由定义 \(\lim _{x \to 0} \frac{\ln ^{\alpha}(1+2 x)}{x}=\lim _{x \to 0} \frac{(2 x)^{\alpha}}{x}=\lim _{x \to 0} 2^{\alpha} x^{\alpha-1}=0\) 所以 \(\alpha-1>0\) ,故 \(\alpha>1\) .当 \(x \to 0^{+}\) 时， \((1-\cos x)^{\frac{1}{\alpha}} \sim \frac{x^{\frac{2}{\alpha}}}{2^{\frac{1}{\alpha}}}\) ，由题意 \(\frac{2}{\alpha}-1>0\) ，即 \(\alpha<2\) ，故选B。

\(\lim _{x \to \infty} \frac{x+\sin \frac{1}{x}}{x}=\lim _{x \to \infty} 1+\lim _{x \to \infty} \frac{\sin \frac{1}{x}}{x}=1+0=1\) ，\(\lim _{x \to \infty}[x+\sin \frac{1}{x}-x]=\lim _{x \to \infty} \sin \frac{1}{x}=0\) ，所以 \(y=x+\sin \frac{1}{x}\) 存在斜渐近线 \(y=x\) ，故选C。

令\(F(x)=g(x)-f(x)=f(0)(1-x)+f(1)x-f(x)\)，则 \(F(0)=F(1)=0\)，\(F'(x)=-f(0)+f(1)-f'(x)\)，\(F''(x)=-f''(x)\)。若 \(f^{\prime \prime}(x) \geq0\) ，则 \(F^{\prime \prime}(x) \leq0\) ，\(F(x)\) 在[0,1]上为凸的，又 \(F(0)=F(1)=0\) ，所以当 \(x \in[0,1]\) 时，\(F(x) \geq0\) ，从而 \(g(x) \geq f(x)\) ，故选D。

\(\left.\frac{d y}{d x}\right|_{t=1}=\left.\frac{2 t+4}{2 t}\right|_{t=1}=3\)，\(\left.\frac{d^{2} y}{d x^{2}}\right|_{t=1}=\left.\frac{d y'}{d x}\right|_{t=1}=\left.\frac{-\frac{2}{t^{2}}}{2 t}\right|_{t=1}=-1\)，\(k=\frac{\left|y''\right|}{\left(1+y^{\prime 2}\right)^{\frac{3}{2}}}=\frac{1}{(1+9)^{\frac{3}{2}}}\)，\(\therefore R=\frac{1}{k}=10 \sqrt{10}\)，故选C。

因为 \(\frac{f(x)}{x}=f'(\xi)=\frac{1}{1+\xi^{2}}\) ，所以 \(\xi^{2}=\frac{x-f(x)}{f(x)}\) ，\(\lim _{x \to 0} \frac{\xi^{2}}{x^{2}}=\lim _{x \to 0} \frac{x-f(x)}{x^{2} f(x)}=\lim _{x \to 0} \frac{x-\arctan x}{x^{2} \arctan x}=\lim _{x \to 0} \frac{1-\frac{1}{1+x^{2}}}{3 x^{2}}=\frac{1}{3}\) ，故选D。

记 \(A=\frac{\partial^{2} u}{\partial x^{2}}\) ，\(B=\frac{\partial^{2} u}{\partial x \partial y}\) ，\(C=\frac{\partial^{2} u}{\partial y^{2}}\) ，由\(B \neq0\) 且\(A+C=0\) ，则 \(\Delta=A C-B^{2}<0\) ，所以 \(u(x, y)\) 在 D 内无极值，则极值在边界处取得，故选A。

由行列式的展开定理展开第一列，\(\begin{aligned} \left|\begin{array}{llll} 0 & a & b & 0 \\ a & 0 & 0 & b \\ 0 & c & d & 0 \\ c & 0 & 0 & d \end{array}\right| & =-a\left|\begin{array}{lll} a & b & 0 \\ c & d & 0 \\ 0 & 0 & d \end{array}\right|-c\left|\begin{array}{lll} a & b & 0 \\ 0 & 0 & b \\ c & d & 0 \end{array}\right| \\ & =-a d(a d-b c)+b c(a d-b c) \\ & =-(a d-b c)^{2} . \end{aligned}\)，故选B。

\(\left(\alpha_{1}+k \alpha_{3},\alpha_{2}+l \alpha_{3}\right)=\left(\begin{array}{lll}\alpha_{1} & \alpha_{2} & \alpha_{3}\end{array}\right)\left(\begin{array}{ll}1 & 0 \\ 0 & 1 \\ k & l\end{array}\right)\)。若\(\alpha_{1}\)，\(\alpha_{2}\)，\(\alpha_{3}\)线性无关，则矩阵秩\(r(\alpha_{1}+k \alpha_{3},\alpha_{2}+l \alpha_{3})=2\)，故向量组线性无关；举反例，令\(\alpha_{3}=0\)，则\(\alpha_{1}\)，\(\alpha_{2}\)线性无关，但\(\alpha_{1}\)，\(\alpha_{2}\)，\(\alpha_{3}\)线性相关。综上所述，对任意常数k,l，向量组线性无关是向量组线性无关的必要非充分条件，故选A。

【解析】

\[\begin{aligned} & \int_{-\infty}^{1} \frac{1}{x^{2}+2 x+5} d x=\int_{-\infty}^{1} \frac{1}{(x+1)^{2}+4} d x=\left.\frac{1}{2} \arctan \frac{x+1}{2}\right|_{-\infty} ^{1} \\ & =\frac{1}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{2}\right)\right]=\frac{3}{8} \pi \end{aligned}\]

【解析】 \(f'(x)=2(x-1)\) ，\(x \in[0,2]\) 且为偶函数

\[则 f'(x)=2(-x-1), x \in[-2,0]\]

又 \(f(x)=-x^{2}-2 x+c\) 且为奇函数，故 \(c=0\)

\[\therefore f(x)=-x^{2}-2 x, x \in[-2,0]\]

又: \(f(x)\) 的周期为4，\(\therefore f(7)=f(-1)=1\)

【解析】对 \(e^{2 y z}+x+y^{2}+z=\frac{7}{4}\) 方程两边同时对 x、y 求偏导

\[\left\{\begin{array}{l} e^{2 y z} \cdot 2 y \cdot \frac{\partial z}{\partial x}+1+\frac{\partial z}{\partial x}=0 \\ e^{2 y z}\left(2 z+2 y \frac{\partial z}{\partial y}\right)+2 y+\frac{\partial z}{\partial y}=0 \end{array}\right.\]

当 \(x=\frac{1}{2}\) ，\(y=\frac{1}{2}\) 时，\(z=0\)

\[故 \left.\frac{\partial z}{\partial x}\right|_{\left(\frac{1}{2}, \frac{1}{2}\right)}=-\frac{1}{2},\left.\frac{\partial z}{\partial y}\right|_{\left(\frac{1}{2}, \frac{1}{2}\right)}=-\frac{1}{2}\]

\[故 \left.d z\right|_{\left(\frac{1}{2}, \frac{1}{2}\right)}=-\frac{1}{2} d x+\left(-\frac{1}{2}\right) d y=-\frac{1}{2}(d x+d y)\]

【解析】由直角坐标和极坐标的关系 \(x=r \cos \theta=\theta \cos \theta\) ，\(y=r \sin \theta=\theta \sin \theta\) ，于是 \((r, \theta)=\left(\frac{\pi}{2}, \frac{\pi}{2}\right)\) 对应于 \((x, y)=\left(0, \frac{\pi}{2}\right)\)

\[切线斜率 \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\left.\frac{\theta \cos \theta+\sin \theta}{\cos \theta-\theta \sin \theta} \therefore \frac{d y}{d x}\right|_{\left(0, \frac{\pi}{2}\right)}=-\frac{2}{\pi}\]

所以切线方程为 \(y-\frac{\pi}{2}=-\frac{2}{\pi}(x-0)\)

\[即 y=-\frac{2}{\pi} x+\frac{\pi}{2}\]

【解析】\(\bar{x}=\frac{\int_{0}^{1} x \rho(x) d x}{\int_{0}^{1} \rho(x) d x}\)

\[\int_{0}^{1} \rho(x) d x=\int_{0}^{1}\left(-x^{2}+2 x+1\right) d x=\left.\left(-\frac{x^{3}}{3}+x^{2}+x\right)\right|_{0} ^{1}=\frac{5}{3}\]

\[\int_{0}^{1} x \rho(x) d x=\int_{0}^{1} x\left(-x^{2}+2 x+1\right) d x=\left.\left(-\frac{x^{4}}{4}+\frac{2}{3} x^{3}+\frac{x^{2}}{2}\right)\right|_{0} ^{1}=\frac{11}{12}\]

\[\overline{x}=\frac{\frac{11}{12}}{\frac{5}{3}}=\frac{11}{20}\]

【解析】配方法：\(f(x_{1}, x_{2}, x_{3})=(x_{1}+a x_{3})^{2}-a^{2} x_{3}^{2}-(x_{2}-2 x_{3})^{2}+4 x_{3}^{2}\)

由于二次型负惯性指数为1，所以 \(4-a^{2} \geq 0\) ，故 \(-2 \leq a \leq 2\) 。

\(\lim _{x \to+\infty} \frac{\int_{1}^{x}\left[t^{2}\left(e^{\frac{1}{t}}-1\right)-t\right] d t}{x^{2} \ln \left(1+\frac{1}{x}\right)}=\lim _{x \to+\infty} \frac{\int_{1}^{x}\left[t^{2}\left(e^{\frac{1}{t}}-1\right)-t\right] d t}{x^{2} \cdot \frac{1}{x}} =\lim _{x \to+\infty}\left[x^{2}\left(e^{\frac{1}{x}}-1\right)-x\right] =lim _{x \to \infty^{+}} \frac{e^{t}-1}{2 t}=lim _{t \to 0^{+}} \frac{t}{2 t}=\frac{1}{2}\)

由 \(x^{2}+y^{2} y'=1-y'\) ，得 \(\left(y^{2}+1\right) y'=1-x^{2}\)

此时上面方程为变量可分离方程，解的通解为 \(\frac{1}{3} y^{3}+y=x-\frac{1}{3} x^{3}+c\)

由 \(y(2)=0\) 得 \(c=\frac{2}{3}\)

又由(1)可得 \(y'(x)=\frac{1-x^{2}}{y^{2}+1}\)

当 \(y'(x)=0\) 时，\(x= \pm 1\) ，且有：

\(x<-1, y'(x)<0\)

\(-10\)

\(x>1, y'(x)<0\)

所以 \(y(x)\) 在 \(x=-1\) 处取得极小值，在 \(x=1\) 处取得极大值

\(y(-1)=0, y(1)=1\)

即：\(y(x)\) 的极大值为1，极小值为0

D关于 \(y=x\) 对称，满足轮换对称性，则：

\begin{align*}
& \iint_{D} \frac{x \sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{x+y} \, dx \, dy = \iint_{D} \frac{y \sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{x+y} \, dx \, dy \\
& \therefore I = \iint_{D} \frac{x \sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{x+y} \, dx \, dy = \frac{1}{2} \iint_{D} \left[ \frac{x \sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{x+y} + \frac{y \sin \left(\pi \sqrt{x^{2}+y^{2}}\right)}{x+y} \right] \, dx \, dy \\
& = \frac{1}{2} \iint_{D} \sin \left(\pi \sqrt{x^{2}+y^{2}}\right) \, dx \, dy \\
& = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} d\theta \int_{1}^{2} \sin(\pi r) \cdot r \, dr \\
& = \frac{\pi}{4} \left( -\frac{1}{\pi} \right) \int_{1}^{2} r \, d(\cos \pi r) \\
& = -\frac{1}{4} \left[ \left. \cos \pi r \cdot r \right|_{1}^{2} - \int_{1}^{2} \cos \pi r \, dr \right] \\
& = -\frac{1}{4} \left[ 2 + 1 - \left. \frac{1}{\pi} \sin \pi r \right|_{1}^{2} \right] \\
& = -\frac{3}{4}
\end{align*}

【解】$\frac{\partial z}{\partial x} = \text{e}^x\cos y \cdot f'$，$\frac{\partial z}{\partial y} = -\text{e}^x\sin y \cdot f'$，
$\frac{\partial^2 z}{\partial x^2} = \text{e}^x\cos y \cdot f' + \text{e}^{2x}\cos^2 y \cdot f''$，$\frac{\partial^2 z}{\partial y^2} = -\text{e}^x\cos y \cdot f' + \text{e}^{2x}\sin^2 y \cdot f''$，
$\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \text{e}^{2x}f''$，
令$u = \text{e}^x\cos y$，由$\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = (4z + \text{e}^x\cos y)\text{e}^{2x}$得
$f''(u) = 4f(u) + u$，或$f''(u) - 4f(u) = u$，
解得$f(u) = C_1\text{e}^{-2u} + C_2\text{e}^{2u} - \frac{1}{4}u$.
由$f(0) = 0$，$f'(0) = 0$得$\begin{cases} C_1 + C_2 = 0, \\ -2C_1 + 2C_2 - \frac{1}{4} = 0, \end{cases}$解得$C_1 = -\frac{1}{16}$，$C_2 = \frac{1}{16}$.
故$f(u) = \frac{1}{16}(\text{e}^{2u} - \text{e}^{-2u}) - \frac{1}{4}u$.

(I)由积分中值定理 \(\int_{a}^{x} g(t) d t=g(\xi)(x-a)\) ，\(\xi \in[a, x]\)

\(\because 0 \leq g(x) \leq 1\) ，\(\therefore 0 \leq g(\xi)(x-a) \leq(x-a)\)

\(\therefore 0 \leq \int_{a}^{x} g(t) d t \leq(x-a)\)

(II)令 \(F(u)=\int_{a}^{u} f(x) g(x) d x-\int_{a}^{a+\int_{a}^{u} g(t) d t} f(x) d x\)

\( \begin{aligned}
     F'(u) &= f(u) g(u) - f\left(a + \int_{a}^{u} g(t) dt\right) \cdot g(u) \\
     &= g(u)\left[f(u) - f\left(a + \int_{a}^{u} g(t) dt\right)\right]
     \end{aligned}\)

由(I)知 \(0 \leq\int_{a}^{u} g(t) d t \leq(u-a)\)

又由于 \(f(x)\) 单增，所以 \(f(u)-f(a+\int_{a}^{u} g(t) d t) \geq 0\) ，\(F'(u) \geq 0\) ，\(F(u)\) 单调不减，\(\therefore F(u) \geq F(a)=0\)

取 \(u=b\) ，得 \(F(b) \geq 0\) ，即(II)成立

\(f_{1}(x)=\frac{x}{1+x}\) ，\(f_{2}(x)=\frac{x}{1+2 x}\) ，\(\cdots\) ，\(f_{n}(x)=\frac{x}{1+n x}\)

\(\begin{aligned}
S_{n} & = \int_{0}^{1} f_{n}(x) \, dx = \int_{0}^{1} \frac{x}{1+nx} \, dx = \int_{0}^{1} \frac{x+\frac{1}{n}-\frac{1}{n}}{1+nx} \, dx \\
& = \frac{1}{n} \int_{0}^{1} 1 \, dx - \frac{1}{n} \int_{0}^{1} \frac{1}{1+nx} \, dx = \frac{1}{n} - \left. \frac{1}{n^{2}} \ln(1+nx) \right|_{0}^{1} \\
& = \frac{1}{n} - \frac{1}{n^{2}} \ln(1+n) \\
\therefore \lim_{n \to \infty} n S_{n} & = 1 - \lim_{n \to \infty} \frac{\ln(1+n)}{n} = 1 - \lim_{x \to \infty} \frac{\ln(1+x)}{x} = 1 - \lim_{x \to \infty} \frac{1}{1+x} = 1 - 0 = 1
\end{aligned}\)

因为 \(\frac{\partial f}{\partial y}=2(y+1)\) ，所以 \(f(x, y)=y^{2}+2 y+\varphi(x)\) ，其中 \(\varphi(x)\) 为待定函数。

又因为 \(f(y, y)=(y+1)^{2}-(2-y) \ln y\) ，则 \(\varphi(y)=1-(2-y) \ln y\) ，从而

\(f(x, y)=y^{2}+2 y+1-(2-x) \ln x=(y+1)^{2}-(2-x) \ln x\)。

令 \(f(x, y)=0\) 可得 \((y+1)^{2}=(2-x) \ln x\) ，当 \(y=-1\) 时，\(x=1\) 或 \(x=2\) ，从而所求的体积为

\(\begin{aligned}
V & = \pi \int_{1}^{2} (y+1)^{2} dx = \pi \int_{1}^{2} (2-x) \ln x dx \\
& = \pi \int_{1}^{2} \ln x d\left(2x - \frac{x^{2}}{2}\right) \\
& = \pi \left[\ln x \left(2x - \frac{x^{2}}{2}\right)\right]_{1}^{2} - \pi \int_{1}^{2} \left(2 - \frac{x}{2}\right) dx \\
& = \pi \cdot 2 \ln 2 - \left.\pi \left(2x - \frac{x^{2}}{4}\right)\right|_{1} ^{2} = \pi \cdot 2 \ln 2 - \pi \cdot \frac{5}{4} = \pi \left(2 \ln 2 - \frac{5}{4}\right)
\end{aligned}\)

【解】（Ⅰ）\( A = \begin{pmatrix} 1 & -2 & 3 & -4 \\ 0 & 1 & -1 & 1 \\ 1 & 2 & 0 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & -2 & 3 & -4 \\ 0 & 1 & -1 & 1 \\ 0 & 4 & -3 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & -2 & 3 & -4 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 1 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & -2 & 0 & 5 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -3 \end{pmatrix} \)，
则方程组\( AX = 0 \)的一个基础解系为\( \xi = (-1, 2, 3, 1)^T \).

（Ⅱ）方法一
由\( (A \vdots E) = \begin{pmatrix} 1 & -2 & 3 & -4 & \vdots & 1 & 0 & 0 \\ 0 & 1 & -1 & 1 & \vdots & 0 & 1 & 0 \\ 1 & 2 & 0 & -3 & \vdots & 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & -2 & 3 & -4 & \vdots & 1 & 0 & 0 \\ 0 & 1 & -1 & 1 & \vdots & 0 & 1 & 0 \\ 0 & 4 & -3 & 1 & \vdots & -1 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & -2 & 0 & -2 & \vdots & 4 & -3 & 1 \\ 0 & 1 & 0 & -2 & \vdots & 1 & -3 & 1 \\ 0 & 0 & 1 & -3 & \vdots & -1 & -4 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & 1 & \vdots & 2 & -6 & -1 \\ 0 & 1 & 0 & -2 & \vdots & 1 & 3 & 1 \\ 0 & 0 & 1 & -3 & \vdots & -1 & -4 & 1 \end{pmatrix} \)，
得\( B = \begin{pmatrix} 2 - k_1 & 6 - k_2 & -1 - k_3 \\ 2k_1 - 1 & 2k_2 - 3 & 2k_3 + 1 \\ 3k_1 - 1 & 3k_2 - 4 & 3k_3 + 1 \\ k_1 & k_2 & k_3 \end{pmatrix} \)（\( k_1, k_2, k_3 \)为任意常数）.

方法二 令\( B = (X_1, X_2, X_3) \)，\( E = (e_1, e_2, e_3) \)，
则\( AB = E \)等价于\( AX_1 = e_1 \)，\( AX_2 = e_2 \)，\( AX_3 = e_3 \)，
方程组\( AX_1 = e_1 \)的通解为\( X_1 = k_1 \begin{pmatrix} -1 \\ 2 \\ 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ -1 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -k_1 + 2 \\ 2k_1 - 1 \\ 3k_1 - 1 \\ k_1 \end{pmatrix} \)（\( k_1 \)为任意常数），
方程组\( AX_2 = e_2 \)的通解为\( X_2 = k_2 \begin{pmatrix} -1 \\ 2 \\ 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 6 \\ -3 \\ -4 \\ 0 \end{pmatrix} = \begin{pmatrix} -k_2 + 6 \\ 2k_2 - 3 \\ 3k_2 - 4 \\ k_2 \end{pmatrix} \)（\( k_2 \)为任意常数），
方程组\( AX_3 = e_3 \)的通解为\( X_3 = k_3 \begin{pmatrix} -1 \\ 2 \\ 3 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -k_3 - 1 \\ 2k_3 + 1 \\ 3k_3 + 1 \\ k_3 \end{pmatrix} \)（\( k_3 \)为任意常数），
故\( B = \begin{pmatrix} -k_1 + 2 & -k_2 + 6 & -k_3 - 1 \\ 2k_1 - 1 & 2k_2 - 3 & 2k_3 + 1 \\ 3k_1 - 1 & 3k_2 - 4 & 3k_3 + 1 \\ k_1 & k_2 & k_3 \end{pmatrix} \)（\( k_1, k_2, k_3 \)为任意常数）.

方法三 令\( B = \begin{pmatrix} x_1 & x_2 & x_3 \\ x_4 & x_5 & x_6 \\ x_7 & x_8 & x_9 \\ x_{10} & x_{11} & x_{12} \end{pmatrix} \)，由\( AB = E \)得
\[
\begin{cases}
x_1 - 2x_4 + 3x_7 - 4x_{10} = 1, \\
x_2 - 2x_5 + 3x_8 - 4x_{11} = 0, \\
x_3 - 2x_6 + 3x_9 - 4x_{12} = 0, \\
x_4 - x_7 + x_{10} = 0, \\
x_5 - x_8 + x_{11} = 1, \\
x_6 - x_9 + x_{12} = 0, \\
x_1 + 2x_4 - 3x_{10} = 0, \\
x_2 + 2x_5 - 3x_{11} = 0, \\
x_3 + 2x_6 - 3x_{12} = 1,
\end{cases}
\]
解得\( \begin{pmatrix} x_1 \\ x_4 \\ x_7 \\ x_{10} \end{pmatrix} = k_1 \begin{pmatrix} -1 \\ 2 \\ 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ -1 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -k_1 + 2 \\ 2k_1 - 1 \\ 3k_1 - 1 \\ k_1 \end{pmatrix} \)，\( \begin{pmatrix} x_2 \\ x_5 \\ x_8 \\ x_{11} \end{pmatrix} = k_2 \begin{pmatrix} -1 \\ 2 \\ 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 6 \\ -3 \\ -4 \\ 0 \end{pmatrix} = \begin{pmatrix} -k_2 + 6 \\ 2k_2 - 3 \\ 3k_2 - 4 \\ k_2 \end{pmatrix} \)，
\( \begin{pmatrix} x_3 \\ x_6 \\ x_9 \\ x_{12} \end{pmatrix} = k_3 \begin{pmatrix} -1 \\ 2 \\ 3 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -k_3 - 1 \\ 2k_3 + 1 \\ 3k_3 + 1 \\ k_3 \end{pmatrix} \)，
故\( B = \begin{pmatrix} -k_1 + 2 & -k_2 + 6 & -k_3 - 1 \\ 2k_1 - 1 & 2k_2 - 3 & 2k_3 + 1 \\ 3k_1 - 1 & 3k_2 - 4 & 3k_3 + 1 \\ k_1 & k_2 & k_3 \end{pmatrix} \)（\( k_1, k_2, k_3 \)为任意常数）.

【详解】证明：设 \( \boldsymbol{A} = \begin{pmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&&\vdots\\1&1&\cdots&1\end{pmatrix} \)，\( \boldsymbol{B} = \begin{pmatrix}0&\cdots&0&1\\0&\cdots&0&2\\\vdots&\vdots&&\vdots\\0&\cdots&0&n\end{pmatrix} \). 

分别求两个矩阵的特征值和特征向量如下： 

\(\vert\lambda \boldsymbol{E} - \boldsymbol{A}\vert = \begin{vmatrix}\lambda - 1&-1&\cdots&-1\\-1&\lambda - 1&\cdots&-1\\\vdots&\vdots&&\vdots\\-1&-1&\cdots&\lambda - 1\end{vmatrix} = (\lambda - n)\lambda^{n - 1}\)， 

所以 \( \boldsymbol{A} \) 的 \( n \) 个特征值为 \( \lambda_1 = n,\lambda_2 = \lambda_3 = \cdots = \lambda_n = 0 \)； 

而且 \( \boldsymbol{A} \) 是实对称矩阵，所以一定可以对角化，且 \( \boldsymbol{A} \sim \begin{pmatrix}\lambda&&\\&0&\\&&\cdots\\&&&0\end{pmatrix} \)； 

\(\vert\lambda \boldsymbol{E} - \boldsymbol{B}\vert = \begin{vmatrix}\lambda&0&\cdots&-1\\0&\lambda&\cdots&-2\\\vdots&\vdots&&\vdots\\0&0&\cdots&\lambda - n\end{vmatrix} = (\lambda - n)\lambda^{n - 1}\) 

所以 \( \boldsymbol{B} \) 的 \( n \) 个特征值也为 \( \lambda_1 = n,\lambda_2 = \lambda_3 = \cdots = \lambda_n = 0 \)； 

对于n - 1重特征值$\boldsymbol{\lambda = 0}$，由于矩阵$\boldsymbol{(0E - B) = -B}$的秩显然为1，所以矩阵$\boldsymbol{B}$对应n - 1重特征值$\boldsymbol{\lambda = 0}$的特征向量应该有n - 1个线性无关，进一步矩阵$\boldsymbol{B}$存在n个线性无关的特征向量，即矩阵$\boldsymbol{B}$一定可以对角化，且$\boldsymbol{B \sim \begin{pmatrix} \lambda \\ & 0 \\ & & \cdots \\ & & & 0 \end{pmatrix}}$ 

从而可知n阶矩阵$\boldsymbol{\begin{pmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix}}$与$\boldsymbol{\begin{pmatrix} 0 & \cdots & 0 & 1 \\ 0 & \cdots & 0 & 2 \\ \vdots & \vdots & & \vdots \\ 0 & \cdots & 0 & n \end{pmatrix}}$相似.