrix}0&m\\-k&m\\k&m\end{pmatrix}.

Multiplying by C\mathbf C,

M=ABC=(mm2m−2k+mk+m−k+2m2k+m−k+mk+2m).\mathbf M=\mathbf A\mathbf B\mathbf C =\begin{pmatrix} m&m&2m\\ -2k+m&k+m&-k+2m\\ 2k+m&-k+m&k+2m \end{pmatrix}.

The three minors needed for expansion along the first row are

D1= (k+m)(k+2m)−(−k+2m)(−k+m)= 6km,D2= (−2k+m)(k+2m)−(−k+2m)(2k+m)= −6km,D3= (−2k+m)(−k+m)−(k+m)(2k+m)= −6km.\begin{align*} D_1=&\,(k+m)(k+2m)-(-k+2m)(-k+m) \\[4mm] =&\,6km, \\[4mm] D_2=&\,(-2k+m)(k+2m)-(-k+2m)(2k+m) \\[4mm] =&\,-6km, \\[4mm] D_3=&\,(-2k+m)(-k+m)-(k+m)(2k+m) \\[4mm] =&\,-6km. \end{align*}

Hence

det⁡M= mD1−mD2+2mD3= 6km2+6km2−12km2=0.\begin{align*} \det\mathbf M=&\,mD_1-mD_2+2mD_3\\[4mm] =&\,6km^2+6km^2-12km^2=0. \end{align*}

Therefore ABC is singular\boxed{\mathbf A\mathbf B\mathbf C\text{ is singular}}.

解法二

思路

展开

另一种简洁验证是观察行之间的线性关系，并用行运算产生零行。

答题过程

展开

We have

AB=(0m−kmkm).\mathbf A\mathbf B=\begin{pmatrix}0&m\\-k&m\\k&m\end{pmatrix}.

Its rows satisfy R2+R3=2R1R_2+R_3=2R_1. Right multiplication by C\mathbf C preserves this relation, so the rows of ABC\mathbf A\mathbf B\mathbf C also satisfy R2+R3=2R1R_2+R_3=2R_1.

The row operation R3↦R3+R2−2R1R_3\mapsto R_3+R_2-2R_1 therefore produces a zero row without changing the determinant. Thus det⁡(ABC)=0\det(\mathbf A\mathbf B\mathbf C)=0, proving that the matrix is singular.