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FP1 2024 Jan Q1

A Level / Edexcel / FP1

IAL 2024 Jan Paper · Question 1

题目

Problem

 M= (2k+1kk+7k+4)\begin{align*} &\,\mathbf M\\ =&\,\begin{pmatrix}2k+1&k\\k+7&k+4\end{pmatrix} \end{align*}

where kk is a constant

(a) Show that M\mathbf{M} is non-singular for all real values of kk.

(3)

(b) Determine M−1\mathbf{M}^{-1} in terms of kk.

(2)
题目中文翻译

已知矩阵

 M= (2k+1kk+7k+4)\begin{align*} &\,\mathbf M\\ =&\,\begin{pmatrix}2k+1&k\\k+7&k+4\end{pmatrix} \end{align*}

其中 kk 为常数。

(a) 证明:对于所有实数 kk,矩阵 M\mathbf{M} 都是非奇异矩阵(non-singular)。

(3)

(b) 用含 kk 的式子表示 M−1\mathbf{M}^{-1}。

(2)

解答

(a)

解法一

思路

展开

矩阵非奇异等价于行列式不为零。先求行列式,再配方为平方项加正常数,从而说明它对每个实数 kk 都严格为正。

答题过程

展开  det⁡M= (2k+1)(k+4) −k(k+7)= 2k2+9k+4 −k2−7k= k2+2k+4= (k+1)2+3.\begin{align*} &\,\det\mathbf M\\ =&\,(2k+1)(k+4)\\ &\,\hspace{2pt}-k(k+7)\\ =&\,2k^2+9k+4\\ &\,\hspace{2pt}-k^2-7k\\ =&\,k^2+2k+4\\ =&\,(k+1)^2+3. \end{align*}

For all real values of kk,

(k+1)2+3⩾3>0.(k+1)^2+3\geqslant3>0.

Therefore, det⁡M≠0\det\mathbf M\ne0, so

the matrix M\mathbf M is non-singular for all real values of kk.

解法二

思路

展开

把行列式视为关于 kk 的二次式。它的判别式为负,因此没有实数零点;又因二次项系数为正,所以行列式始终为正。

答题过程

展开

As in Method 1,

det⁡M=k2+2k+4.\det\mathbf M=k^2+2k+4.

The discriminant of this quadratic is

22−4(1)(4)=−12<0.2^2-4(1)(4)=-12<0.

Hence det⁡M\det\mathbf M has no real zeros. Since its leading coefficient is positive,

det⁡M>0\det\mathbf M>0

for all real kk. Therefore,

the matrix M\mathbf M is non-singular for all real values of kk.

解法三

思路

展开

也可以用最小值证明。对行列式关于 kk 求导,找出唯一驻点,再以二阶导数确认这是最小值;最小值仍大于零。

答题过程

展开

Let

D(k)= det⁡M= k2+2k+4.\begin{align*} D(k)=&\,\det\mathbf M\\ =&\,k^2+2k+4. \end{align*}

Then

D′(k)=2k+2.D'(k)=2k+2.

At a stationary point,

2k+2=0,2k+2=0,

so k=−1k=-1. Since

D′′(k)=2>0,D''(k)=2>0,

this gives the minimum value of DD. Moreover,

 D(−1)= (−1)2+2(−1)+4= 1−2+4= 3>0.\begin{align*} &\,D(-1)\\ =&\,(-1)^2+2(-1)+4\\ =&\,1-2+4\\ =&\,3>0. \end{align*}

Thus det⁡M\det\mathbf M is never zero, and hence

the matrix M\mathbf M is non-singular for all real values of kk.

(b)

解法一

思路

展开

使用二阶矩阵的逆矩阵公式:交换主对角元素,并把副对角元素变号,再除以 (a) 求得的行列式。

答题过程

展开

Using the inverse formula for a 2×22\times2 matrix and writing −(k+7)=−k−7-(k+7)=-k-7, we obtain

 M−1= 1k2+2k+4 ×(k+4−k−k−72k+1).\begin{align*} &\,\mathbf M^{-1}\\ =&\,\frac1{k^2+2k+4}\\ &\,\hspace{2pt}\times \begin{pmatrix} k+4&-k\\ -k-7&2k+1 \end{pmatrix}. \end{align*}

for all real kk.