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FP3 2025 June Q3

A Level / Edexcel / FP3

IAL 2025 June Paper · Question 3

题目

Problem

3.

M= (1−143aba1b)\begin{align*} M =&\, \begin{pmatrix} 1 & -1 & 4 \\ 3 & a & b \\ a & 1 & b \end{pmatrix} \end{align*}

where aa and bb are constants.

Given that i+2j+k\mathbf{i} + 2\mathbf{j} + \mathbf{k} is an eigenvector of MM,

(a) determine the corresponding eigenvalue.

(2)

(b) Hence determine the value of aa and the value of bb.

(2)

(c) Determine

(i) the other eigenvalues of MM,

(3)

(ii) eigenvectors which correspond to these eigenvalues.

(3)
题目中文翻译 M= (1−143aba1b)\begin{align*} M =&\, \begin{pmatrix} 1 & -1 & 4 \\ 3 & a & b \\ a & 1 & b \end{pmatrix} \end{align*}

其中 aa 和 bb 为常数。

已知 i+2j+k\mathbf{i} + 2\mathbf{j} + \mathbf{k} 是矩阵 MM 的一个特征向量，

(a) 求对应的特征值。

(2)

(b) 从而求 aa 和 bb 的值。

(2)

(c) 求

(i) MM 的其他特征值，

(3)

(ii) 与这些特征值对应的特征向量。

(3)

解答

(a)

解法一

思路

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设给定特征向量为 v=(1,2,1)T\mathbf v=(1,2,1)^T。由 Mv=λvM\mathbf v=\lambda\mathbf v，只比较第一行即可求出 λ\lambda，因为第一行不含未知常数 a,ba,b。

答题过程

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Let

v=(121).\mathbf v= \begin{pmatrix} 1\\ 2\\ 1 \end{pmatrix}.

Since Mv=λvM\mathbf v=\lambda\mathbf v, comparison of the first components gives

1(1)+(−1)(2)+4(1)=λ(1).1(1)+(-1)(2)+4(1)=\lambda(1).

Therefore,

λ=3.\boxed{\lambda=3}.

(b)

解法一

思路

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承接 (a) 的 λ=3\lambda=3，分别比较 Mv=3vM\mathbf v=3\mathbf v 的第二、第三个分量，得到关于 a,ba,b 的两个一次方程并联立求解。

答题过程

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Using λ=3\lambda=3, comparison of the second components gives

3+2a+b=6.3+2a+b=6.

Comparison of the third components gives

a+2+b=3.a+2+b=3.

Hence

2a+b= 3,a+b= 1.\begin{align*} 2a+b=&\,3,\\ a+b=&\,1. \end{align*}

Subtracting the second equation from the first,

a=2.a=2.

Then 2+b=12+b=1, so

a=2,b=−1.\boxed{a=2,\qquad b=-1}.

(c)(i)

解法一

思路

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把 (b) 的 a,ba,b 代回矩阵，再求特征方程。为使三次式首项为正，展开 det⁡(λI−M)\det(\lambda I-M)；已知的特征值 33 应成为一个因子，余下二次因式给出另外两个特征值。

答题过程

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With a=2a=2 and b=−1b=-1,

M=(1−1432−121−1).M= \begin{pmatrix} 1&-1&4\\ 3&2&-1\\ 2&1&-1 \end{pmatrix}.

The characteristic equation is

det⁡(λI−M)=0.\det(\lambda I-M)=0.

Expanding along the first row,

det⁡(λI−M)= ∣λ−11−4−3λ−21−2−1λ+1∣= (λ−1)((λ−2)(λ+1)+1) +(3λ+1)−4(2λ−1)= λ3−2λ2−5λ+6= (λ−3)(λ2+λ−2)= (λ−3)(λ−1)(λ+2).\begin{align*} \det(\lambda I-M) =&\, \begin{vmatrix} \lambda-1&1&-4\\ -3&\lambda-2&1\\ -2&-1&\lambda+1 \end{vmatrix}\\ =&\,(\lambda-1) \big((\lambda-2)(\lambda+1)+1\big)\\ &\,\hspace{2pt}+(3\lambda+1)-4(2\lambda-1)\\ =&\,\lambda^3-2\lambda^2-5\lambda+6\\ =&\,(\lambda-3)(\lambda^2+\lambda-2)\\ =&\,(\lambda-3)(\lambda-1)(\lambda+2). \end{align*}

Therefore, besides the known eigenvalue 33, the other eigenvalues are

λ=1andλ=−2.\boxed{\lambda=1\quad\text{and}\quad\lambda=-2}.

(c)(ii)

解法一

思路

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对 (c)(i) 的每个新特征值分别解 (M−λI)x=0(M-\lambda I)\mathbf x=\mathbf0。每组齐次方程都有一个自由变量，取便于计算的值即可得到对应特征向量；任意非零倍数都同样正确。

答题过程

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Let an eigenvector be

(xyz).\begin{pmatrix} x\\ y\\ z \end{pmatrix}.

For λ=1\lambda=1, the first two equations from (M−I)x=0(M-I)\mathbf x=\mathbf0 are

−y+4z= 0,3x+y−z= 0.\begin{align*} -y+4z=&\,0,\\ 3x+y-z=&\,0. \end{align*}

Thus y=4zy=4z and x=−zx=-z. Taking z=1z=1 gives the eigenvector

(−141)(λ=1).\boxed{ \begin{pmatrix} -1\\ 4\\ 1 \end{pmatrix}} \qquad(\lambda=1).

For λ=−2\lambda=-2, the first two equations from (M+2I)x=0(M+2I)\mathbf x=\mathbf0 are

3x−y+4z= 0,3x+4y−z= 0.\begin{align*} 3x-y+4z=&\,0,\\ 3x+4y-z=&\,0. \end{align*}

The first equation gives y=3x+4zy=3x+4z. Substituting this into the second,

15x+15z=0,15x+15z=0,

so x=−zx=-z and y=zy=z. Taking z=1z=1 gives the eigenvector

(−111)(λ=−2).\boxed{ \begin{pmatrix} -1\\ 1\\ 1 \end{pmatrix}} \qquad(\lambda=-2).

解法二

思路

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特征向量必须与 M−λIM-\lambda I 的每一行都垂直。因此，可以先将其中两个不平行的行向量作叉积，再检查所得非零向量是否也与第三行垂直，便得到对应特征向量。

答题过程

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An eigenvector for λ\lambda satisfies (M−λI)v=0(M-\lambda I)\mathbf v=\mathbf0, so it is perpendicular to every row of M−λIM-\lambda I.

For λ=1\lambda=1, the first two rows are (0,−1,4)(0,-1,4) and (3,1,−1)(3,1,-1). Their vector product is

∣ijk0−1431−1∣= ((−1)(−1)−4(1))i −(0(−1)−4(3))j +(0(1)−(−1)(3))k= −3i+12j+3k.\begin{align*} \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 0&-1&4\\ 3&1&-1 \end{vmatrix} =&\,\big((-1)(-1)-4(1)\big)\mathbf i\\[2mm] &\,\hspace{2pt}-\big(0(-1)-4(3)\big)\mathbf j\\[2mm] &\,\hspace{4pt}+\big(0(1)-(-1)(3)\big)\mathbf k\\[2mm] =&\,-3\mathbf i+12\mathbf j+3\mathbf k. \end{align*}

Taking one third of this vector gives v1=(−1,4,1)T\mathbf v_1=(-1,4,1)^T. The remaining row is (2,1,−2)(2,1,-2), and

(2,1,−2)⋅(−1,4,1)= −2+4−2=0.\begin{align*} (2,1,-2)\mathbin{\boldsymbol\cdot}(-1,4,1) =&\,-2+4-2=0. \end{align*}

Thus

v1=(−141)(λ=1).\boxed{\mathbf v_1=\begin{pmatrix}-1\\4\\1\end{pmatrix}} \qquad(\lambda=1).

For λ=−2\lambda=-2, the first two rows of M+2IM+2I are (3,−1,4)(3,-1,4) and (3,4,−1)(3,4,-1). Hence

∣ijk3−1434−1∣= ((−1)(−1)−4(4))i −(3(−1)−4(3))j +(3(4)−(−1)(3))k= −15i+15j+15k.\begin{align*} \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 3&-1&4\\ 3&4&-1 \end{vmatrix} =&\,\big((-1)(-1)-4(4)\big)\mathbf i\\[2mm] &\,\hspace{2pt}-\big(3(-1)-4(3)\big)\mathbf j\\[2mm] &\,\hspace{4pt}+\big(3(4)-(-1)(3)\big)\mathbf k\\[2mm] =&\,-15\mathbf i+15\mathbf j+15\mathbf k. \end{align*}

Taking one fifteenth gives v−2=(−1,1,1)T\mathbf v_{-2}=(-1,1,1)^T. The third row is (2,1,1)(2,1,1), and

(2,1,1)⋅(−1,1,1)= −2+1+1=0.\begin{align*} (2,1,1)\mathbin{\boldsymbol\cdot}(-1,1,1) =&\,-2+1+1=0. \end{align*}

Therefore,

v−2=(−111)(λ=−2).\boxed{\mathbf v_{-2}=\begin{pmatrix}-1\\1\\1\end{pmatrix}} \qquad(\lambda=-2).

Any non-zero scalar multiple of either vector is also a corresponding eigenvector.