部分习题参考答案

原书第 220 页

部分习题参考答案

习题一

(1) $ 1 $; (2) $ \sin 2x - \cos^{2}x $; (3) 0; (4) 48; (5) $ (x + 2y)(x - y)^{2} $.

(1) $ x_{1}=6, x_{2}=2, x_{3}=7 $; (2) $ x_{1}=3, x_{2}=6, x_{3}=15 $.

(1) i = 3, j = 4; (2) i = 3, j = 5.

$ \frac{n(n-1)}{2}-m. $

(1) 逆序数为 13，奇；(2) 逆序数为 $ \frac{n(n-1)}{2} $，当 n = 4k 或者 4k + 1 时为偶，当 n = 4k + 2 或者 4k + 3 时为奇；(3) 逆序数为 $ \frac{n(3n-1)}{2} $，当 n = 4k 或者 n = 4k + 3 时为偶，当 n = 4k + 1 或者 n = 4k + 2 时为奇.

$a_{12}a_{23}a_{35}a_{41}a_{54}, -a_{12}a_{24}a_{35}a_{41}a_{53}$.

(1) $ abe(d - c) $; (2) 0; (3) $ (a^2 - b^2)^2 $; (4) $ (-1)^{\frac{(n-2)(n-1)}{2}} a_{11} a_{2n} a_{3,n-1} \cdots a_{n2} $; (5) $ (-1)^{n-1} a_n $.

(1) 5 300; (2) 68; (3) 160; (4) -16 772 800; (5) $ (-1)^{\frac{n(n-1)}{2}}(a-x)^{n-1}[(n-1)x+a] $;

(6) 当 a=0 时, $ D=\{\begin{aligned}&x,&n=1,\\&x^{2},&n=2,\\&0,&n\geqslant3,\end{aligned}. $; 当 $ a\neq0 $ 时, $ D=\{\begin{aligned}&x,&n=1,\\&D=(-1)^{\frac{n(n-1)}{2}}&n\geqslant2,\\&a^{n-2}[a^{2}-(n-1)x^{2}],\end{aligned}. $

(1) 1 960; (2) $ a_{11}a_{22}a_{33}a_{44} - a_{11}a_{22}a_{34}a_{43} - a_{12}a_{21}a_{33}a_{44} + a_{12}a_{21}a_{34}a_{43} $;

$$ x^{n}+(-1)^{n+1}y^{n};(4)\begin{cases}\frac{\left[(x+\sqrt{x^{2}-4yz})^{n+1}-(x-\sqrt{x^{2}-4yz})^{n+1}\right]}{2^{n+1}\sqrt{x^{2}-4yz}},&x^{2}\neq4yz,\ $ n+1)\left(\frac{x}{2}\right)^{n},&x^{2}=4yz.\end{cases} $$

60(11- $ \sqrt{2} $).

$ (-1)^{\frac{n(n+1)}{2}} \prod_{i=1}^{n} i! $

(1) $ x_{1} = -\frac{3}{25} $, $ x_{2} = \frac{8}{5} $, $ x_{3} = \frac{29}{25} $; (2) $ x_{1} = \frac{1}{3} $, $ x_{2} = \frac{1}{3} $, $ x_{3} = -\frac{2}{3} $, $ x_{4} = 0 $;

(3) $ x_{i} = 0 $, $ i = 1, 2, 3, 4, 5 $.

原书第 221 页

$ \lambda = 1 $ 或 $ \frac{1}{2} $.

习题二

$$ \begin{pmatrix}7&6&4&6\\5&7&10&4\\4&8&8&5\\9&8&6&7\end{pmatrix};(2)\begin{pmatrix}26&24\\24&21\end{pmatrix}; $$

(3)

$$ \begin{pmatrix}a^{2}+b^{2}+c^{2}&ac+ab+bc&a+b+c&a+b+c\\ac+ab+bc&a^{2}+b^{2}+c^{2}&a+b+c&a+b+c\\a+b+c&a+b+c&3&3\end{pmatrix}; $$

(4)

$$ \begin{pmatrix}\lambda^{3}&3\lambda^{2}&3\lambda\\0&\lambda^{3}&3\lambda^{2}\\0&0&\lambda^{3}\end{pmatrix},\begin{pmatrix}\lambda^{n}&C_{n}^{1}\lambda^{n-1}&C_{n}^{2}\lambda^{n-2}\\0&\lambda^{n}&C_{n}^{1}\lambda^{n-1}\\0&0&\lambda^{n}\end{pmatrix}; $$

(5)

$$ \begin{pmatrix}\cos n\alpha&-\sin n\alpha\\\sin n\alpha&\cos n\alpha\end{pmatrix} $$

$$ \begin{pmatrix}a_{1}b_{11}&a_{1}b_{1\dot{2}}&\cdots&a_{1}b_{1p}\\a_{2}b_{21}&a_{2}b_{22}&\cdots&a_{2}b_{2p}\\\vdots&\vdots&&\vdots\\a_{n}b_{n1}&a_{n}b_{n2}&\cdots&a_{n}b_{np}\end{pmatrix}. $$

$$ \boldsymbol{A}^{2}=\begin{pmatrix}0&0&1\\-2&0&1\\-2&-2&1\end{pmatrix},\boldsymbol{A}^{3}=\begin{pmatrix}-2&0&1\\-2&-2&1\\-2&-2&-1\end{pmatrix},f(\boldsymbol{A})=\begin{pmatrix}0&-2&-2\\4&0&-4\\8&4&-4\end{pmatrix}; $$

(2)

$$ \boldsymbol{A}^{5}=\begin{pmatrix}-2&-2&-1\\2&-2&-3\\6&2&-5\end{pmatrix},\boldsymbol{A}^{6}=\begin{pmatrix}2&-2&-3\\6&2&-5\\10&6&-3\end{pmatrix},g(\boldsymbol{A})=\begin{pmatrix}11&-2&-6\\12&11&-8\\16&12&3\end{pmatrix}. $$

$ \begin{pmatrix}a&b\\-2b&a-b\end{pmatrix} $，其中a,b为任意复数.

$$ \begin{pmatrix}1&3&5&-1\\0&-7&-13&6\\0&0&0&-4\\0&0&0&0\end{pmatrix};(2)\begin{pmatrix}1&0&0&0&-26\\0&1&0&0&-\frac{203}{2}\\0&0&1&0&-\frac{19}{2}\\0&0&0&1&\frac{285}{2}\end{pmatrix}. $$

所有的矩阵都是满秩.

原书第 222 页

(1)

$$ \begin{pmatrix}\frac{1}{2}&0&0\\0&\frac{5}{13}&\frac{1}{13}\\0&-\frac{3}{13}&\frac{2}{13}\end{pmatrix};(2)\begin{pmatrix}\frac{31}{224}&-\frac{1}{56}&\frac{3}{32}&\frac{17}{112}\\\frac{11}{56}&\frac{1}{14}&-\frac{1}{8}&-\frac{3}{28}\\\frac{1}{224}&\frac{9}{56}&-\frac{3}{32}&\frac{15}{112}\\\frac{17}{56}&-\frac{1}{14}&-\frac{3}{8}&\frac{3}{28}\end{pmatrix}; $$

(3)

$$ \begin{pmatrix}-\frac{5}{4}&-\frac{1}{4}&\frac{1}{4}&\frac{7}{2}\\\frac{9}{8}&\frac{1}{8}&-\frac{13}{8}&-\frac{3}{4}\\-\frac{1}{8}&-\frac{1}{8}&\frac{5}{8}&-\frac{1}{4}\\\frac{1}{8}&\frac{1}{8}&\frac{3}{8}&-\frac{3}{4}\end{pmatrix};(4)\frac{1}{3}\begin{pmatrix}-2&1&1&1\\1&-2&1&1\\1&1&-2&1\\1&1&1&-2\end{pmatrix}; $$

(5)

$$ )\frac{1}{11}\begin{pmatrix}-11&-11&11&0\\-19&-18&16&2\\15&20&-8&-1\\30&18&-27&-2\end{pmatrix};(6)\frac{1}{54}\begin{pmatrix}9&0&-63&-162\\2&6&-20&-42\\-3&72&3&-18\\1&-24&17&60\end{pmatrix}. $$

16.

$$ \left(\begin{array}{c c c c c c}{0}&{0}&{0}&{\cdots}&{0}&{\frac{1}{a_{n}}}\\ {\frac{1}{a_{1}}}&{0}&{0}&{\cdots}&{0}&{0}\\ {0}&{\frac{1}{a_{2}}}&{0}&{\cdots}&{0}&{0}\\ {\vdots}&{\vdots}&{\vdots}&&{\vdots}&{\vdots}\\ {0}&{0}&{0}&{\cdots}&{\frac{1}{a_{n-1}}}&{0}\end{array}\right). $$

17.

$$ \begin{pmatrix}{{{1}}}&{{{-1}}}&{{{-1}}}&{{{0}}}&{{{\cdots}}}&{{{0}}}&{{{0}}}&{{{0}}} \\{{{1}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{\cdots}}}&{{{0}}}&{{{0}}}&{{{0}}} \\{{{0}}}&{{{1}}}&{{{1}}}&{{{-1}}}&{{{\cdots}}}&{{{0}}}&{{{0}}}&{{{0}}} \\{{{\vdots}}}&{{{\vdots}}}&{{{\vdots}}}&{{{\vdots}}}&{{{\vdots}}}&{{{\vdots}}}&{{{\vdots}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{\cdots}}}&{{{1}}}&{{{1}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{\cdots}}}&{{{0}}}&{{{1}}}&{{{2}}}\end{pmatrix}. $$

(1) $ x_1 = -\frac{17}{20} $, $ x_2 = \frac{1}{6} $, $ x_3 = \frac{9}{10} $; (2) $ x_1 = -\frac{75}{61} $, $ x_2 = -\frac{50}{61} $, $ x_3 = \frac{76}{61} $.

$$ \begin{pmatrix}{{{-1}}}&{{{-11}}}&{{{5}}} \\{{{-10}}}&{{{-72}}}&{{{33}}} \\{{{-24}}}&{{{-159}}}&{{{73}}}\end{pmatrix};(2)\begin{pmatrix}{{{0}}}&{{{1}}}&{{{0}}} \\{{{0}}}&{{{0}}}&{{{1}}} \\{{{1}}}&{{{0}}}&{{{0}}}\end{pmatrix}. $$

原书第 223 页

20.

$$ \begin{pmatrix}2&13&-6&0&0&0\\5&28&-13&0&0&0\\1&-5&2&0&0&0\\0&0&0&11&14&0\\0&0&0&7&18&0\\0&0&0&0&0&9\end{pmatrix}. $$

21.

$$ \begin{pmatrix}\boldsymbol{E}_{k}&-\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{E}_{l}\end{pmatrix}. $$

22.

$$ \begin{pmatrix}-\boldsymbol{B}^{-1}\boldsymbol{C}\boldsymbol{A}^{-1}&\boldsymbol{B}^{-1}\\\boldsymbol{A}^{-1}&\boldsymbol{0}\end{pmatrix}. $$

习题三

(1) a 与 b 垂直; (2) a 与 b 同向.

利用向量的线性运算证明.

点 A 关于 xOy 面对称点的坐标为 $ (2,4,1) $; 点 B 关于 y 轴对称点的坐标为 $ (2,4,-1) $.

$ 4i + j - 2k $

$$ \left|\boldsymbol{a}\right|=3,\cos\alpha=\frac{1}{3},\cos\beta=\frac{2}{3},\cos\gamma=-\frac{2}{3}. $$

(1) 20; (2) -6.

$ \left|r\right| = \sqrt{n^{2} + l^{2} + m^{2}} $.

$ a \cdot b = 6 $, 夹角为 $ \arccos \frac{1}{5} $.

(1) $ -3(a \times b) $; (2) $ -7(a \times b) $.

10.

$$ \frac{15}{2}. $$

$ x = -10i + 5j + 5k $.

(1) 不共线; (2) 共线.

验证三个向量 $ \overrightarrow{AB} $, $ \overrightarrow{AC} $, $ \overrightarrow{AD} $ 的混合积为 0.

(1) $ 3i - 2j + 5k $; (2) i - j.

$ 2x + 9y - 6z - 121 = 0 $.

x - 3y - 2z = 0.

3x - 7y + 5z - 4 = 0.

$ x + y - 3z - 4 = 0 $

9y - z - 2 = 0.

2x - y = 0.

(1) C = -6 且 $ D \neq -3 $; (2) C = -6 且 D = -3.

$ 2x + 3y + z - 6 = 0 $.

原书第 224 页

利用平面的三点式方程和行列式的性质证明.

(1) $ \frac{x-1}{3} = \frac{y}{1} = \frac{z+2}{2} $; (2) $ \frac{x-1}{0} = \frac{y-1}{1} = \frac{z-3}{4} $;

(3) $ \frac{x-2}{-1} = \frac{y-3}{3} = \frac{z+5}{4} $; (4) $ \frac{x-3}{-2} = \frac{y-1}{1} = \frac{z-2}{1} $.

(1) 相交，交点为 $ (3, -1, -2) $; (2) 直线在平面上.

(1) $ 2x - 4y - z + 5 = 0 $; (2) x - y - z + 1 = 0; (3) $ 4x + 8y - 2z + 1 = 0 $.

2x - 16y - 13y + 31 = 0.

29.0.

(1) 2; (2) $ \frac{5}{6} $.

$ x + 2y - 2z - 7 = 0 $ 或 $ x + 2y - 2z + 5 = 0 $.

习题四

(1) $ (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (1, 1, 1, 1, 2) $; (2) $ (x_{1}, x_{2}, x_{3}, x_{4}) = (1, -1, -1, 1) $.

$ \beta = \frac{5}{4}\alpha_1 + \frac{1}{4}\alpha_2 - \frac{1}{4}\alpha_3 - \frac{1}{4}\alpha_4 $.

$$ \begin{pmatrix}\frac{5}{8}&\frac{1}{8}&-\frac{1}{8}\\\frac{3}{2}&\frac{1}{2}&\frac{3}{2}\\\frac{1}{8}&\frac{5}{8}&\frac{3}{8}\end{pmatrix}. $$

(1) $ \begin{pmatrix} -1 & -1 & 2 \\ -1 & 2 & -1 \\ 3 & 1 & 2 \end{pmatrix} $;

(2) $ (1,-1,1) $, $ \left(\frac{1}{9},-\frac{2}{9},\frac{4}{9}\right) $.

(33, -82, 154).

$ \gamma = (1,1,1,-1) $.

$ \left(\begin{array}{ccc}\frac{1}{2}&2&0\\\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}\\0&\frac{1}{2}&-\frac{1}{2}\end{array}\right) $, (1,1,3).

(2) $ \xi_{1}, \xi_{2}, \xi_{4} $ 为极大线性无关组.

秩为3，且 $ \alpha_{1}, \alpha_{2}, \alpha_{3} $ 为一个极大线性无关组.

(1) $ \eta_{1}=(-2,1,0,0) $, $ \eta_{2}=(1,0,0,1) $ 为一个基础解系. 通解为 $ \eta=k_{1}(-2,1,0,0)+k_{2}(1,0,0,1) $，其中 $ k_{1}, k_{2} $ 为任意常数;

原书第 225 页

(2) $ \eta_{1}=(1,-2,1,0,0),\eta_{2}=(1,-2,0,1,0),\eta_{3}=(5,-6,0,0,1) $为一个基础解系.通解为 $ \eta=k_{1}(1,-2,1,0,0)+k_{2}(1,-2,0,1,0)+k_{3}(5,-6,0,0,1) $，其中 $ k_{1},k_{2},k_{3} $为任意常数；

(3) $ \eta_{1}=(-3,2,1,0,0),\eta_{2}=(-5,3,0,0,1) $为一个基础解系.通解为 $ \eta=k_{1}(-3,2,1,0,0)+k_{2}(-5,3,0,0,1) $，其中 $ k_{1},k_{2} $为任意常数.

$ \eta_{1}=\left(-\frac{11}{7},-\frac{1}{7},1,0\right) $， $ \eta_{2}=(0,2,0,1) $为一个基础解系。取特解 $ \gamma_{0}=\left(\frac{1}{7},\frac{2}{7},0,0\right) $，则通解为 $ \gamma=\left(\frac{1}{7},\frac{2}{7},0,0\right)+k_{1}\left(-\frac{11}{7},-\frac{1}{7},1,0\right)+k_{2}(0,2,0,1) $，其中 $ k_{1} $， $ k_{2} $为任意常数。

(1) 通解为 $ \gamma=\left(\frac{5}{7},-\frac{25}{7},\frac{69}{7},0,0\right)+k_{1}\left(-\frac{6}{7},\frac{2}{7},-\frac{3}{7},1,0\right)+k_{2}\left(-\frac{1}{7},\frac{5}{7},-\frac{11}{7},0,1\right) $，其中 $ k_{1}, k_{2} $ 为任意常数；

(2) 通解为 $ \gamma = (-16, 23, 0, 0, 0) + k_1 (1, -2, 1, 0, 0) + k_2 (1, -2, 0, 1, 0) + k_3 (5, -6, 0, 0, 1) $，其中 $ k_1, k_2, k_3 $ 为任意常数；

(3) 通解为 $ \gamma = \left( \frac{20}{9}, -\frac{5}{3}, -\frac{1}{9}, 0, 0 \right) + k_{1} \left( 1, -\frac{5}{2}, 0, 1, 0 \right) + k_{2} \left( -\frac{53}{18}, \frac{5}{6}, \frac{2}{9}, 0, 1 \right) $，其中 $ k_{1}, k_{2} $ 为任意常数；

(4) 通解为 $ \gamma = \left( \frac{1}{8}, -\frac{1}{4}, 0, 1 \right) + k\left( -\frac{9}{8}, \frac{1}{4}, 1, 0 \right) $，其中 k 为任意常数.

$ t \neq -2 $ 时无解；t = -2, p \neq -8 有解，通解为 $ \gamma = (-1,1,0,0) + k(-1,-2,0,1) $，其中 k 为任意常数；t = -2, p = -8 有解，通解为 $ \gamma = (-1,1,0,0) + k_{1}(4,-2,1,0) + k_{2}(-1,2,0,1) $，其中 $ k_{1}, k_{2} $ 为任意常数.

当 s 为偶数时， $ t_{1} \neq \pm t_{2} $; 当 s 为奇数时， $ t_{1} \neq -t_{2} $.

(1) 当 $ \lambda = 1 $ 时，有无穷多解， $ \gamma = (1,0,0) + k_1(-1,1,0) + k_2(-1,0,1) $，其中 $ k_1, k_2 $ 为任意常数；当 $ \lambda = -2 $ 时，无解；当 $ \lambda \neq 1 $ 且 $ \lambda \neq -2 $ 时，有唯一解 $ x_1 = \frac{1}{\lambda + 2} $， $ x_2 = \frac{1}{\lambda + 2} $， $ x_3 = \frac{1}{\lambda + 2} $。

(2) 当 $ \lambda = 0 $ 或 -3 时，无解；当 $ \lambda \neq 0 $ 且 $ \lambda \neq -3 $ 时，有唯一解 $ x_{1} = \frac{-\lambda^{2} + 2}{\lambda^{2} + 3\lambda} $， $ x_{2} = \frac{2\lambda - 1}{\lambda^{2} + 3\lambda} $， $ x_{3} = \frac{\lambda^{3} + 2\lambda^{2} - \lambda - 1}{\lambda^{2} + 3\lambda} $.

(3) 当 $ \lambda = 1 $ 时，有无穷多解， $ \gamma = (1,0,0) + k_{1}(-1,1,0) + k_{2}(-1,0,1) $，其中 $ k_{1}, k_{2} $ 为任意常数；当 $ \lambda = 3 $ 时，无解；当 $ \lambda \neq 1 $ 且 $ \lambda \neq 3 $ 时，有唯一解 $ x_{1} = -1 $， $ x_{2} = \frac{\lambda - 4}{\lambda - 3} $， $ x_{3} = -\frac{1}{\lambda - 3} $。

在有解的情况下，一般解为 $ \gamma=(a_{1}+a_{2}+a_{3}+a_{4},a_{2}+a_{3}+a_{4},a_{3}+a_{4},a_{4},0)+k(1,1,1,1,1) $，其中 k 为任意常数.

原书第 226 页

习题五

(1) 属于特征值 1 的特征向量为 $ k_{1}(0,1,1)^{\mathrm{T}} $ ( $ k_{1} \neq 0 $)；属于特征值 2（二重）的特征向量为 $ k_{2}(0,1,0)^{\mathrm{T}} + k_{3}(1,0,1)^{\mathrm{T}} $ ( $ k_{2}, k_{3} $ 为任意非零常数);

(2) 属于特征值 -4 的特征向量为 $ k_{1}(1, -2, 3)^{\mathrm{T}} $ ( $ k_{1} $ 为任意非零常数); 属于特征值 2 (二重) 的特征向量为 $ k_{2}(1, 0, 1)^{\mathrm{T}} + k_{3}(0, 1, 2)^{\mathrm{T}} $ ( $ k_{2}, k_{3} $ 为任意非零常数);

(3) 属于特征值 0 的特征向量为 $ k_{1}(17,7,-5)^{\mathrm{T}} $ ( $ k_{1} \neq 0 $); 属于特征值 1 (二重) 的特征向量为 $ k_{2}(2,1,-1)^{\mathrm{T}} $ ( $ k_{2} $ 为任意非零常数);

（4）属于特征值1（二重）的特征向量为 $ k_{1}(1,0,0,1)^{\mathrm{T}}+k_{2}(0,1,1,0)^{\mathrm{T}} $， $ (k_{1},k_{2} $ 为任意非零常数 $ ) $；属于特征值-1（二重）的特征向量为 $ k_{3}(1,0,0,-1)^{\mathrm{T}}+k_{4}(0,1,-1,0)^{\mathrm{T}} $， $ (k_{3},k_{4} $ 为任意非零常数 $ ) $

(1) 属于特征值 n-1 的特征向量为 $ k_{1}(1,1,1,\cdots,1)^{\mathrm{T}} $ ( $ k_{1} $ 为任意非零常数); 属于特征值 -1 (n-1 重) 的特征向量为 $ k_{2}(1,-1,0,\cdots,0)^{\mathrm{T}} + k_{3}(1,0,-1,0,\cdots,0)^{\mathrm{T}} + \cdots + k_{n}(1,0,0,0,\cdots,-1)^{\mathrm{T}} $ ( $ k_{i} $ 为任意非零常数, i=2,3,\cdots,n);

(2) 属于特征值 n 的特征向量为 $ k_{1}(1,1,1,\cdots,1)^{\mathrm{T}} $ ( $ k_{1} $ 为任意非零常数); 属于特征值 0 (n-1 重) 的特征向量为 $ k_{2}(1,-1,0,\cdots,0)^{\mathrm{T}}+k_{3}(1,0,-1,0,\cdots,0)^{\mathrm{T}}+\cdots+k_{n}(1,0,0,0,\cdots,-1)^{\mathrm{T}} $ ( $ k_{i} $ 为任意非零常数, i=2,\cdots,n).

$$ 4.\ \boldsymbol{A}^{k}=\begin{pmatrix}1&2\cdot5^{k-1}[1+(-1)^{k+1}]&5^{k-1}[4+(-1)^{k}]-1\\0&5^{k-1}[1+4\cdot(-1)^{k}]&2\cdot5^{k-1}[1+(-1)^{k+1}]\\0&2\cdot5^{k-1}[1+(-1)^{k+1}]&5^{k-1}[4+(-1)^{k}]\end{pmatrix}. $$

特征值为 3, 9, 9. 特征向量为 $ k_{1}(0,1,1)^{\mathrm{T}} $ ( $ k_{1} \neq 0 $); $ k_{2}(-1,1,0)^{\mathrm{T}} + k_{3}(1,1,-1)^{\mathrm{T}} $ ( $ k_{2}, k_{3} $ 为任意非零常数).

若 A 仅与自身相似，则对任意的可逆矩阵 P，有 $ P^{-1}AP = A $，即 AP = PA。这样有 A = kE (k 为任意非零常数).

(1) $ T = \begin{pmatrix} -1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix} $, $ T^{-1}AT = \mathrm{diag}(1, 2, 2) $;

$$ \boldsymbol{T}=\begin{pmatrix}0&1&1\\1&0&0\\0&1&-1\end{pmatrix},\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=diag(1,1,-1); $$

(3) 特征值为 -1 (三重)，线性无关特征向量只有一个，不可对角化；

$$ \boldsymbol{T}=\left(\begin{matrix}{2}&{\cdot}&{1+\mathrm{i}}&{1-\mathrm{i}}\\ {4}&{2+3\mathrm{i}}&{2-3\mathrm{i}}\\ {-1}&{-\mathrm{i}}&{\mathrm{i}}\\ \end{matrix}\right),\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=\mathrm{d i a g}(1,\mathrm{i},-\mathrm{i}). $$

$$ 10.a=-2,b=2. $$

原书第 227 页

$$ \begin{array}{r}{11.\;t=1,\boldsymbol{T}=\left(\begin{array}{r r r}{1}&{1}&{1}\\ {-1}&{0}&{1}\\ {0}&{-1}&{1}\end{array}\right),\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=\mathrm{d i a g}(1,1,4).}\end{array} $$

$$ \begin{array}{r}{\boldsymbol{T}=\left(\begin{array}{c c c}{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{6}}&{\frac{2}{3}}\\ {0}&{-\frac{2\sqrt{2}}{3}}&{\frac{1}{3}}\\ {-\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{6}}&{\frac{2}{3}}\end{array}\right),\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=\mathrm{d i a g}(2,2,11);}\end{array} $$

$$ \begin{array}{r}{\boldsymbol{T}=\left(\begin{array}{c c c}{0}&{\displaystyle\frac{2\sqrt{2}}{3}}&{\displaystyle\frac{1}{3}}\\ {\displaystyle\frac{\sqrt{2}}{2}}&{-\displaystyle\frac{\sqrt{2}}{6}}&{\displaystyle\frac{2}{3}}\\ {\displaystyle\frac{\sqrt{2}}{2}}&{\displaystyle\frac{\sqrt{2}}{6}}&{-\frac{2}{3}}\end{array}\right),\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=\mathrm{d i a g}(1,1,10);}\end{array} $$

$$ \begin{aligned}\boldsymbol{T}&=\begin{pmatrix}\frac{\sqrt{2}}{2}&\frac{\sqrt{6}}{6}&-\frac{\sqrt{3}}{6}&-\frac{1}{2}\\\frac{\sqrt{2}}{2}&-\frac{\sqrt{6}}{6}&\frac{\sqrt{3}}{6}&\frac{1}{2}\\0&\frac{\sqrt{6}}{3}&\frac{\sqrt{3}}{6}&\frac{1}{2}\\0&0&\frac{\sqrt{3}}{2}&-\frac{1}{2}\end{pmatrix},\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=diag(3,3,3,7);\end{aligned} $$

$$ \begin{aligned}\boldsymbol{T}&=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}\end{pmatrix},\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=diag(1,-1,-3,7).\end{aligned} $$

$$ 16.a=-2;\boldsymbol{T}=\begin{pmatrix}\frac{\sqrt{3}}{3}&\frac{\sqrt{2}}{2}&\frac{\sqrt{6}}{6}\\\frac{\sqrt{3}}{3}&0&-\frac{\sqrt{6}}{3}\\\frac{\sqrt{3}}{3}&-\frac{\sqrt{2}}{2}&\frac{\sqrt{6}}{6}\end{pmatrix},\boldsymbol{T}^{-1}\boldsymbol{A}\boldsymbol{T}=diag(0,3,-3). $$

习题六

(1) $ f=(x_{1},x_{2},x_{3}) $ $ \begin{pmatrix}a^{2}&ab&ac\\ab&b^{2}&bc\\ac&bc&b^{2}\end{pmatrix} $ $ \begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix} $;

原书第 228 页

(2)

$$ f=\left(x_{1},x_{2},x_{3}\right)\begin{pmatrix}1&-1&-2\\-1&-1&\frac{3}{2}\\-2&\frac{3}{2}&-7\end{pmatrix}\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}; $$

(3)

$$ f=\left(x_{1},x_{2},x_{3},x_{4}\right)\left(\begin{array}{r r r r}{0}&{0}&{-1}&{0}\\ {0}&{0}&{3}&{0}\\ {-1}&{3}&{0}&{0}\\ {0}&{0}&{0}&{2}\end{array}\right)\left(\begin{matrix}{x_{1}}\\ {x_{2}}\\ {x_{3}}\\ {x_{4}}\end{matrix}\right); $$

(4)

$$ f=\left(x_{1},x_{2},x_{3},x_{4}\right)\left(\begin{array}{r r r r}{1}&{-1}&{2}&{3}\\ {-1}&{2}&{-2}&{0}\\ {2}&{-2}&{3}&{-1}\\ {3}&{0}&{-1}&{4}\end{array}\right)\left(\begin{matrix}{x_{1}}\\ {x_{2}}\\ {x_{3}}\\ {x_{4}}\end{matrix}\right). $$

$$ 2.\ \left(1\right)f=x_{1}^{2}+2x_{2}^{2}-3x_{3}^{2}+4x_{1}x_{3}+6x_{2}x_{3};\left(2\right)f=-4x_{1}^{2}+2x_{3}^{2}+6x_{1}x_{2}+4x_{1}x_{3}+12x_{2}x_{3}. $$

$$ y_{1}^{2}-3y_{2}^{2}+\frac{7}{3}y_{3}^{2};(2)y_{1}^{2}-3y_{2}^{2}-\frac{2}{3}y_{3}^{2};(3)z_{1}^{2}-z_{2}^{2}+\frac{1}{4}z_{3}^{2}-z_{4}^{2}. $$

$$ \left(1\right)f=y_{1}^{2}+y_{2}^{2}-y_{3}^{2},\left(\begin{matrix}x_{1}\\ x_{2}\\ x_{3}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\\ 0&1&0\\ \frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\end{matrix}\right)\left(\begin{matrix}y_{1}\\ y_{2}\\ y_{3}\end{matrix}\right); $$

$$ (2)f=3y_{1}^{2}+6y_{2}^{2}+9y_{3}^{2},\begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\end{pmatrix}=\frac{1}{3}\begin{pmatrix}-2&-1&2\\ -2&2&-1\\ 1&2&2\end{pmatrix}\begin{pmatrix}y_{1}\\ y_{2}\\ y_{3}\end{pmatrix}; $$

$$ (3)f=-3y_{2}^{2}+2y_{3}^{2}+2y_{4}^{2},\begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}&0\\\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}&0\\0&\frac{1}{\sqrt{5}}&0&-\frac{2}{\sqrt{5}}\\0&\frac{2}{\sqrt{5}}&0&\frac{1}{\sqrt{5}}\end{pmatrix}\begin{pmatrix}y_{1}\\ y_{2}\\ y_{3}\\ y_{4}\end{pmatrix}. $$

(1) 2,1; (2) 1,2; (3) 3,3.

(1) 正定; (2) $ -\frac{4}{5} \lt \lambda \lt 0 $ 时, 正定; (3) 不论 $ \lambda $ 取何值, 二次型均不正定; (4) 正定.

正定.

利用定理 2.2.

9.（1）a＞2时正定；（2）a＜-1时负定.

10—13. 用定义或定理 2.1.

球心坐标为 $ (1, -2, 0) $，半径为 3.

(1) 母线平行于 z 轴的椭圆柱面;

原书第 229 页

(2) 母线平行于 x 轴的双曲柱面；

(3) 母线平行于 z 轴的抛物柱面.

(1) $ y^{2} + z^{2} = 2x $; (2) $ \frac{x^{2}}{9} + \frac{z^{2}}{9} + \frac{y^{2}}{4} = 1 $.

表示过点 $ (0,3,0) $，平行于 z 轴的直线.

$ 2\left(x-\frac{1}{2}\right)^{2}+y^{2}=\frac{11}{2} $

24.（1）椭球面；（2）单叶双曲面；（3）二次锥面；（4）双曲抛物面；（5）椭圆抛物面.

(1) $ 4x_{1}^{2}-4y_{1}^{2}-8z_{1}^{2}=5 $，双叶双曲面；

(2) $ x_{1}^{2}-y_{1}^{2}=1 $，双曲柱面；

(3) $ x_{1}^{2}+y_{1}^{2}-2z_{1}^{2}=1 $，单叶双曲面；

(4) $ 2x_{1}^{2}+4y_{1}^{2}+4z_{1}^{2}=1 $，椭球面.

习题七

(1) 是; (2) 不是; (3) 是.

(1) $ \left(\frac{5}{4}, \frac{1}{4}, -\frac{1}{4}, -\frac{1}{4}\right) $; (2) $ (1, 0, -1, 0) $.

$$ \left(\begin{array}{r r r r}{{2}}&{{0}}&{{5}}&{{6}}\\ {{1}}&{{3}}&{{3}}&{{6}}\\ {{-1}}&{{1}}&{{2}}&{{3}}\\ {{1}}&{{0}}&{{1}}&{{3}}\end{array}\right);\left(2\right)\left(\begin{array}{r r r r}{{1}}&{{0}}&{{0}}&{{1}}\\ {{1}}&{{1}}&{{0}}&{{1}}\\ {{0}}&{{1}}&{{1}}&{{1}}\\ {{0}}&{{0}}&{{1}}&{{0}}\end{array}\right). $$

4.（1）是；(2) 是.

$$ \boldsymbol{A}=\begin{pmatrix}a_{11}&0&a_{21}&0\\0&a_{11}&0&a_{21}\\a_{12}&0&a_{22}&0\\0&a_{12}&0&a_{22}\end{pmatrix}. $$

(1) $ \begin{pmatrix} a_{33} & a_{31} & a_{32} \\ a_{13} & a_{11} & a_{12} \\ a_{23} & a_{21} & a_{22} \end{pmatrix} $; (2) $ \begin{pmatrix} a_{11} & 5a_{12} & -6a_{13} \\ \frac{1}{5}a_{22} & -\frac{6}{5}a_{11} & a_{12} \\ -\frac{1}{6}a_{31} & -\frac{5}{6}a_{32} & a_{23} \end{pmatrix} $;

(3)

$$ \begin{pmatrix}a_{11}+a_{12}&a_{12}&a_{13}\\a_{21}+a_{22}-a_{11}-a_{12}&a_{22}-a_{12}&a_{23}-a_{13}\\a_{31}+a_{32}&a_{32}&a_{33}\end{pmatrix}. $$

原书第 230 页

$$ \begin{pmatrix}0&1&0&\cdots&0\\0&0&1&\cdots&0\\\vdots&\vdots&\vdots&&\vdots\\0&0&0&\cdots&1\\0&0&0&\cdots&0\end{pmatrix}. $$

(1) $ \begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{pmatrix} $; (2) $ (0, -12, 7) $.

$$ \begin{pmatrix}1&3&6\\-2&-1&0\\1&0&2\end{pmatrix}. $$