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FP3 2020 Oct Q6

A Level / Edexcel / FP3

IAL 2020 Oct Paper · Question 6

题目

Problem

6.

The matrix A\mathbf{A} is given by

A=(1−1111112a)\mathbf{A} = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & a \end{pmatrix}

where a≠1a \ne 1.

(a) Find A−1\mathbf{A}^{-1} in terms of aa.

(4)
B=(1−11111124)\mathbf{B} = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix}

The straight line l1l_1 is mapped onto the straight line l2l_2 by the transformation represented by the matrix B\mathbf{B}.

The equation of l2l_2 is

(r−(12i+4j+6k))×(−6i+2j+3k)=0.\bigl(\mathbf{r} - (12\mathbf{i} + 4\mathbf{j} + 6\mathbf{k})\bigr) \times (-6\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = \mathbf{0}.

(b) Find a vector equation for the line l1l_1.

(4)
题目中文翻译

矩阵 A\mathbf{A} 定义为

A=(1−1111112a)\mathbf{A} = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & a \end{pmatrix}

其中 a≠1a \ne 1。

(a) 用 aa 表示求 A−1\mathbf{A}^{-1}。

(4)
B=(1−11111124)\mathbf{B} = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix}

由矩阵 B\mathbf{B} 表示的变换把直线 l1l_1 映射到直线 l2l_2。

直线 l2l_2 的方程为

(r−(12i+4j+6k))×(−6i+2j+3k)=0\bigl(\mathbf{r} - (12\mathbf{i} + 4\mathbf{j} + 6\mathbf{k})\bigr) \times (-6\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = \mathbf{0}

(b) 求直线 l1l_1 的向量方程。

(4)

解答

(a)

解法一

思路

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使用 A−1=1det⁡Aadj⁡AA^{-1}=\dfrac{1}{\det A}\operatorname{adj}A。先求行列式，再列出代数余子式矩阵并转置。题设 a≠1a\ne1 正好保证分母 2a−22a-2 不为零。

答题过程

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Expanding the determinant,

det⁡A= (a−2)+(a−1)+(2−1)= 2a−2.\begin{align*} \det A =&\,(a-2)+(a-1)+(2-1)\\[4mm] =&\,2a-2. \end{align*}

The nine cofactors are

C11= 1(a)−1(2)=a−2,C12= −(1(a)−1(1))=1−a,C13= 1(2)−1(1)=1,C21= −((−1)(a)−1(2))=a+2,C22= 1(a)−1(1)=a−1,C23= −(1(2)−(−1)(1))=−3,C31= (−1)(1)−1(1)=−2,C32= −(1(1)−1(1))=0,C33= 1(1)−(−1)(1)=2.\begin{align*} C_{11}=&\,1(a)-1(2)=a-2,\\ C_{12}=&\,-\big(1(a)-1(1)\big)=1-a,\\ C_{13}=&\,1(2)-1(1)=1,\\ C_{21}=&\,-\big((-1)(a)-1(2)\big)=a+2,\\ C_{22}=&\,1(a)-1(1)=a-1,\\ C_{23}=&\,-\big(1(2)-(-1)(1)\big)=-3,\\ C_{31}=&\,(-1)(1)-1(1)=-2,\\ C_{32}=&\,-\big(1(1)-1(1)\big)=0,\\ C_{33}=&\,1(1)-(-1)(1)=2. \end{align*}

The cofactor matrix is

C=(a−21−a1a+2a−1−3−202).C=\begin{pmatrix} a-2&1-a&1\\ a+2&a-1&-3\\ -2&0&2 \end{pmatrix}.

Hence

adj⁡A=CT=(a−2a+2−21−aa−101−32).\operatorname{adj}A=C^{\mathsf T} =\begin{pmatrix} a-2&a+2&-2\\ 1-a&a-1&0\\ 1&-3&2 \end{pmatrix}.

Since a≠1a\ne1,

A−1=12a−2(a−2a+2−21−aa−101−32).\boxed{ A^{-1}=\frac{1}{2a-2} \begin{pmatrix} a-2&a+2&-2\\ 1-a&a-1&0\\ 1&-3&2 \end{pmatrix}}.

(b)

解法一

思路

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矩阵 BB 就是 (a) 中取 a=4a=4 的 AA，所以可直接得到 B−1B^{-1}。先把 l2l_2 写成参数式，再用 B−1B^{-1} 同时逆变换直线上的一点与方向向量，即得到原像直线 l1l_1。

答题过程

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Putting a=4a=4 in the result from part (a),

B−1=16(26−2−3301−32).B^{-1}=\frac16 \begin{pmatrix} 2&6&-2\\ -3&3&0\\ 1&-3&2 \end{pmatrix}.

The equation of l2l_2 is equivalent to

r2=(1246)+λ(−623).\mathbf{r}_2 =\begin{pmatrix}12\\4\\6\end{pmatrix} +\lambda\begin{pmatrix}-6\\2\\3\end{pmatrix}.

The corresponding point on l1l_1 is

B−1(1246)= 16(26−2−3301−32)(1246)= 16(2(12)+6(4)−2(6)−3(12)+3(4)12−3(4)+2(6))= 16(36−2412)= (6−42).\begin{align*} B^{-1}\begin{pmatrix}12\\4\\6\end{pmatrix} =&\,\frac16 \begin{pmatrix} 2&6&-2\\ -3&3&0\\ 1&-3&2 \end{pmatrix} \begin{pmatrix}12\\4\\6\end{pmatrix}\\[4mm] =&\,\frac16\begin{pmatrix} 2(12)+6(4)-2(6)\\ -3(12)+3(4)\\ 12-3(4)+2(6) \end{pmatrix}\\ =&\,\frac16\begin{pmatrix}36\\-24\\12\end{pmatrix}\\ =&\,\begin{pmatrix}6\\-4\\2\end{pmatrix}. \end{align*}

The corresponding direction vector is

B−1(−623)= 16(26−2−3301−32)(−623)= 16(2(−6)+6(2)−2(3)−3(−6)+3(2)−6−3(2)+2(3))= 16(−624−6)= (−14−1).\begin{align*} B^{-1}\begin{pmatrix}-6\\2\\3\end{pmatrix} =&\,\frac16 \begin{pmatrix} 2&6&-2\\ -3&3&0\\ 1&-3&2 \end{pmatrix} \begin{pmatrix}-6\\2\\3\end{pmatrix}\\[4mm] =&\,\frac16\begin{pmatrix} 2(-6)+6(2)-2(3)\\ -3(-6)+3(2)\\ -6-3(2)+2(3) \end{pmatrix}\\ =&\,\frac16\begin{pmatrix}-6\\24\\-6\end{pmatrix}\\ =&\,\begin{pmatrix}-1\\4\\-1\end{pmatrix}. \end{align*}

Therefore a vector equation of l1l_1 is

r=(6−42)+λ(−14−1),λ∈R.\boxed{ \mathbf{r} =\begin{pmatrix}6\\-4\\2\end{pmatrix} +\lambda\begin{pmatrix}-1\\4\\-1\end{pmatrix}}, \qquad \lambda\in\mathbb{R}.