线性代数（第六版）同济大学习题一（1-4题）个人解答

线性代数（第六版）同济大学习题一（1-4题）

1. 利用对角线法则计算下列三阶行列式： \begin{aligned}&1. \ 利用对角线法则计算下列三阶行列式：&\end{aligned} 1. 利用对角线法则计算下列三阶行列式：

( 1 ) ∣ 2 0 1 1 − 4 − 1 − 1 8 3 ∣ ； ( 2 ) ∣ a b c b c a c a b ∣ ； ( 3 ) ∣ 1 1 1 a b c a 2 b 2 c 2 ∣ ； ( 4 ) ∣ x y x + y y x + y x x + y x y ∣ . \begin{aligned} &\ \ (1)\ \ \left|\begin{array}{cccc}2 &0 &1\\\\1 &-4 &-1\\\\-1 &8 &3\end{array}\right|；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \left|\begin{array}{cccc}a &b &c\\\\b &c &a\\\\c &a &b\end{array}\right|；\\\\ &\ \ (3)\ \ \left|\begin{array}{cccc}1 &1 &1\\\\a &b &c\\\\a^2 &b^2 &c^2\end{array}\right|；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ \left|\begin{array}{cccc}x &y &x+y\\\\y &x+y &x\\\\x+y &x &y\end{array}\right|. & \end{aligned} (1) 21−10−481−13 ； (2) abcbcacab ； (3) 1aa21bb21cc2 ； (4) xyx+yyx+yxx+yxy .

解：

( 1 ) ∣ 2 0 1 1 − 4 − 1 − 1 8 3 ∣ = 2 × ( − 4 ) × 3 + 0 × ( − 1 ) × ( − 1 ) + 1 × 1 × 8 − 2 × ( − 1 ) × 8 − 0 × 1 × 3 − 1 × ( − 4 ) × ( − 1 ) = − 24 + 0 + 8 − ( − 16 ) − 0 − 4 = − 4. ( 2 ) ∣ a b c b c a c a b ∣ = a c b + b a c + c b a − a a a − b b b − c c c = 3 a b c − a 3 − b 3 − c 3 ( 3 ) ∣ 1 1 1 a b c a 2 b 2 c 2 ∣ = b c 2 + c a 2 + a b 2 − c b 2 − a c 2 − a 2 b = c 2 ( b − a ) − c ( b + a ) ( b − a ) + a b ( b − a ) = ( b − a ) ( c 2 − c b − c a + a b ) = ( b − a ) ( c ( c − b ) − a ( c − b ) ) = ( b − a ) ( c − b ) ( c − a ) ( 4 ) ∣ x y x + y y x + y x x + y x y ∣ = x ( x + y ) y + y x ( x + y ) + ( x + y ) y x − x 3 − y 3 − ( x + y ) 3 = 3 x y ( x + y ) − ( x 3 + y 3 ) − ( x + y ) 3 = 3 x y ( x + y ) − ( x + y ) ( x 2 + y 2 − x y ) − ( x + y ) 3 = ( x + y ) ( 3 x y − x 2 − y 2 + x y − ( x 2 + 2 x y + y 2 ) ) = ( x + y ) ( 3 x y − x 2 − y 2 + x y − x 2 − 2 x y − y 2 ) = ( x + y ) ( 2 x y − 2 x 2 − 2 y 2 ) = − 2 ( x + y ) ( x 2 + y 2 − x y ) = − 2 ( x 3 + y 3 ) . \begin{aligned} &\ \ (1)\ \left|\begin{array}{cccc}2 &0 &1\\\\1 &-4 &-1\\\\-1 &8 &3\end{array}\right|=2\times(-4)\times3+0\times(-1)\times(-1)+1\times1\times8-2\times(-1)\times8-0\times1\times3-1\times(-4)\times(-1)\\\\ &\ \ \ \ \ \ \ \ =-24+0+8-(-16)-0-4=-4.\\\\ &\ \ (2)\ \left|\begin{array}{cccc}a &b &c\\\\b &c &a\\\\c &a &b\end{array}\right|=acb+bac+cba-aaa-bbb-ccc=3abc-a^3-b^3-c^3\\\\ &\ \ (3)\ \left|\begin{array}{cccc}1 &1 &1\\\\a &b &c\\\\a^2 &b^2 &c^2\end{array}\right|=bc^2+ca^2+ab^2-cb^2-ac^2-a^2b=c^2(b-a)-c(b+a)(b-a)+ab(b-a)\\\\ &\ \ \ \ \ \ \ \ =(b-a)(c^2-cb-ca+ab)=(b-a)(c(c-b)-a(c-b))=(b-a)(c-b)(c-a)\\\\ &\ \ (4)\ \left|\begin{array}{cccc}x &y &x+y\\\\y &x+y &x\\\\x+y &x &y\end{array}\right|=x(x+y)y+yx(x+y)+(x+y)yx-x^3-y^3-(x+y)^3\\\\ &\ \ \ \ \ \ \ \ =3xy(x+y)-(x^3+y^3)-(x+y)^3=3xy(x+y)-(x+y)(x^2+y^2-xy)-(x+y)^3\\\\ &\ \ \ \ \ \ \ \ =(x+y)(3xy-x^2-y^2+xy-(x^2+2xy+y^2))=(x+y)(3xy-x^2-y^2+xy-x^2-2xy-y^2)\\\\ &\ \ \ \ \ \ \ \ =(x+y)(2xy-2x^2-2y^2)=-2(x+y)(x^2+y^2-xy)=-2(x^3+y^3). & \end{aligned} (1) 21−10−481−13 =2×(−4)×3+0×(−1)×(−1)+1×1×8−2×(−1)×8−0×1×3−1×(−4)×(−1) =−24+0+8−(−16)−0−4=−4. (2) abcbcacab =acb+bac+cba−aaa−bbb−ccc=3abc−a3−b3−c3 (3) 1aa21bb21cc2 =bc2+ca2+ab2−cb2−ac2−a2b=c2(b−a)−c(b+a)(b−a)+ab(b−a) =(b−a)(c2−cb−ca+ab)=(b−a)(c(c−b)−a(c−b))=(b−a)(c−b)(c−a) (4) xyx+yyx+yxx+yxy =x(x+y)y+yx(x+y)+(x+y)yx−x3−y3−(x+y)3 =3xy(x+y)−(x3+y3)−(x+y)3=3xy(x+y)−(x+y)(x2+y2−xy)−(x+y)3 =(x+y)(3xy−x2−y2+xy−(x2+2xy+y2))=(x+y)(3xy−x2−y2+xy−x2−2xy−y2) =(x+y)(2xy−2x2−2y2)=−2(x+y)(x2+y2−xy)=−2(x3+y3).

2. 按自然数从小到大为标准次序，求下列各排列的逆序数： \begin{aligned}&2. \ 按自然数从小到大为标准次序，求下列各排列的逆序数：&\end{aligned} 2. 按自然数从小到大为标准次序，求下列各排列的逆序数：

( 1 ) 1234 ； ( 2 ) 4132 ； ( 3 ) 3421 ； ( 4 ) 2413 ； ( 5 ) 13 ⋅ ⋅ ⋅ ( 2 n − 1 ) 24 ⋅ ⋅ ⋅ ( 2 n ) ； ( 6 ) 13 ⋅ ⋅ ⋅ ( 2 n − 1 ) ( 2 n ) ( 2 n − 2 ) ⋅ ⋅ ⋅ 2. \begin{aligned} &\ \ (1)\ \ 1234；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ 4132；\\\\ &\ \ (3)\ \ 3421；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ 2413；\\\\ &\ \ (5)\ \ 13\cdot\cdot\cdot(2n-1)24\cdot\cdot\cdot(2n)；\ \ \ \ \ \ \ (6)\ \ 13\cdot\cdot\cdot(2n-1)(2n)(2n-2)\cdot\cdot\cdot2. & \end{aligned} (1) 1234； (2) 4132； (3) 3421； (4) 2413； (5) 13⋅⋅⋅(2n−1)24⋅⋅⋅(2n)； (6) 13⋅⋅⋅(2n−1)(2n)(2n−2)⋅⋅⋅2.

解：

( 1 ) t = ∑ t = 1 4 t i = 0 + 0 + 0 + 0 = 0. ( 2 ) t = ∑ t = 1 4 t i = 0 + 1 + 1 + 2 = 4. ( 3 ) t = ∑ t = 1 4 t i = 0 + 0 + 2 + 3 = 5. ( 4 ) t = ∑ t = 1 4 t i = 0 + 0 + 2 + 1 = 3. ( 5 ) t = ∑ t = 1 2 n t i = ( n − 1 ) + ( n − 2 ) + ⋅ ⋅ ⋅ + 0 = n ( n − 1 ) 2 . ( 6 ) t = ∑ t = 1 2 n t i = ( 2 n − 2 ) + ( 2 n − 4 ) + ⋅ ⋅ ⋅ + 2 = 2 ( n − 1 ) + 2 ( n − 2 ) + ⋅ ⋅ ⋅ + 2 = 2 [ ( n − 1 ) + ( n − 2 ) + ⋅ ⋅ ⋅ + 1 ] = 2 × n ( n − 1 ) 2 = n ( n − 1 ) . \begin{aligned} &\ \ (1)\ t=\sum_{t=1}^{4}t_i=0+0+0+0=0.\\\\ &\ \ (2)\ t=\sum_{t=1}^{4}t_i=0+1+1+2=4.\\\\ &\ \ (3)\ t=\sum_{t=1}^{4}t_i=0+0+2+3=5.\\\\ &\ \ (4)\ t=\sum_{t=1}^{4}t_i=0+0+2+1=3.\\\\ &\ \ (5)\ t=\sum_{t=1}^{2n}t_i=(n-1)+(n-2)+\cdot\cdot\cdot+0=\frac{n(n-1)}{2}.\\\\ &\ \ (6)\ t=\sum_{t=1}^{2n}t_i=(2n-2)+(2n-4)+\cdot\cdot\cdot+2=2(n-1)+2(n-2)+\cdot\cdot\cdot+2=2[(n-1)+(n-2)+\cdot\cdot\cdot+1]\\\\ &\ \ \ \ \ \ \ \ =2\times\frac{n(n-1)}{2}=n(n-1).\\\\ & \end{aligned} (1) t=t=1∑4ti=0+0+0+0=0. (2) t=t=1∑4ti=0+1+1+2=4. (3) t=t=1∑4ti=0+0+2+3=5. (4) t=t=1∑4ti=0+0+2+1=3. (5) t=t=1∑2nti=(n−1)+(n−2)+⋅⋅⋅+0=2n(n−1). (6) t=t=1∑2nti=(2n−2)+(2n−4)+⋅⋅⋅+2=2(n−1)+2(n−2)+⋅⋅⋅+2=2[(n−1)+(n−2)+⋅⋅⋅+1] =2×2n(n−1)=n(n−1).

3. 写出四阶行列式中含有因子 a 11 a 23 的项 . \begin{aligned}&3. \ 写出四阶行列式中含有因子a_{11}a_{23}的项.&\end{aligned} 3. 写出四阶行列式中含有因子a11a23的项.

解：

四阶行列式为 ∣ a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 ∣ = ∑ ( − 1 ) t a 1 p 1 a 2 p 2 a 3 p 3 a 4 p 4 ，含有因子 a 11 a 23 的项的一般形式为 ( − 1 ) t a 11 a 23 a 3 p 3 a 4 p 4 ，因此， p 3 和 p 4 由 2 和 4 构成，分别为 ( − 1 ) t a 11 a 23 a 32 a 44 = ( − 1 ) 1 a 11 a 23 a 32 a 44 = − a 11 a 23 a 32 a 44 ， ( − 1 ) t a 11 a 23 a 34 a 42 = ( − 1 ) 2 a 11 a 23 a 34 a 42 = a 11 a 23 a 34 a 42 \begin{aligned} &\ \ 四阶行列式为\left|\begin{array}{cccc}a_{11} &a_{12} &a_{13} &a_{14}\\\\a_{21} &a_{22} &a_{23} &a_{24}\\\\a_{31} &a_{32} &a_{33} &a_{34}\\\\a_{41} &a_{42} &a_{43} &a_{44}\end{array}\right|=\sum(-1)^ta_{1p_1}a_{2p_2}a_{3p_3}a_{4p_4}，\\\\ &\ \ 含有因子a_{11}a_{23}的项的一般形式为(-1)^ta_{11}a_{23}a_{3p_3}a_{4p_4}，\\\\ &\ \ 因此，p_3和p_4由2和4构成，分别为(-1)^ta_{11}a_{23}a_{32}a_{44}=(-1)^1a_{11}a_{23}a_{32}a_{44}=-a_{11}a_{23}a_{32}a_{44}，\\\\ &\ \ (-1)^ta_{11}a_{23}a_{34}a_{42}=(-1)^2a_{11}a_{23}a_{34}a_{42}=a_{11}a_{23}a_{34}a_{42} & \end{aligned} 四阶行列式为 a11a21a31a41a12a22a32a42a13a23a33a43a14a24a34a44 =∑(−1)ta1p1a2p2a3p3a4p4，含有因子a11a23的项的一般形式为(−1)ta11a23a3p3a4p4，因此，p3和p4由2和4构成，分别为(−1)ta11a23a32a44=(−1)1a11a23a32a44=−a11a23a32a44， (−1)ta11a23a34a42=(−1)2a11a23a34a42=a11a23a34a42

4. 计算下列各行列式： \begin{aligned}&4. \ 计算下列各行列式：&\end{aligned} 4. 计算下列各行列式：

( 1 ) ∣ 4 1 2 4 1 2 0 2 10 5 2 0 0 1 1 7 ∣ ； ( 2 ) ∣ 2 1 4 1 3 − 1 2 1 1 2 3 2 5 0 6 2 ∣ ； ( 3 ) ∣ − a b a c a e b d − c d d e b f c f − e f ∣ ； ( 4 ) ∣ 1 1 1 a b c b + c c + a a + b ∣ ； ( 5 ) ∣ a 1 0 0 − 1 b 1 0 0 − 1 c 1 0 0 − 1 d ∣ ； ( 6 ) ∣ 1 2 3 4 1 3 4 1 1 4 1 2 1 1 2 3 ∣ \begin{aligned} &\ \ (1)\ \ \left|\begin{array}{cccc}4 &1 &2 &4\\\\1 &2 &0 &2\\\\10 &5 &2 &0\\\\0 &1 &1 &7\end{array}\right|；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \left|\begin{array}{cccc}2 &1 &4 &1\\\\3 &-1 &2 &1\\\\1 &2 &3 &2\\\\5 &0 &6 &2\end{array}\right|；\\\\ &\ \ (3)\ \ \left|\begin{array}{cccc}-ab &ac &ae\\\\bd &-cd &de\\\\bf &cf &-ef\end{array}\right|；\ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ \left|\begin{array}{cccc}1 &1 &1\\\\a &b &c\\\\b+c &c+a &a+b\end{array}\right|；\\\\ &\ \ (5)\ \ \left|\begin{array}{cccc}a &1 &0 &0\\\\-1 &b &1 &0\\\\0 &-1 &c &1\\\\0 &0 &-1 &d\end{array}\right|；\ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ \left|\begin{array}{cccc}1 &2 &3 &4\\\\1 &3 &4 &1\\\\1 &4 &1 &2\\\\1 &1 &2 &3\end{array}\right| & \end{aligned} (1) 41100125120214207 ； (2) 23151−12042361122 ； (3) −abbdbfac−cdcfaede−ef ； (4) 1ab+c1bc+a1ca+b ； (5) a−1001b−1001c−1001d ； (6) 1111234134124123

解：

( 1 ) ∣ 4 1 2 4 1 2 0 2 10 5 2 0 0 1 1 7 ∣ = r 1 − r 3 ∣ − 6 − 4 0 4 1 2 0 2 10 5 2 0 0 1 1 7 ∣ = r 1 − 2 r 2 ∣ − 8 − 8 0 0 1 2 0 2 10 5 2 0 0 1 1 7 ∣ = c 1 − 2 c 2 ∣ 8 − 8 0 0 − 3 2 0 2 0 5 2 0 − 2 1 1 7 ∣ = c 2 + c 1 ∣ 8 0 0 0 − 3 − 1 0 2 0 5 2 0 − 2 − 1 1 7 ∣ = c 4 + 2 c 2 ∣ 8 0 0 0 − 3 − 1 0 0 0 5 2 10 − 2 − 1 1 5 ∣ = r 3 − 2 r 4 ∣ 8 0 0 0 − 3 − 1 0 0 4 7 0 0 − 2 − 1 1 5 ∣ = 0 ( 2 ) ∣ 2 1 4 1 3 − 1 2 1 1 2 3 2 5 0 6 2 ∣ = r 1 + r 2 ∣ 5 0 6 2 3 − 1 2 1 1 2 3 2 5 0 6 2 ∣ = r 1 = r 4 ，根据行列式性质 2 推论 0 ( 3 ) ∣ − a b a c a e b d − c d d e b f c f − e f ∣ = − a b c d e f + a b c d e f + a b c d e f − ( − a b c d e f ) − ( − a b c d e f ) − ( − a b c d e f ) = 4 a b c d e f ( 4 ) ∣ 1 1 1 a b c b + c c + a a + b ∣ = c 2 − c 1 ∣ 1 0 1 a − ( a − b ) c b + c a − b a + b ∣ = c 3 − c 1 ∣ 1 0 0 a − ( a − b ) − ( a − c ) b + c a − b a − c ∣ = r 2 + r 3 ∣ 1 0 0 a + b + c 0 0 b + c a − b a − c ∣ = 0 ( 5 ) ∣ a 1 0 0 − 1 b 1 0 0 − 1 c 1 0 0 − 1 d ∣ = r 2 − b r 1 ∣ a 1 0 0 − 1 − a b 0 1 0 0 − 1 c 1 0 0 − 1 d ∣ = r 3 + r 1 ∣ a 1 0 0 − 1 − a b 0 1 0 a 0 c 1 0 0 − 1 d ∣ = 1 × ( − 1 ) 1 + 2 ∣ − 1 − a b 1 0 a c 1 0 − 1 d ∣ = − ( − c d − a b c d − 1 − a b − a d ) = a b c d + a b + a d + c d + 1 ( 6 ) ∣ 1 2 3 4 1 3 4 1 1 4 1 2 1 1 2 3 ∣ = c 2 − 2 c 1 ∣ 1 0 3 4 1 1 4 1 1 2 1 2 1 − 1 2 3 ∣ = c 3 − 3 c 1 ∣ 1 0 0 4 1 1 1 1 1 2 − 2 2 1 − 1 − 1 3 ∣ = c 4 − 4 c 1 ∣ 1 0 0 0 1 1 1 − 3 1 2 − 2 − 2 1 − 1 − 1 − 1 ∣ = r 2 + r 4 ∣ 1 0 0 0 2 0 0 − 4 1 2 − 2 − 2 1 − 1 − 1 − 1 ∣ = r 2 − 2 r 3 ∣ 1 0 0 0 0 − 4 4 0 1 2 − 2 − 2 1 − 1 − 1 − 1 ∣ = c 3 + c 2 ∣ 1 0 0 0 0 − 4 0 0 1 2 0 − 2 1 − 1 − 2 − 1 ∣ = r 3 − 2 r 4 ∣ 1 0 0 0 0 − 4 0 0 − 1 4 4 0 1 − 1 − 2 − 1 ∣ = 16 \begin{aligned} &\ \ (1)\ \left|\begin{array}{cccc}4 &1 &2 &4\\\\1 &2 &0 &2\\\\10 &5 &2 &0\\\\0 &1 &1 &7\end{array}\right|\xlongequal{r_1-r_3}\left|\begin{array}{cccc}-6 &-4 &0 &4\\\\1 &2 &0 &2\\\\10 &5 &2 &0\\\\0 &1 &1 &7\end{array}\right|\xlongequal{r_1-2r_2}\left|\begin{array}{cccc}-8 &-8 &0 &0\\\\1 &2 &0 &2\\\\10 &5 &2 &0\\\\0 &1 &1 &7\end{array}\right|\xlongequal{c_1-2c_2}\left|\begin{array}{cccc}8 &-8 &0 &0\\\\-3 &2 &0 &2\\\\0 &5 &2 &0\\\\-2 &1 &1 &7\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_2+c_1}\left|\begin{array}{cccc}8 &0 &0 &0\\\\-3 &-1 &0 &2\\\\0 &5 &2 &0\\\\-2 &-1 &1 &7\end{array}\right|\xlongequal{c_4+2c_2}\left|\begin{array}{cccc}8 &0 &0 &0\\\\-3 &-1 &0 &0\\\\0 &5 &2 &10\\\\-2 &-1 &1 &5\end{array}\right|\xlongequal{r_3-2r_4}\left|\begin{array}{cccc}8 &0 &0 &0\\\\-3 &-1 &0 &0\\\\4 &7 &0 &0\\\\-2 &-1 &1 &5\end{array}\right|=0\\\\ &\ \ (2)\ \left|\begin{array}{cccc}2 &1 &4 &1\\\\3 &-1 &2 &1\\\\1 &2 &3 &2\\\\5 &0 &6 &2\end{array}\right|\xlongequal{r_1+r_2}\left|\begin{array}{cccc}5 &0 &6 &2\\\\3 &-1 &2 &1\\\\1 &2 &3 &2\\\\5 &0 &6 &2\end{array}\right|\xlongequal{r_1=r_4，根据行列式性质2推论}0\\\\ &\ \ (3)\ \left|\begin{array}{cccc}-ab &ac &ae\\\\bd &-cd &de\\\\bf &cf &-ef\end{array}\right|=-abcdef+abcdef+abcdef-(-abcdef)-(-abcdef)-(-abcdef)=4abcdef\\\\ &\ \ (4)\ \left|\begin{array}{cccc}1 &1 &1\\\\a &b &c\\\\b+c &c+a &a+b\end{array}\right|\xlongequal{c_2-c_1}\left|\begin{array}{cccc}1 &0 &1\\\\a &-(a-b) &c\\\\b+c &a-b &a+b\end{array}\right|\xlongequal{c_3-c_1}\left|\begin{array}{cccc}1 &0 &0\\\\a &-(a-b) &-(a-c)\\\\b+c &a-b &a-c\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{r_2+r_3}\left|\begin{array}{cccc}1 &0 &0\\\\a+b+c &0 &0\\\\b+c &a-b &a-c\end{array}\right|=0\\\\ &\ \ (5)\ \left|\begin{array}{cccc}a &1 &0 &0\\\\-1 &b &1 &0\\\\0 &-1 &c &1\\\\0 &0 &-1 &d\end{array}\right|\xlongequal{r_2-br_1}\left|\begin{array}{cccc}a &1 &0 &0\\\\-1-ab &0 &1 &0\\\\0 &-1 &c &1\\\\0 &0 &-1 &d\end{array}\right|\xlongequal{r_3+r_1}\left|\begin{array}{cccc}a &1 &0 &0\\\\-1-ab &0 &1 &0\\\\a &0 &c &1\\\\0 &0 &-1 &d\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ =1\times(-1)^{1+2}\left|\begin{array}{cccc}-1-ab &1 &0\\\\a &c &1\\\\0 &-1 &d\end{array}\right|=-(-cd-abcd-1-ab-ad)=abcd+ab+ad+cd+1\\\\ &\ \ (6)\ \left|\begin{array}{cccc}1 &2 &3 &4\\\\1 &3 &4 &1\\\\1 &4 &1 &2\\\\1 &1 &2 &3\end{array}\right|\xlongequal{c_2-2c_1}\left|\begin{array}{cccc}1 &0 &3 &4\\\\1 &1 &4 &1\\\\1 &2 &1 &2\\\\1 &-1 &2 &3\end{array}\right|\xlongequal{c_3-3c_1}\left|\begin{array}{cccc}1 &0 &0 &4\\\\1 &1 &1 &1\\\\1 &2 &-2 &2\\\\1 &-1 &-1 &3\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_4-4c_1}\left|\begin{array}{cccc}1 &0 &0 &0\\\\1 &1 &1 &-3\\\\1 &2 &-2 &-2\\\\1 &-1 &-1 &-1\end{array}\right|\xlongequal{r_2+r_4}\left|\begin{array}{cccc}1 &0 &0 &0\\\\2 &0 &0 &-4\\\\1 &2 &-2 &-2\\\\1 &-1 &-1 &-1\end{array}\right|\xlongequal{r_2-2r_3}\left|\begin{array}{cccc}1 &0 &0 &0\\\\0 &-4 &4 &0\\\\1 &2 &-2 &-2\\\\1 &-1 &-1 &-1\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_3+c_2}\left|\begin{array}{cccc}1 &0 &0 &0\\\\0 &-4 &0 &0\\\\1 &2 &0 &-2\\\\1 &-1 &-2 &-1\end{array}\right|\xlongequal{r_3-2r_4}\left|\begin{array}{cccc}1 &0 &0 &0\\\\0 &-4 &0 &0\\\\-1 &4 &4 &0\\\\1 &-1 &-2 &-1\end{array}\right|=16 & \end{aligned} (1) 41100125120214207 r1−r3 −61100−425100214207 r1−2r2 −81100−825100210207 c1−2c2 8−30−2−825100210207 c2+c1 8−30−20−15−100210207 c4+2c2 8−30−20−15−1002100105 r3−2r4 8−34−20−17−100010005 =0 (2) 23151−12042361122 r1+r2 53150−12062362122 r1=r4，根据行列式性质2推论 0 (3) −abbdbfac−cdcfaede−ef =−abcdef+abcdef+abcdef−(−abcdef)−(−abcdef)−(−abcdef)=4abcdef (4) 1ab+c1bc+a1ca+b c2−c1 1ab+c0−(a−b)a−b1ca+b c3−c1 1ab+c0−(a−b)a−b0−(a−c)a−c r2+r3 1a+b+cb+c00a−b00a−c =0 (5) a−1001b−1001c−1001d r2−br1 a−1−ab0010−1001c−1001d r3+r1 a−1−aba0100001c−1001d =1×(−1)1+2 −1−aba01c−101d =−(−cd−abcd−1−ab−ad)=abcd+ab+ad+cd+1 (6) 1111234134124123 c2−2c1 1111012−134124123 c3−3c1 1111012−101−2−14123 c4−4c1 1111012−101−2−10−3−2−1 r2+r4 1211002−100−2−10−4−2−1 r2−2r3 10110−42−104−2−100−2−1 c3+c2 10110−42−1000−200−2−1 r3−2r4 10−110−44−1004−2000−1 =16