题目

Problem

4. (i)

A=\begin{pmatrix}-3&8\\-3&k\end{pmatrix}

where $k$ is a constant

The transformation represented by $A$ transforms triangle $T$ to triangle $T'$

The area of triangle $T'$ is three times the area of triangle $T$

Determine the possible values of $k$

(4)

(ii)

B=\begin{pmatrix}a&-4\\2&3\end{pmatrix}

and

BC=\begin{pmatrix}2&5&1\\1&4&2\end{pmatrix}

where $a$ is a constant

Determine, in terms of $a$ , the matrix $C$

(4)

题目中文翻译

A=\begin{pmatrix}-3&8\\-3&k\end{pmatrix}

其中 $k$ 是常数。

矩阵 $A$ 表示的变换将三角形 $T$ 变成三角形 $T'$ 。

三角形 $T'$ 的面积是三角形 $T$ 面积的三倍。

求 $k$ 的可能值。

(4)

(ii)

B=\begin{pmatrix}a&-4\\2&3\end{pmatrix}

且

BC=\begin{pmatrix}2&5&1\\1&4&2\end{pmatrix}

其中 $a$ 是常数。

用 $a$ 表示矩阵 $C$ 。

(4)

解答

(i)

解法一

思路

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二维线性变换把面积放大为原来的 $|\det A|$ 倍。题目只给面积倍数，没有限定是否发生翻转，因此行列式可为 $3$ 或 $-3$ ，两种情况都要保留。

答题过程

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The area scale factor is $|\det A|$ . Now

\begin{align*} \det A =&\,(-3)k-8(-3)\\ =&\,24-3k. \end{align*}

Since the area of $T'$ is three times the area of $T$ ,

|24-3k|=3.

For the positive determinant,

\begin{align*} 24-3k=&\,3,\\ 3k=&\,21,\\ k=&\,7. \end{align*}

For the negative determinant,

\begin{align*} 24-3k=&\,-3,\\ 3k=&\,27,\\ k=&\,9. \end{align*}

Thus

\boxed{k=7\text{ or }k=9}.

(ii)

解法一

思路

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由 $BC$ 已知，在等式左侧乘 $B^{-1}$ ，得到 $C=B^{-1}(BC)$ 。先求二阶矩阵的逆矩阵，再按正确次序与右侧的 $2\times3$ 矩阵相乘。用 $D=3a+8$ 简记共同分母，逐一计算六个元素。

答题过程

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First,

\begin{align*} &\,\det B\\ =&\,3a-(-4)(2)\\ =&\,3a+8. \end{align*}

Write $D=3a+8$ . Then

\begin{align*} &\,B^{-1}\\ =&\,\frac1D\\ &\,\hspace{2pt}\times \begin{pmatrix} 3&4\\ -2&a \end{pmatrix}. \end{align*}

Multiplying on the left gives

C=B^{-1}(BC).

Let $M$ denote the product of the two matrices below:

\begin{align*} &\,M\\ =&\, \begin{pmatrix} 3&4\\ -2&a \end{pmatrix}\\ &\,\hspace{2pt}\times \begin{pmatrix} 2&5&1\\ 1&4&2 \end{pmatrix}. \end{align*}

Each entry is a row-by-column product:

\begin{align*} m_{11}=&\,3(2)+4(1)\\ =&\,10,\\ m_{12}=&\,3(5)+4(4)\\ =&\,31,\\ m_{13}=&\,3(1)+4(2)\\ =&\,11. \end{align*}

For the second row,

\begin{align*} &\,m_{21}\\ =&\,(-2)(2)+a(1)\\ =&\,a-4,\\ &\,m_{22}\\ =&\,(-2)(5)+a(4)\\ =&\,4a-10,\\ &\,m_{23}\\ =&\,(-2)(1)+a(2)\\ =&\,2a-2. \end{align*}

Therefore $C=D^{-1}M$ , so

C=

\boxed{ \begin{pmatrix} \frac{10}{D}&\frac{31}{D}&\frac{11}{D}\\[2mm] \frac{a-4}{D}&\frac{4a-10}{D}&\frac{2a-2}{D} \end{pmatrix} },

where $D=3a+8$ .

解法二

思路

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也可以直接设出 $C$ 的六个元素。矩阵乘法后，每一列各产生两个一次方程；逐列消元即可求出对应的两个元素，不必显式写出 $B^{-1}$ 。

答题过程

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Let

C=\begin{pmatrix}p&q&r\\s&t&u\end{pmatrix}.

Matrix multiplication gives three pairs of equations:

\begin{align*} ap-4s=&\,2,\\ 2p+3s=&\,1; \end{align*}

\begin{align*} aq-4t=&\,5,\\ 2q+3t=&\,4; \end{align*}

\begin{align*} ar-4u=&\,1,\\ 2r+3u=&\,2. \end{align*}

For each pair, multiply the first equation by $3$ and the second by $4$ , then add. For the first column,

\begin{align*} 3ap-12s=&\,6,\\ 8p+12s=&\,4. \end{align*}

Adding eliminates $s$ :

(3a+8)p=10.

For the second column,

\begin{align*} 3aq-12t=&\,15,\\ 8q+12t=&\,16, \end{align*}

(3a+8)q=31.

For the third column,

\begin{align*} 3ar-12u=&\,3,\\ 8r+12u=&\,8, \end{align*}

(3a+8)r=11.

Writing $D=3a+8$ ,

\begin{align*} p=&\,\frac{10}{D},\\ q=&\,\frac{31}{D},\\ r=&\,\frac{11}{D}. \end{align*}

Substitute into the second equation of each pair. From $2p+3s=1$ ,

\begin{align*} s=&\,\frac{1-2p}{3}\\ =&\,\frac{1-20/D}{3}\\ =&\,\frac{D-20}{3D}\\ =&\,\frac{3a+8-20}{3D}\\ =&\,\frac{3a-12}{3D}\\ =&\,\frac{a-4}{D}. \end{align*}

From $2q+3t=4$ ,

\begin{align*} t=&\,\frac{4-2q}{3}\\ =&\,\frac{4-62/D}{3}\\ =&\,\frac{4D-62}{3D}\\ =&\,\frac{12a+32-62}{3D}\\ =&\,\frac{12a-30}{3D}\\ =&\,\frac{4a-10}{D}. \end{align*}

From $2r+3u=2$ ,

\begin{align*} u=&\,\frac{2-2r}{3}\\ =&\,\frac{2-22/D}{3}\\ =&\,\frac{2D-22}{3D}\\ =&\,\frac{6a+16-22}{3D}\\ =&\,\frac{6a-6}{3D}\\ =&\,\frac{2a-2}{D}. \end{align*}

Therefore

C=

\boxed{ \begin{pmatrix} \frac{10}{D}&\frac{31}{D}&\frac{11}{D}\\[2mm] \frac{a-4}{D}&\frac{4a-10}{D}&\frac{2a-2}{D} \end{pmatrix} },

where $D=3a+8$ .

FP1 2023 May Q4

题目

解答

(i)

解法一

思路

答题过程

(ii)

解法一

思路

答题过程

解法二

思路

答题过程