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FP3 2023 June Q2

A Level / Edexcel / FP3

IAL 2023 June Paper · Question 2

题目

Problem

2.

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

M=(2000143−2−3)M=\begin{pmatrix} 2&0&0\\ 0&1&4\\ 3&-2&-3 \end{pmatrix}

(a) Determine M−1M^{-1}

(3)

The transformation represented by MM maps the plane Π1\Pi_1 to the plane Π2\Pi_2

The point (x,y,z)(x,y,z) on Π1\Pi_1 maps to the point (u,v,w)(u,v,w) on Π2\Pi_2

(b) Determine xx, yy and zz in terms of uu, vv and ww as appropriate.

(3)

The plane Π1\Pi_1 has equation

3x−7y+2z=−33x-7y+2z=-3

(c) Find a Cartesian equation for Π2\Pi_2

Give your answer in the form au+bv+cw=dau+bv+cw=d where aa, bb, cc and dd are integers to be determined.

(2)
题目中文翻译

2.

本题必须写出完整的解题步骤。

不接受完全依赖计算器技术的解答。

M=(2000143−2−3)M=\begin{pmatrix} 2&0&0\\ 0&1&4\\ 3&-2&-3 \end{pmatrix}

(a) 求 M−1M^{-1}

(3)

矩阵 MM 所表示的变换把平面 Π1\Pi_1 映射到平面 Π2\Pi_2

平面 Π1\Pi_1 上的点 (x,y,z)(x,y,z) 映射到平面 Π2\Pi_2 上的点 (u,v,w)(u,v,w)

(b) 结合 u,v,wu,v,w，求 x,y,zx,y,z

(3)

平面 Π1\Pi_1 的方程为

3x−7y+2z=−33x-7y+2z=-3

(c) 求 Π2\Pi_2 的笛卡尔方程

答案写成 au+bv+cw=dau+bv+cw=d 的形式，其中 a,b,c,da,b,c,d 为待定整数。

(2)

解答

(a)

解法一：使用伴随矩阵

思路

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使用 M−1=1det⁡Madj⁡MM^{-1}=\frac{1}{\det M}\operatorname{adj}M。先求行列式，再逐项计算余子式并按棋盘符号得到代数余子式矩阵；转置后得到伴随矩阵。

答题过程

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Expanding the determinant along the first row,

det⁡M= 2∣14−2−3∣= 2((−3)−(−8))= 10.\begin{align*} \det M =&\,2 \begin{vmatrix} 1&4\\ -2&-3 \end{vmatrix} \\ =&\,2\bigl((-3)-(-8)\bigr) \\ =&\,10. \end{align*}

The minors are

m11= 1(−3)−4(−2)=5,m12= 0(−3)−4(3)=−12,m13= 0(−2)−1(3)=−3,m21= 0(−3)−0(−2)=0,m22= 2(−3)−0(3)=−6,m23= 2(−2)−0(3)=−4,m31= 0(4)−0(1)=0,m32= 2(4)−0(0)=8,m33= 2(1)−0(0)=2.\begin{align*} m_{11}=&\,1(-3)-4(-2)=5, \\ m_{12}=&\,0(-3)-4(3)=-12, \\ m_{13}=&\,0(-2)-1(3)=-3, \\ m_{21}=&\,0(-3)-0(-2)=0, \\ m_{22}=&\,2(-3)-0(3)=-6, \\ m_{23}=&\,2(-2)-0(3)=-4, \\ m_{31}=&\,0(4)-0(1)=0, \\ m_{32}=&\,2(4)-0(0)=8, \\ m_{33}=&\,2(1)-0(0)=2. \end{align*}

Thus the matrix of minors is

(5−12−30−6−4082).\begin{pmatrix} 5&-12&-3\\ 0&-6&-4\\ 0&8&2 \end{pmatrix}.

Applying the alternating signs gives the cofactor matrix

(512−30−640−82).\begin{pmatrix} 5&12&-3\\ 0&-6&4\\ 0&-8&2 \end{pmatrix}.

Hence

adj⁡M=(50012−6−8−342).\operatorname{adj}M= \begin{pmatrix} 5&0&0\\ 12&-6&-8\\ -3&4&2 \end{pmatrix}.

Therefore,

M−1=110(50012−6−8−342).\boxed{ M^{-1}=\frac1{10} \begin{pmatrix} 5&0&0\\ 12&-6&-8\\ -3&4&2 \end{pmatrix} }.

(b)

解法一：使用逆矩阵

思路

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变换关系是 (u,v,w)T=M(x,y,z)T(u,v,w)^T=M(x,y,z)^T。左乘 (a) 求得的 M−1M^{-1}，即可一次得到 x,y,zx,y,z 关于 u,v,wu,v,w 的表达式。

答题过程

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The transformation gives

(uvw)=M(xyz).\begin{pmatrix}u\\v\\w\end{pmatrix} =M\begin{pmatrix}x\\y\\z\end{pmatrix}.

Thus

(xyz)= M−1(uvw)= (12u65u−35v−45w−310u+25v+15w).\begin{align*} \begin{pmatrix}x\\y\\z\end{pmatrix} =&\,M^{-1} \begin{pmatrix}u\\v\\w\end{pmatrix} \\ =&\, \begin{pmatrix} \frac12u\\ \frac65u-\frac35v-\frac45w\\ -\frac3{10}u+\frac25v+\frac15w \end{pmatrix}. \end{align*}

Therefore,

x=12u,\boxed{x=\frac12u}, y=65u−35v−45w,\boxed{y=\frac65u-\frac35v-\frac45w},

and

z=−310u+25v+15w.\boxed{z=-\frac3{10}u+\frac25v+\frac15w}.

解法二：直接联立变换方程

思路

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直接把矩阵乘法拆成三个标量方程。先由第一式求 xx，再由第二式表示 yy，代入第三式求 zz，这样无需显式使用逆矩阵。

答题过程

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Comparing components in

M(xyz)=(uvw)M\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}u\\v\\w\end{pmatrix}

gives

2x=u,y+4z=v,3x−2y−3z=w.2x=u, \qquad y+4z=v, \qquad 3x-2y-3z=w.

The first two equations give

x=12u,y=v−4z.x=\frac12u, \qquad y=v-4z.

Substituting these into the third equation,

32u−2(v−4z)−3z= w,5z= w−32u+2v.\begin{align*} \frac32u-2(v-4z)-3z=&\,w, \\ 5z=&\,w-\frac32u+2v. \end{align*}

Hence

z=−310u+25v+15w.z=-\frac3{10}u+\frac25v+\frac15w.

Then

y= v−4(−310u+25v+15w)= 65u−35v−45w.\begin{align*} y =&\,v-4\biggl( -\frac3{10}u+\frac25v+\frac15w \biggr) \\ =&\,\frac65u-\frac35v-\frac45w. \end{align*}

Therefore,

x=12u,\boxed{x=\frac12u}, y=65u−35v−45w,\boxed{y=\frac65u-\frac35v-\frac45w},

and

z=−310u+25v+15w.\boxed{z=-\frac3{10}u+\frac25v+\frac15w}.

(c)

解法一

思路

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承接 (b)，把 x,y,zx,y,z 关于 u,v,wu,v,w 的表达式代入 Π1\Pi_1。合并同类项后得到 Π2\Pi_2 的方程，再整体乘以 22，使所有系数成为整数。

答题过程

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Using the expressions from part (b), the equation of Π1\Pi_1 becomes

3(12u) −7(65u−35v−45w) +2(−310u+25v+15w)=−3.\begin{align*} 3\biggl(\frac12u\biggr) &\,-7\biggl( \frac65u-\frac35v-\frac45w \biggr) \\ &\,+2\biggl( -\frac3{10}u+\frac25v+\frac15w \biggr)=-3. \end{align*}

Collecting like terms,

−152u+5v+6w=−3.-\frac{15}{2}u+5v+6w=-3.

Multiplying by 22, a Cartesian equation for Π2\Pi_2 is

−15u+10v+12w=−6.\boxed{-15u+10v+12w=-6}.