【答案】(D).
【解】 方法一 对$f(x)=\cos\sqrt{|x|}$,
$\lim\limits_{x \to 0}\frac{f(x)-f(0)}{x}=\lim\limits_{x \to 0}\frac{\cos\sqrt{|x|}-1}{x}=-\frac{1}{2}\lim\limits_{x \to 0}\frac{|x|}{x}$不存在,即$f(x)=\cos\sqrt{|x|}$在$x = 0$处不可导,应选(D).
方法二
当$f(x)=|x|\sin|x|$时,$f(x)-f(0)\sim x^2$,$f(x)$在$x = 0$处可导且导数为0,不选(A);
当$f(x)=|x|\sin\sqrt{|x|}$时,$f(x)-f(0)\sim |x|^{\frac{3}{2}}$,$f(x)$在$x = 0$处可导且导数为0,不选(B);
当$f(x)=\cos|x|$时,$f(x)-f(0)\sim -\frac{1}{2}x^2$,$f(x)$在$x = 0$处可导且导数为0,不选(C),
应选(D).

【答案】(B).
【解】 设切点为 \( (x_0, y_0, z_0) \)，则
\[
\begin{cases}
z_0 = x_0^2 + y_0^2, \\
(2x_0, 2y_0, -1) \cdot (1, -1, 0) = 0, \\
(2x_0, 2y_0, -1) \cdot (x_0 - 1, y_0, z_0) = 0,
\end{cases}
\]
解之得 \(\begin{cases} x_0 = 0, \\ y_0 = 0, \\ z_0 = 0, \end{cases}\) 或 \(\begin{cases} x_0 = 1, \\ y_0 = 1, \\ z_0 = 2, \end{cases}\)
故所求切平面为 \( z = 0 \) 或 \( 2x + 2y - z = 2 \). 应选(B)

【答案】 (B).
【解】 $\sum\limits_{n=0}^{\infty}(-1)^n\frac{2n + 3}{(2n + 1)!} = \sum\limits_{n=0}^{\infty}(-1)^n\frac{1}{(2n)!} + 2\sum\limits_{n=0}^{\infty}(-1)^n\frac{1}{(2n + 1)!}$
$= \cos1 + 2\sin1$，
应选(B).

【答案】 (C).
【解】\( M = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1 + x)^2}{1 + x^2}dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(1 + \frac{2x}{1 + x^2}\right)dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1dx = \pi \)，
当\( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \)时，\( 1 + \sqrt{\cos x} > 1 > \frac{1 + x}{e^x} \)，
\( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \sqrt{\cos x})dx > \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1dx > \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + x}{e^x}dx \)，即\( K > M > N \).应选(C).

【答案】（A）.
【解】 方法一
令$M=\begin{pmatrix} 1&1&0 \\ 0&1&1 \\ 0&0&1 \end{pmatrix}$，$A=\begin{pmatrix} 1&1&-1 \\ 0&1&1 \\ 0&0&1 \end{pmatrix}$，$B=\begin{pmatrix} 1&0&-1 \\ 0&1&1 \\ 0&0&1 \end{pmatrix}$，
$C=\begin{pmatrix} 1&1&-1 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$，$D=\begin{pmatrix} 1&0&-1 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$，
显然矩阵$A$，$B$，$C$，$D$的特征值都是$\lambda_1 = \lambda_2 = \lambda_3 = 1$，
$E - M = \begin{pmatrix} 0&-1&0 \\ 0&0&-1 \\ 0&0&0 \end{pmatrix}$，$E - A = \begin{pmatrix} 0&-1&1 \\ 0&0&-1 \\ 0&0&0 \end{pmatrix}$，$E - B = \begin{pmatrix} 0&0&1 \\ 0&0&-1 \\ 0&0&0 \end{pmatrix}$，
$E - C = \begin{pmatrix} 0&-1&1 \\ 0&0&0 \\ 0&0&0 \end{pmatrix}$，$E - D = \begin{pmatrix} 0&0&1 \\ 0&0&0 \\ 0&0&0 \end{pmatrix}$，
因为$r(E - M) = r(E - A) = 2$，所以应选（A）.

方法二
取$P = \begin{pmatrix} 1&-1&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$，则$P^{-1} = \begin{pmatrix} 1&1&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$，
因为$P^{-1}\begin{pmatrix} 1&1&-1 \\ 0&1&1 \\ 0&0&1 \end{pmatrix}P = \begin{pmatrix} 1&1&0 \\ 0&1&1 \\ 0&0&1 \end{pmatrix}$，所以$\begin{pmatrix} 1&1&0 \\ 0&1&1 \\ 0&0&1 \end{pmatrix}$与$\begin{pmatrix} 1&1&-1 \\ 0&1&1 \\ 0&0&1 \end{pmatrix}$相似，应选（A）.

【答案】(A).
【解】\( (A,AB)=A(E,B) \),
显然\( r(A,AB)=r[A(E,B)] \leq r(A) \),
又\( r(A,AB) \geq r(A) \),
于是\( r(A,AB)=r(A) \),应选(A).

【答案】(A).
【解】$\int_{0}^{1}f(x)dx = \int_{1}^{2}f(x)dx = \frac{1}{2}\int_{0}^{2}f(x)dx = 0.3$，
$P(X < 0) = \int_{-\infty}^{0}f(x)dx = \int_{-\infty}^{1}f(x)dx - \int_{0}^{1}f(x)dx = 0.5 - 0.3 = 0.2$，应选(A).

【答案】 (D).
【解】 若\( \sigma^{2} \)已知，则假设\( H_{0} \)的接受域：\( |u| < u_{\frac{\alpha}{2}} \)，其中\( u_{\frac{\alpha}{2}} \)为正态分布的\( \frac{\alpha}{2} \)（上）分位数。
若\( \sigma^{2} \)未知，则假设\( H_{0} \)的接受域：\( |t| < t_{\frac{\alpha}{2}}(n - 1) \)，其中\( t_{\frac{\alpha}{2}}(n - 1) \)为自由度是\( n - 1 \)的\( t \)分布的\( \frac{\alpha}{2} \)（上）分位数。显然检验水平\( \alpha \)变小，接受域都变大。应选(D)。

【答案】$-2$.
【解】 因为$\lim\limits _{x→0}\frac {1}{\sin kx}\cdot (\frac {1 - \tan x}{1 + \tan x} - 1) = \lim\limits _{x→0}\frac {1}{\sin kx}\cdot \frac {-2\tan x}{1 + \tan x} = \lim\limits _{x→0}\frac {1}{kx}\cdot \frac {-2x}{1 + \tan x} = \frac {-2}{k}$，
故$\lim\limits _{x→0}(\frac {1 - \tan x}{1 + \tan x})^{\frac {1}{\sin kx}} = e^{\frac {-2}{k}} = e$，即$k = -2$．

【答案】 $2(\ln 2 - 1)$.
【解】 $f'(1) = (2^x)'|_{x=1} = 2\ln 2$,
$\int_{0}^{1}xf''(x)\text{d}x = \int_{0}^{1}x\text{d}f'(x) = xf'(x)|_{0}^{1} - \int_{0}^{1}f'(x)\text{d}x = f'(1) - f(1) + f(0)$
$= 2\ln 2 - 2 = 2(\ln 2 - 1)$.

【答案】$\boldsymbol{i} - \boldsymbol{k}$.
【解】$\text{rot}\ \boldsymbol{F}(1,1,0) = \begin{vmatrix}
\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
xy & -yz & zx
\end{vmatrix}_{(1,1,0)} = (y\boldsymbol{i} - z\boldsymbol{j} - x\boldsymbol{k})\big|_{(1,1,0)} = \boldsymbol{i} - \boldsymbol{k}$.

【答案】 $-\frac{\pi}{3}$．
【解】$\oint_{L}xyds=\frac{1}{3}\oint_{L}(xy+yz+xz)ds=\frac{1}{6}\oint_{L}[(x+y+z)^{2}-(x^{2}+y^{2}+z^{2})]ds$
$=-\frac{1}{6}\oint_{L}ds=-\frac{\pi}{3}$．

因为$\alpha_{1}$，$\alpha_{2}$线性无关，所以$\alpha_{1} + \alpha_{2} \neq 0$。由$A^{2}(\alpha_{1} + \alpha_{2}) = \alpha_{1} + \alpha_{2}$可知，$\alpha_{1} + \alpha_{2}$是$A^{2}$对应特征值$1$的特征向量。又因为$A$有两个不同的特征值，设为$\lambda_{1}$，$\lambda_{2}$，则$A^{2}$的特征值为$\lambda_{1}^{2}$，$\lambda_{2}^{2}$。由于$A^{2}$有特征值$1$，且$\alpha_{1}$，$\alpha_{2}$是$A$的特征向量，所以$\lambda_{1}^{2} = 1$，$\lambda_{2}^{2} = 1$，解得$\lambda_{1} = \pm 1$，$\lambda_{2} = \pm 1$。又因为$\lambda_{1} \neq \lambda_{2}$，所以$\lambda_{1} = 1$，$\lambda_{2} = -1$，则$|A| = \lambda_{1} \lambda_{2} = -1$。

【答案】$\frac{1}{4}$。
【解】由$BC = \varnothing$，得$P(BC) = 0$，因为$ABC \subset BC$，所以$P(ABC) = 0$。
$P(AC | AB \cup C) = \frac{P[AC(AB \cup C)]}{P(AB \cup C)} = \frac{P[(ABC) \cup (AC)]}{P(AB \cup C)} = \frac{P(AC)}{P(AB \cup C)}$
$= \frac{P(A)P(C)}{P(A)P(B) + P(C) - P(ABC)} = \frac{\frac{1}{2}P(C)}{\frac{1}{2} \times \frac{1}{2} + P(C) - 0} = \frac{1}{4}$，
解得$P(C) = \frac{1}{4}$。

【解】$\int \mathrm{e}^{2x}\arctan\sqrt{\mathrm{e}^{x}-1}\mathrm{d}x$
$=\frac{1}{2}\int \arctan\sqrt{\mathrm{e}^{x}-1}\mathrm{d}\mathrm{e}^{2x} = \frac{1}{2}\mathrm{e}^{2x}\arctan\sqrt{\mathrm{e}^{x}-1} - \frac{1}{2}\int \mathrm{e}^{2x}\mathrm{d}\arctan\sqrt{\mathrm{e}^{x}-1}$
$=\frac{1}{2}\mathrm{e}^{2x}\arctan\sqrt{\mathrm{e}^{x}-1} - \frac{1}{4}\int \frac{\mathrm{e}^{2x}}{\sqrt{\mathrm{e}^{x}-1}}\mathrm{d}x = \frac{1}{2}\mathrm{e}^{2x}\arctan\sqrt{\mathrm{e}^{x}-1} - \frac{1}{2}\int \mathrm{e}^{x}\mathrm{d}\sqrt{\mathrm{e}^{x}-1}$
$=\frac{1}{2}\mathrm{e}^{2x}\arctan\sqrt{\mathrm{e}^{x}-1} - \frac{1}{2}\left(\mathrm{e}^{x}\sqrt{\mathrm{e}^{x}-1} - \int \sqrt{\mathrm{e}^{x}-1}\mathrm{d}\mathrm{e}^{x}\right)$
$=\frac{1}{2}\mathrm{e}^{2x}\arctan\sqrt{\mathrm{e}^{x}-1} - \frac{1}{2}\left[\mathrm{e}^{x}\sqrt{\mathrm{e}^{x}-1} - \frac{2}{3}\left(\mathrm{e}^{x}-1\right)^{\frac{3}{2}}\right] + C$
$=\frac{1}{2}\mathrm{e}^{2x}\arctan\sqrt{\mathrm{e}^{x}-1} - \frac{1}{6}\left(\mathrm{e}^{x} + 2\right)\sqrt{\mathrm{e}^{x}-1} + C$.

【解】 设铁丝分成的三段长分别是x,y,z,则x+y+z=2,且依次围成的圆、正方形与正三角形三个图形的面积之和为
$f(x,y,z)=π(\frac {x}{2π})^{2}+(\frac {y}{4})^{2}+\frac {\sqrt {3}}{4}(\frac {z}{3})^{2}=\frac {x^{2}}{4π}+\frac {y^{2}}{16}+\frac {\sqrt {3}}{36}z^{2},$
构造拉格朗日函数:$L(x,y,z,λ)=\frac {x^{2}}{4π}+\frac {y^{2}}{16}+\frac {\sqrt {3}}{36}z^{2}+λ(x+y+z-2)$
令$\begin{cases} L'_{x}=\frac {x}{2π}+λ=0, \\ L'_{y}=\frac {y}{8}+λ=0, \\ L'_{z}=\frac {\sqrt {3}}{18}z+λ=0, \\ L'_{λ}=x+y+z-2=0, \end{cases}$ 得$\begin{cases} x_{0}=\frac {2π}{π+4+3\sqrt {3}}, \\ y_{0}=\frac {8}{π+4+3\sqrt {3}}, \\ z_{0}=\frac {6\sqrt {3}}{π+4+3\sqrt {3}}, \end{cases}$ 此时$f(x_{0},y_{0},z_{0})=\frac {1}{π+4+3\sqrt {3}}.$
又当$x+y+z=2,xyz=0$时,$f(x,y,z)$的最小值为$f(0,\frac {8}{4+3\sqrt {3}},\frac {6\sqrt {3}}{4+3\sqrt {3}})=\frac {1}{4+3\sqrt {3}}.$
所以三个图形的面积之和存在最小值,最小值为$f(x_{0},y_{0},z_{0})=\frac {1}{π+4+3\sqrt {3}}.$

【解】补曲面\( \Sigma_1:\begin{cases}x = 0, \\ 3y^2 + 3z^2 = 1\end{cases} \)，取后侧，Ω为Σ与\( \Sigma_1 \)所围成的立体. 利用高斯公式可得
\( I = \iiint_{\Sigma + \Sigma_1} x\text{d}y\text{d}z + (y^3 + 2)\text{d}z\text{d}x + z^3\text{d}x\text{d}y - \iint_{\Sigma_1} x\text{d}y\text{d}z + (y^3 + 2)\text{d}z\text{d}x + z^3\text{d}x\text{d}y \)
\( = \iiint_{\Omega} (1 + 3y^2 + 3z^2)\text{d}x\text{d}y\text{d}z - 0 = \iiint_{\Omega} (1 + 3y^2 + 3z^2)\text{d}x\text{d}y\text{d}z \)
利用柱坐标变换\( y = r\cos\theta, z = r\sin\theta, x = x \)得
\( \iiint_{\Omega} (1 + 3y^2 + 3z^2)\text{d}x\text{d}y\text{d}z = \int_{0}^{2\pi} \text{d}\theta \int_{0}^{\frac{\sqrt{3}}{3}} \text{d}r \int_{0}^{\sqrt{1 - 3r^2}} (1 + 3r^2)r\text{d}x \)
\( = 2\pi \int_{0}^{\frac{\sqrt{3}}{3}} \sqrt{1 - 3r^2}(1 + 3r^2)r\text{d}r = \frac{2\pi}{3} \int_{0}^{1} (2 - t^2)t^2\text{d}t = \frac{14\pi}{45} \).

【解】(Ⅰ) 当 \( f(x) = x \) 时，方程化为 \( y' + y = x \)，其通解为
\( y = \mathrm{e}^{-\int dx} \left( C + \int x \mathrm{e}^{\int dx} dx \right) = \mathrm{e}^{-x} \left( C + x \mathrm{e}^{x} - \mathrm{e}^{x} \right) = C \mathrm{e}^{-x} + x - 1 \).

(Ⅱ) 方程 \( y' + y = f(x) \) 的通解为 \( y = \mathrm{e}^{-\int_{0}^{x} dt} \left[ C + \int_{0}^{x} \mathrm{e}^{\int_{0}^{t} ds} f(t) dt \right] \)，
即 \( y = \mathrm{e}^{-x} \left[ C + \int_{0}^{x} \mathrm{e}^{t} f(t) dt \right] \).

得 \( y(x + T) - y(x) = \mathrm{e}^{-x} \left[ \left( \frac{1}{\mathrm{e}^{T}} - 1 \right) C + \frac{1}{\mathrm{e}^{T}} \int_{0}^{x + T} \mathrm{e}^{t} f(t) dt - \int_{0}^{x} \mathrm{e}^{t} f(t) dt \right] \)
若 \( f(x) \) 是周期为 \( T \) 的连续函数，则
\( \frac{1}{\mathrm{e}^{T}} \int_{0}^{x + T} \mathrm{e}^{t} f(t) dt = \frac{1}{\mathrm{e}^{T}} \int_{0}^{T} \mathrm{e}^{t} f(t) dt + \frac{1}{\mathrm{e}^{T}} \int_{T}^{x + T} \mathrm{e}^{t} f(t) dt \)
\( = \frac{1}{\mathrm{e}^{T}} \int_{0}^{T} \mathrm{e}^{t} f(t) dt + \frac{1}{\mathrm{e}^{T}} \int_{0}^{x} \mathrm{e}^{u + T} f(u + T) du \)
\( = \frac{1}{\mathrm{e}^{T}} \int_{0}^{T} \mathrm{e}^{t} f(t) dt + \frac{1}{\mathrm{e}^{T}} \int_{0}^{x} \mathrm{e}^{T} \mathrm{e}^{u} f(u) du = \frac{1}{\mathrm{e}^{T}} \int_{0}^{T} \mathrm{e}^{t} f(t) dt + \int_{0}^{x} \mathrm{e}^{u} f(u) du \)
于是 \( y(x + T) - y(x) = \mathrm{e}^{-x} \left[ \left( \frac{1}{\mathrm{e}^{T}} - 1 \right) C + \frac{1}{\mathrm{e}^{T}} \int_{0}^{T} \mathrm{e}^{t} f(t) dt \right] \)，
因此当且仅当 \( C = \frac{1}{\mathrm{e}^{T} - 1} \int_{0}^{T} \mathrm{e}^{t} f(t) dt \) 时，\( y(x + T) - y(x) = 0 \)，即方程存在唯一的以 \( T \) 为周期的解.

【证明】 因为$x_{1} \neq 0$，所以$\text{e}^{x_{2}} = \frac{\text{e}^{x_{1}} - 1}{x_{1}}$．
由微分中值定理，存在$\xi \in (0, x_{1})$，使得$\frac{\text{e}^{x_{1}} - 1}{x_{1}} = \text{e}^{\xi}$，即$\text{e}^{x_{2}} = \text{e}^{\xi}$，因此$0 < x_{2} < x_{1}$．
假设$0 < x_{n+1} < x_{n}$，则$\text{e}^{x_{n+2}} = \frac{\text{e}^{x_{n+1}} - 1}{x_{n+1}} = \text{e}^{\eta}$（$0 < \eta < x_{n+1}$），得$0 < x_{n+2} < x_{n+1}$．故$\{ x_{n}\}$是单调减少的数列，且有下界，从而$\{ x_{n}\}$收敛．
设$\lim\limits_{n \to \infty} x_{n} = a$，在等式$x_{n}\text{e}^{x_{n+1}} = \text{e}^{x_{n}} - 1$两边取极限，得$a\text{e}^{a} = \text{e}^{a} - 1$，显然$a = 0$为其解．
又令$f(x) = x\text{e}^{x} - \text{e}^{x} + 1$，则$f'(x) = x\text{e}^{x}$．
当$x > 0$时，$f'(x) = x\text{e}^{x} > 0$，函数$f(x)$在$[0, +\infty)$上单调增加，
所以$a = 0$是方程$a\text{e}^{a} = \text{e}^{a} - 1$在$[0, +\infty)$上的唯一解，故$\lim\limits_{n \to \infty} x_{n} = 0$．

【解】(Ⅰ)$f(x_1,x_2,x_3)=(x_1 - x_2 + x_3)^2 + (x_2 + x_3)^2 + (x_1 + ax_3)^2=0$的充分必要条件是$\begin{cases}x_1 - x_2 + x_3=0,\\x_2 + x_3=0,\\x_1 + ax_3=0.\end{cases}$
对齐次线性方程组的系数矩阵作初等行变换得
$\boldsymbol{A}=\begin{pmatrix}1&-1&1\\0&1&1\\1&0&a\end{pmatrix}\to\begin{pmatrix}1&-1&1\\0&1&1\\0&1&a - 1\end{pmatrix}\to\begin{pmatrix}1&-1&1\\0&1&1\\0&0&a - 2\end{pmatrix}$，
$a\neq2$时，$f(x_1,x_2,x_3)=0$只有零解$x=(x_1,x_2,x_3)^{\mathrm{T}}=(0,0,0)^{\mathrm{T}}$，
$a=2$时，$\boldsymbol{A}\to\begin{pmatrix}1&0&2\\0&1&1\\0&0&0\end{pmatrix}$，
$f(x_1,x_2,x_3)=0$有非零解$x=(x_1,x_2,x_3)^{\mathrm{T}}=k(-2,-1,1)^{\mathrm{T}}$，其中$k\neq0$.

(Ⅱ)$a\neq2$时，令$\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}=\begin{pmatrix}1&-1&1\\0&1&1\\1&0&a\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$，
因$\begin{vmatrix}1&-1&1\\0&1&1\\1&0&a\end{vmatrix}=\begin{vmatrix}1&-1&1\\0&1&1\\0&0&a - 2\end{vmatrix}=a - 2\neq0$，则矩阵$\begin{pmatrix}1&-1&1\\0&1&1\\1&0&a\end{pmatrix}$可逆，
所以$f(x_1,x_2,x_3)$的规范形为$f(y_1,y_2,y_3)=y_1^2 + y_2^2 + y_3^2$.

$a=2$时，
$f(x_1,x_2,x_3)=(x_1 - x_2 + x_3)^2 + (x_2 + x_3)^2 + (x_1 + ax_3)^2$
$=2x_1^2 + 2x_2^2 + 6x_3^2 - 2x_1x_2 + 6x_1x_3$
$=2(x_1 - x_2 + x_3)^2 + \frac{3}{2}x_2^2 + \frac{3}{2}x_3^2 + 3x_2x_3$
$=2(x_1 - x_2 + x_3)^2 + \frac{3}{2}(x_2 + x_3)^2$
所以$f(x_1,x_2,x_3)$的规范形为$f(y_1,y_2,y_3)=y_1^2 + y_2^2$.

【解】(Ⅰ)显然\( r(A)=2 \)，因为初等变换不改变矩阵的秩，所以\( r(B)=2 \)，
而\( B = \begin{pmatrix} 1 & a & 2 \\ 0 & 1 & 1 \\ -1 & 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & a & 2 \\ 0 & 1 & 1 \\ 0 & a + 1 & 3 \end{pmatrix} \to \begin{pmatrix} 1 & a & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 - a \end{pmatrix} \)，故\( a = 2 \)。

(Ⅱ)\( A = \begin{pmatrix} 1 & 2 & 2 \\ 1 & 3 & 0 \\ 2 & 7 & -2 \end{pmatrix} \)，\( B = \begin{pmatrix} 1 & 2 & 2 \\ 0 & 1 & 1 \\ -1 & 1 & 1 \end{pmatrix} \)，
令\( P = (X_1, X_2, X_3) \)，
由\( (A \vdots B) = \begin{pmatrix} 1 & 2 & 2 & \vdots & 1 & 2 & 2 \\ 1 & 3 & 0 & \vdots & 0 & 1 & 1 \\ 2 & 7 & -2 & \vdots & -1 & 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 6 & \vdots & 3 & 4 & 4 \\ 0 & 1 & -2 & \vdots & -1 & -1 & -1 \\ 0 & 0 & 0 & \vdots & 0 & 0 & 0 \end{pmatrix} \)得
\( X_1 = k_1\begin{pmatrix} -6 \\ 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -6k_1 + 3 \\ 2k_1 - 1 \\ k_1 \end{pmatrix} \)，\( X_2 = k_2\begin{pmatrix} -6 \\ 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -6k_2 + 4 \\ 2k_2 - 1 \\ k_2 \end{pmatrix} \)，
\( X_3 = k_3\begin{pmatrix} -6 \\ 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -6k_3 + 4 \\ 2k_3 - 1 \\ k_3 \end{pmatrix} \)，
则所求的可逆矩阵为\( P = \begin{pmatrix} -6k_1 + 3 & -6k_2 + 4 & -6k_3 + 4 \\ 2k_1 - 1 & 2k_2 - 1 & 2k_3 - 1 \\ k_1 & k_2 & k_3 \end{pmatrix} \)（\( k_1, k_2, k_3 \)为任意常数且\( k_2 \neq k_3 \)）。

【解】(Ⅰ) 因为\( E(X)=0 \)，\( E(X^2)=1 \)，\( E(Y)=\lambda \)，以及X，Y相互独立，故
\( \text{Cov}(X,Z)=\text{Cov}(X,XY)=E(X^2Y)-E(X)E(XY)=E(X^2)E(Y)-E^2(X)E(Y)=\lambda \).
(Ⅱ) 由Y服从参数为\( \lambda \)的泊松分布，即\( P(Y=j)=\frac{\lambda^j}{j!}\text{e}^{-\lambda} (j=0,1,2,\cdots) \)，于是，Z的所有可能取值为全体整数. 故Z的概率分布为
1) 当k为正整数时，有
\( P(Z=k)=P(XY=k)=P(X=1,Y=k)=P(X=1)P(Y=k) \)
\( =\frac{1}{2} \cdot \frac{\lambda^k}{k!}\text{e}^{-\lambda} (k=1,2,3,\cdots) \),
2) 当k为负整数时，有
\( P(Z=k)=P(XY=k)=P(X=-1,Y=-k)=P(X=-1)P(Y=-k) \)
\( =\frac{1}{2} \cdot \frac{\lambda^{-k}}{(-k)!}\text{e}^{-\lambda} (k=-1,-2,-3,\cdots) \),
3) 当k为0时，有
\( P(Z=0)=P(XY=0)=P(X=-1,Y=0)+P(X=1)P(Y=0)=P(X=-1,Y=0) \)
\( =\frac{1}{2}\text{e}^{-\lambda} + \frac{1}{2}\text{e}^{-\lambda} = \text{e}^{-\lambda} \).
故\( P(Z=k)= \begin{cases}
\frac{1}{2} \cdot \frac{\lambda^k}{k!}\text{e}^{-\lambda}, & k=1,2,3,\cdots \\
\text{e}^{-\lambda}, & k=0, \\
\frac{1}{2} \cdot \frac{\lambda^{-k}}{(-k)!}\text{e}^{-\lambda}, & k=-1,-2,-3,\cdots
\end{cases} \)

【解】(Ⅰ)似然函数为
$$L(x_1,x_2,\cdots,x_n;\sigma) = \prod_{i=1}^n f(x_i;\sigma) = \frac{1}{2^n \sigma^n}\text{e}^{-\frac{1}{\sigma}\sum_{i=1}^n |x_i|},-\infty < x_i < +\infty,i=1,2,\cdots$$
于是$\ln L = -n\ln 2 - n\ln \sigma - \frac{1}{\sigma}\sum_{i=1}^n |x_i|$.
令$\frac{\text{d}\ln L}{\text{d}\sigma} = -\frac{n}{\sigma} + \frac{1}{\sigma^2}\sum_{i=1}^n |x_i| = 0$,得$\sigma$的最大似然估计量为$\hat{\sigma} = \frac{1}{n}\sum_{i=1}^n |X_i|$.

(Ⅱ)$E(\hat{\sigma}) = \frac{1}{n}\sum_{i=1}^n E(|X_i|) = E(|X|) = \int_{-\infty}^{+\infty} |x| \cdot \frac{1}{2\sigma}\text{e}^{-\frac{|x|}{\sigma}}\text{d}x = \int_{0}^{+\infty} \frac{x}{\sigma}\text{e}^{-\frac{x}{\sigma}}\text{d}x$
$= -\int_{0}^{+\infty} x\text{d}\text{e}^{-\frac{x}{\sigma}} = -x\text{e}^{-\frac{x}{\sigma}}\big|_0^{+\infty} + \int_{0}^{+\infty} \text{e}^{-\frac{x}{\sigma}}\text{d}x = \sigma$.

$D(\hat{\sigma}) = \frac{1}{n^2}\sum_{i=1}^n D(|X_i|) = \frac{D(|X|)}{n} = \frac{E[|X|^2] - E^2(|X|)}{n}$
$= \frac{1}{n}\left(\int_{-\infty}^{+\infty} |x|^2 \cdot \frac{1}{2\sigma}\text{e}^{-\frac{|x|}{\sigma}}\text{d}x - \sigma^2\right) = \frac{1}{n}\left(\int_{0}^{+\infty} \frac{x^2}{\sigma}\text{e}^{-\frac{x}{\sigma}}\text{d}x - \sigma^2\right)$
$= \frac{1}{n}\left(\int_{0}^{+\infty} -x^2\text{d}\text{e}^{-\frac{x}{\sigma}} - \sigma^2\right)$
$= \frac{1}{n}\left(\int_{0}^{+\infty} 2x\text{e}^{-\frac{x}{\sigma}}\text{d}x - \sigma^2\right) = \frac{1}{n}(2\sigma^2 - \sigma^2) = \frac{\sigma^2}{n}$.