线性代数（第六版）同济大学 习题一（5-6题）
5. 求解下列方程： \begin{aligned}&5. \ 求解下列方程：&\end{aligned} 5. 求解下列方程：
( 1 ) ∣ x + 1 2 − 1 2 x + 1 1 − 1 1 x + 1 ∣ = 0 ； ( 2 ) ∣ 1 1 1 1 x a b c x 2 a 2 b 2 c 2 x 3 a 3 b 3 c 3 ∣ = 0 ， 其中 a ， b ， c 互不相等 . \begin{aligned} &\ \ (1)\ \ \left|\begin{array}{cccc}x+1 &2 &-1\\\\2 &x+1 &1\\\\-1 &1 &x+1\end{array}\right|=0；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \left|\begin{array}{cccc}1 &1 &1 &1\\\\x &a &b &c\\\\x^2 &a^2 &b^2 &c^2\\\\x^3 &a^3 &b^3 &c^3\end{array}\right|=0，\\\\ &\ \ \ \ \ \ \ \ 其中a，b，c互不相等. & \end{aligned} (1) x+12−12x+11−11x+1 =0； (2) 1xx2x31aa2a31bb2b31cc2c3 =0， 其中a，b，c互不相等.
解：
( 1 ) ∣ x + 1 2 − 1 2 x + 1 1 − 1 1 x + 1 ∣ = ( x + 1 ) 3 − 2 − 2 − ( x + 1 ) − 4 ( x + 1 ) − ( x + 1 ) = x 3 + 3 x 2 − 3 x − 9 = ( x 2 − 3 ) ( x + 3 ) ，解方程 ( x 2 − 3 ) ( x + 3 ) = 0 ，得 x 1 = 3 ， x 2 = − 3 ， x 3 = − 3 ( 2 ) ∣ 1 1 1 1 x a b c x 2 a 2 b 2 c 2 x 3 a 3 b 3 c 3 ∣ = ∣ 1 1 1 1 0 a − x b − x c − x 0 a ( a − x ) b ( b − x ) c ( c − x ) 0 a 2 ( a − x ) b 2 ( b − x ) c 2 ( c − x ) ∣ = ( a − x ) ( b − a ) ( b − x ) ( c − b ) ( c − a ) ( c − x ) = ( a − b ) ( b − c ) ( a − c ) ( x − a ) ( x − b ) ( x − c ) ， 解方程 ( a − b ) ( b − c ) ( a − c ) ( x − a ) ( x − b ) ( x − c ) = 0 ，得 x 1 = a ， x 2 = b ， x 3 = c . \begin{aligned} &\ \ (1)\ \left|\begin{array}{cccc}x+1 &2 &-1\\\\2 &x+1 &1\\\\-1 &1 &x+1\end{array}\right|=(x+1)^3-2-2-(x+1)-4(x+1)-(x+1)\\\\ &\ \ \ \ \ \ \ \ =x^3+3x^2-3x-9=(x^2-3)(x+3)，解方程(x^2-3)(x+3)=0，得x_1=\sqrt{3}，x_2=-\sqrt{3}，x_3=-3\\\\ &\ \ (2)\ \left|\begin{array}{cccc}1 &1 &1 &1\\\\x &a &b &c\\\\x^2 &a^2 &b^2 &c^2\\\\x^3 &a^3 &b^3 &c^3\end{array}\right|=\left|\begin{array}{cccc}1 &1 &1 &1\\\\0 &a-x &b-x &c-x\\\\0 &a(a-x) &b(b-x) &c(c-x)\\\\0 &a^2(a-x) &b^2(b-x) &c^2(c-x)\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ =(a-x)(b-a)(b-x)(c-b)(c-a)(c-x)=(a-b)(b-c)(a-c)(x-a)(x-b)(x-c)，\\\\ &\ \ \ \ \ \ \ \ 解方程(a-b)(b-c)(a-c)(x-a)(x-b)(x-c)=0，得x_1=a，x_2=b，x_3=c. & \end{aligned} (1) x+12−12x+11−11x+1 =(x+1)3−2−2−(x+1)−4(x+1)−(x+1) =x3+3x2−3x−9=(x2−3)(x+3)，解方程(x2−3)(x+3)=0，得x1=3 ，x2=−3 ，x3=−3 (2) 1xx2x31aa2a31bb2b31cc2c3 = 10001a−xa(a−x)a2(a−x)1b−xb(b−x)b2(b−x)1c−xc(c−x)c2(c−x) =(a−x)(b−a)(b−x)(c−b)(c−a)(c−x)=(a−b)(b−c)(a−c)(x−a)(x−b)(x−c)， 解方程(a−b)(b−c)(a−c)(x−a)(x−b)(x−c)=0，得x1=a，x2=b，x3=c.
6. 证明： \begin{aligned}&6. \ 证明：&\end{aligned} 6. 证明：
( 1 ) ∣ a 2 a b b 2 2 a a + b 2 b 1 1 1 ∣ = ( a − b ) 3 ； ( 2 ) ∣ a x + b y a y + b z a z + b x a y + b z a z + b x a x + b y a z + b x a x + b y a y + b z ∣ = ( a 3 + b 3 ) ∣ x y z y z x z x y ∣ ； ( 3 ) ∣ a 2 ( a + 1 ) 2 ( a + 2 ) 2 ( a + 3 ) 2 b 2 ( b + 1 ) 2 ( b + 2 ) 2 ( b + 3 ) 2 c 2 ( c + 1 ) 2 ( c + 2 ) 2 ( c + 3 ) 2 d 2 ( d + 1 ) 2 ( d + 2 ) 2 ( d + 3 ) 2 ∣ = 0 ； ( 4 ) ∣ 1 1 1 1 a b c d a 2 b 2 c 2 d 2 a 4 b 4 c 4 d 4 ∣ = ( a − b ) ( a − c ) ( a − d ) ( b − c ) ( b − d ) ( c − d ) ( a + b + c + d ) ； ( 5 ) ∣ x − 1 0 0 0 x − 1 0 0 0 x − 1 a 0 a 1 a 2 a 3 ∣ = a 3 x 3 + a 2 x 2 + a 1 x + a 0 . \begin{aligned} &\ \ (1)\ \ \left|\begin{array}{cccc}a^2 &ab &b^2\\\\2a &a+b &2b\\\\1 &1 &1\end{array}\right|=(a-b)^3；\\\\ &\ \ (2)\ \ \left|\begin{array}{cccc}ax+by &ay+bz &az+bx\\\\ay+bz &az+bx &ax+by\\\\az+bx &ax+by &ay+bz\end{array}\right|=(a^3+b^3)\left|\begin{array}{cccc}x &y &z\\\\y &z &x\\\\z &x &y\end{array}\right|；\\\\ &\ \ (3)\ \ \left|\begin{array}{cccc}a^2 &(a+1)^2 &(a+2)^2 &(a+3)^2\\\\b^2 &(b+1)^2 &(b+2)^2 &(b+3)^2\\\\c^2 &(c+1)^2 &(c+2)^2 &(c+3)^2\\\\d^2 &(d+1)^2 &(d+2)^2 &(d+3)^2\end{array}\right|=0；\\\\ &\ \ (4)\ \ \left|\begin{array}{cccc}1 &1 &1 &1\\\\a &b &c &d\\\\a^2 &b^2 &c^2 &d^2\\\\a^4 &b^4 &c^4 &d^4\end{array}\right|=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d)；\\\\ &\ \ (5)\ \ \left|\begin{array}{cccc}x &-1 &0 &0\\\\0 &x &-1 &0\\\\0 &0 &x &-1\\\\a_0 &a_1 &a_2 &a_3\end{array}\right|=a_3x^3+a_2x^2+a_1x+a_0. & \end{aligned} (1) a22a1aba+b1b22b1 =(a−b)3； (2) ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz =(a3+b3) xyzyzxzxy ； (3) a2b2c2d2(a+1)2(b+1)2(c+1)2(d+1)2(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2(c+3)2(d+3)2 =0； (4) 1aa2a41bb2b41cc2c41dd2d4 =(a−b)(a−c)(a−d)(b−c)(b−d)(c−d)(a+b+c+d)； (5) x00a0−1x0a10−1xa200−1a3 =a3x3+a2x2+a1x+a0.
解：
( 1 ) ∣ a 2 a b b 2 2 a a + b 2 b 1 1 1 ∣ = a 2 ( a + b ) + 2 a b 2 + 2 a b 2 − 2 a 2 b − 2 a 2 b − b 2 ( a + b ) = ( a 2 − b 2 ) ( a + b ) + 4 a b 2 − 4 a 2 b = ( a + b ) 2 ( a − b ) − 4 a b ( a − b ) = ( a − b ) ( a 2 − 2 a b + b 2 ) = ( a − b ) 3 ( 2 ) ∣ a x + b y a y + b z a z + b x a y + b z a z + b x a x + b y a z + b x a x + b y a y + b z ∣ = 根据性质 5 ∣ a x a y a z a y a z a x a z a x a y ∣ + ∣ a x a y a z a y a z a x b x b y b z ∣ + ∣ a x a y a z b z b x b y a z a x a y ∣ + ∣ a x a y a z b z b x b y b x b y b z ∣ + ∣ b y b z b x a y a z a x a z a x a y ∣ + ∣ b y b z b x a y a z a x b x b y b z ∣ + ∣ b y b z b x b z b x b y a z a x a y ∣ + ∣ b y b z b x b z b x b y b x b y b z ∣ = 根据性质 3 a 3 ∣ x y z y z x z x y ∣ + a 2 b ∣ x y z y z x x y z ∣ + a 2 b ∣ x y z z x y z x y ∣ + a b 2 ∣ x y z z x y x y z ∣ + a 2 b ∣ y z x y z x z x y ∣ + a b 2 ∣ y z x y z x x y z ∣ + a b 2 ∣ y z x z x y z x y ∣ + b 3 ∣ y z x z x y x y z ∣ = 根据性质 4 a 3 ∣ x y z y z x z x y ∣ + b 3 ∣ y z x z x y x y z ∣ = 根据性质 2 ( a 3 + b 3 ) ∣ x y z y z x z x y ∣ ( 3 ) ∣ a 2 ( a + 1 ) 2 ( a + 2 ) 2 ( a + 3 ) 2 b 2 ( b + 1 ) 2 ( b + 2 ) 2 ( b + 3 ) 2 c 2 ( c + 1 ) 2 ( c + 2 ) 2 ( c + 3 ) 2 d 2 ( d + 1 ) 2 ( d + 2 ) 2 ( d + 3 ) 2 ∣ = c 2 − c 1 ∣ a 2 2 a + 1 ( a + 2 ) 2 ( a + 3 ) 2 b 2 2 b + 1 ( b + 2 ) 2 ( b + 3 ) 2 c 2 2 c + 1 ( c + 2 ) 2 ( c + 3 ) 2 d 2 2 d + 1 ( d + 2 ) 2 ( d + 3 ) 2 ∣ = c 3 − c 1 ∣ a 2 2 a + 1 4 a + 4 ( a + 3 ) 2 b 2 2 b + 1 4 b + 4 ( b + 3 ) 2 c 2 2 c + 1 4 c + 4 ( c + 3 ) 2 d 2 2 d + 1 4 d + 4 ( d + 3 ) 2 ∣ = c 4 − c 1 ∣ a 2 2 a + 1 4 a + 4 6 a + 9 b 2 2 b + 1 4 b + 4 6 b + 9 c 2 2 c + 1 4 c + 4 6 c + 9 d 2 2 d + 1 4 d + 4 6 d + 9 ∣ = c 3 − c 2 ∣ a 2 2 a + 1 2 a + 3 6 a + 9 b 2 2 b + 1 2 b + 3 6 b + 9 c 2 2 c + 1 2 c + 3 6 c + 9 d 2 2 d + 1 2 d + 3 6 d + 9 ∣ = 3 ∣ a 2 2 a + 1 2 a + 3 2 a + 3 b 2 2 b + 1 2 b + 3 2 b + 3 c 2 2 c + 1 2 c + 3 2 c + 3 d 2 2 d + 1 2 d + 3 2 d + 3 ∣ = c 3 = c 4 ，根据行列式性质 2 推论 0 ( 4 ) ∣ 1 1 1 1 a b c d a 2 b 2 c 2 d 2 a 4 b 4 c 4 d 4 ∣ = c 2 − c 1 ∣ 1 0 1 1 a b − a c d a 2 ( b − a ) ( b + a ) c 2 d 2 a 4 ( b − a ) ( b + a ) ( b 2 + a 2 ) c 4 d 4 ∣ = c 3 − c 1 ∣ 1 0 0 1 a b − a c − a d a 2 ( b − a ) ( b + a ) ( c − a ) ( c + a ) d 2 a 4 ( b − a ) ( b + a ) ( b 2 + a 2 ) ( c − a ) ( c + a ) ( c 2 + a 2 ) d 4 ∣ = c 4 − c 1 ∣ 1 0 0 0 a b − a c − a d − a a 2 ( b − a ) ( b + a ) ( c − a ) ( c + a ) ( d − a ) ( d + a ) a 4 ( b − a ) ( b + a ) ( b 2 + a 2 ) ( c − a ) ( c + a ) ( c 2 + a 2 ) ( d − a ) ( d + a ) ( d 2 + a 2 ) ∣ = ( b − a ) ( c − a ) ( d − a ) ∣ 1 1 1 b + a c + a d + a ( b + a ) ( b 2 + a 2 ) ( c + a ) ( c 2 + a 2 ) ( d + a ) ( d 2 + a 2 ) ∣ = c 2 − c 1 ( b − a ) ( c − a ) ( d − a ) ∣ 1 0 1 b + a c − b d + a ( b + a ) ( b 2 + a 2 ) ( c + a ) ( c 2 + a 2 ) − ( b + a ) ( b 2 + a 2 ) ( d + a ) ( d 2 + a 2 ) ∣ = c 3 − c 1 ( b − a ) ( c − a ) ( d − a ) ∣ 1 0 0 b + a c − b d − b ( b + a ) ( b 2 + a 2 ) ( c − b ) ( a 2 + b 2 + c 2 + a b + b c + a c ) ( d − b ) ( a 2 + b 2 + d 2 + a b + b d + a d ) ∣ = ( b − a ) ( c − a ) ( d − a ) ∣ c − b d − b ( c − b ) ( a 2 + b 2 + c 2 + a b + b c + a c ) ( d − b ) ( a 2 + b 2 + d 2 + a b + b d + a d ) ∣ = ( b − a ) ( c − a ) ( d − a ) ( c − b ) ( d − b ) ∣ 1 1 a 2 + b 2 + c 2 + a b + b c + a c a 2 + b 2 + d 2 + a b + b d + a d ∣ = ( b − a ) ( c − a ) ( d − a ) ( c − b ) ( d − b ) ( d − c ) ( a + b + c + d ) = ( a − b ) ( a − c ) ( a − d ) ( b − c ) ( b − d ) ( c − d ) ( a + b + c + d ) . ( 5 ) ∣ x − 1 0 0 0 x − 1 0 0 0 x − 1 a 0 a 1 a 2 a 3 ∣ = c 4 + 1 x c 3 ∣ x − 1 0 0 0 x − 1 − 1 x 0 0 x 0 a 0 a 1 a 2 a 3 + a 2 x ∣ = x ⋅ ( − 1 ) 3 + 3 ∣ x − 1 0 0 x − 1 x a 0 a 1 a 3 + a 2 x ∣ = x ( a 3 x 2 + a 2 x + a 0 x + a 1 ) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 \begin{aligned} &\ \ (1)\ \left|\begin{array}{cccc}a^2 &ab &b^2\\\\2a &a+b &2b\\\\1 &1 &1\end{array}\right|=a^2(a+b)+2ab^2+2ab^2-2a^2b-2a^2b-b^2(a+b)\\\\ &\ \ \ \ \ \ \ \ =(a^2-b^2)(a+b)+4ab^2-4a^2b=(a+b)^2(a-b)-4ab(a-b)=(a-b)(a^2-2ab+b^2)=(a-b)^3\\\\ &\ \ (2)\ \left|\begin{array}{cccc}ax+by &ay+bz &az+bx\\\\ay+bz &az+bx &ax+by\\\\az+bx &ax+by &ay+bz\end{array}\right|\xlongequal{根据性质5}\left|\begin{array}{cccc}ax &ay &az\\\\ay &az &ax\\\\az &ax &ay\end{array}\right|+\left|\begin{array}{cccc}ax &ay &az\\\\ay &az &ax\\\\bx &by &bz\end{array}\right|+\left|\begin{array}{cccc}ax &ay &az\\\\bz &bx &by\\\\az &ax &ay\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ +\left|\begin{array}{cccc}ax &ay &az\\\\bz &bx &by\\\\bx &by &bz\end{array}\right|+\left|\begin{array}{cccc}by &bz &bx\\\\ay &az &ax\\\\az &ax &ay\end{array}\right|+\left|\begin{array}{cccc}by &bz &bx\\\\ay &az &ax\\\\bx &by &bz\end{array}\right|+\left|\begin{array}{cccc}by &bz &bx\\\\bz &bx &by\\\\az &ax &ay\end{array}\right|+\left|\begin{array}{cccc}by &bz &bx\\\\bz &bx &by\\\\bx &by &bz\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{根据性质3}a^3\left|\begin{array}{cccc}x &y &z\\\\y &z &x\\\\z &x &y\end{array}\right|+a^2b\left|\begin{array}{cccc}x &y &z\\\\y &z &x\\\\x &y &z\end{array}\right|+a^2b\left|\begin{array}{cccc}x &y &z\\\\z &x &y\\\\z &x &y\end{array}\right|+ab^2\left|\begin{array}{cccc}x &y &z\\\\z &x &y\\\\x &y &z\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ +a^2b\left|\begin{array}{cccc}y &z &x\\\\y &z &x\\\\z &x &y\end{array}\right|+ab^2\left|\begin{array}{cccc}y &z &x\\\\y &z &x\\\\x &y &z\end{array}\right|+ab^2\left|\begin{array}{cccc}y &z &x\\\\z &x &y\\\\z &x &y\end{array}\right|+b^3\left|\begin{array}{cccc}y &z &x\\\\z &x &y\\\\x &y &z\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{根据性质4}a^3\left|\begin{array}{cccc}x &y &z\\\\y &z &x\\\\z &x &y\end{array}\right|+b^3\left|\begin{array}{cccc}y &z &x\\\\z &x &y\\\\x &y &z\end{array}\right|\xlongequal{根据性质2}(a^3+b^3)\left|\begin{array}{cccc}x &y &z\\\\y &z &x\\\\z &x &y\end{array}\right|\\\\ &\ \ (3)\ \left|\begin{array}{cccc}a^2 &(a+1)^2 &(a+2)^2 &(a+3)^2\\\\b^2 &(b+1)^2 &(b+2)^2 &(b+3)^2\\\\c^2 &(c+1)^2 &(c+2)^2 &(c+3)^2\\\\d^2 &(d+1)^2 &(d+2)^2 &(d+3)^2\end{array}\right|\xlongequal{c_2-c_1}\left|\begin{array}{cccc}a^2 &2a+1 &(a+2)^2 &(a+3)^2\\\\b^2 &2b+1 &(b+2)^2 &(b+3)^2\\\\c^2 &2c+1 &(c+2)^2 &(c+3)^2\\\\d^2 &2d+1 &(d+2)^2 &(d+3)^2\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_3-c_1}\left|\begin{array}{cccc}a^2 &2a+1 &4a+4 &(a+3)^2\\\\b^2 &2b+1 &4b+4 &(b+3)^2\\\\c^2 &2c+1 &4c+4 &(c+3)^2\\\\d^2 &2d+1 &4d+4 &(d+3)^2\end{array}\right|\xlongequal{c_4-c_1}\left|\begin{array}{cccc}a^2 &2a+1 &4a+4 &6a+9\\\\b^2 &2b+1 &4b+4 &6b+9\\\\c^2 &2c+1 &4c+4 &6c+9\\\\d^2 &2d+1 &4d+4 &6d+9\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_3-c_2}\left|\begin{array}{cccc}a^2 &2a+1 &2a+3 &6a+9\\\\b^2 &2b+1 &2b+3 &6b+9\\\\c^2 &2c+1 &2c+3 &6c+9\\\\d^2 &2d+1 &2d+3 &6d+9\end{array}\right|=3\left|\begin{array}{cccc}a^2 &2a+1 &2a+3 &2a+3\\\\b^2 &2b+1 &2b+3 &2b+3\\\\c^2 &2c+1 &2c+3 &2c+3\\\\d^2 &2d+1 &2d+3 &2d+3\end{array}\right|\xlongequal{c_3=c_4，根据行列式性质2推论}0\\\\ &\ \ (4)\ \left|\begin{array}{cccc}1 &1 &1 &1\\\\a &b &c &d\\\\a^2 &b^2 &c^2 &d^2\\\\a^4 &b^4 &c^4 &d^4\end{array}\right|\xlongequal{c_2-c_1}\left|\begin{array}{cccc}1 &0 &1 &1\\\\a &b-a &c &d\\\\a^2 &(b-a)(b+a) &c^2 &d^2\\\\a^4 &(b-a)(b+a)(b^2+a^2) &c^4 &d^4\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_3-c_1}\left|\begin{array}{cccc}1 &0 &0 &1\\\\a &b-a &c-a &d\\\\a^2 &(b-a)(b+a) &(c-a)(c+a) &d^2\\\\a^4 &(b-a)(b+a)(b^2+a^2) &(c-a)(c+a)(c^2+a^2) &d^4\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_4-c_1}\left|\begin{array}{cccc}1 &0 &0 &0\\\\a &b-a &c-a &d-a\\\\a^2 &(b-a)(b+a) &(c-a)(c+a) &(d-a)(d+a)\\\\a^4 &(b-a)(b+a)(b^2+a^2) &(c-a)(c+a)(c^2+a^2) &(d-a)(d+a)(d^2+a^2)\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ =(b-a)(c-a)(d-a)\left|\begin{array}{cccc}1 &1 &1\\\\b+a &c+a &d+a\\\\(b+a)(b^2+a^2) &(c+a)(c^2+a^2) &(d+a)(d^2+a^2)\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_2-c_1}(b-a)(c-a)(d-a)\left|\begin{array}{cccc}1 &0 &1\\\\b+a &c-b &d+a\\\\(b+a)(b^2+a^2) &(c+a)(c^2+a^2)-(b+a)(b^2+a^2) &(d+a)(d^2+a^2)\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ \xlongequal{c_3-c_1}(b-a)(c-a)(d-a)\left|\begin{array}{cccc}1 &0 &0\\\\b+a &c-b &d-b\\\\(b+a)(b^2+a^2) &(c-b)(a^2+b^2+c^2+ab+bc+ac) &(d-b)(a^2+b^2+d^2+ab+bd+ad)\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ =(b-a)(c-a)(d-a)\left|\begin{array}{cccc}c-b &d-b\\\\(c-b)(a^2+b^2+c^2+ab+bc+ac) &(d-b)(a^2+b^2+d^2+ab+bd+ad)\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ =(b-a)(c-a)(d-a)(c-b)(d-b)\left|\begin{array}{cccc}1 &1\\\\a^2+b^2+c^2+ab+bc+ac &a^2+b^2+d^2+ab+bd+ad\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ =(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)(a+b+c+d)\\\\ &\ \ \ \ \ \ \ \ =(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d).\\\\ &\ \ (5)\ \left|\begin{array}{cccc}x &-1 &0 &0\\\\0 &x &-1 &0\\\\0 &0 &x &-1\\\\a_0 &a_1 &a_2 &a_3\end{array}\right|\xlongequal{c_4+\frac{1}{x}c_3}\left|\begin{array}{cccc}x &-1 &0 &0\\\\0 &x &-1 &-\frac{1}{x}\\\\0 &0 &x &0\\\\a_0 &a_1 &a_2 &a_3+\frac{a_2}{x}\end{array}\right|=x\cdot(-1)^{3+3}\left|\begin{array}{cccc}x &-1 &0\\\\0 &x &-\frac{1}{x}\\\\a_0 &a_1 &a_3+\frac{a_2}{x}\end{array}\right|\\\\ &\ \ \ \ \ \ \ \ =x(a_3x^2+a_2x+\frac{a_0}{x}+a_1)=a_3x^3+a_2x^2+a_1x+a_0 & \end{aligned} (1) a22a1aba+b1b22b1 =a2(a+b)+2ab2+2ab2−2a2b−2a2b−b2(a+b) =(a2−b2)(a+b)+4ab2−4a2b=(a+b)2(a−b)−4ab(a−b)=(a−b)(a2−2ab+b2)=(a−b)3 (2) ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz 根据性质5 axayazayazaxazaxay + axaybxayazbyazaxbz + axbzazaybxaxazbyay + axbzbxaybxbyazbybz + byayazbzazaxbxaxay + byaybxbzazbybxaxbz + bybzazbzbxaxbxbyay + bybzbxbzbxbybxbybz 根据性质3 a3 xyzyzxzxy +a2b xyxyzyzxz +a2b xzzyxxzyy +ab2 xzxyxyzyz +a2b yyzzzxxxy +ab2 yyxzzyxxz +ab2 yzzzxxxyy +b3 yzxzxyxyz 根据性质4 a3 xyzyzxzxy +b3 yzxzxyxyz 根据性质2 (a3+b3) xyzyzxzxy (3) a2b2c2d2(a+1)2(b+1)2(c+1)2(d+1)2(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2(c+3)2(d+3)2 c2−c1 a2b2c2d22a+12b+12c+12d+1(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2(c+3)2(d+3)2 c3−c1 a2b2c2d22a+12b+12c+12d+14a+44b+44c+44d+4(a+3)2(b+3)2(c+3)2(d+3)2 c4−c1 a2b2c2d22a+12b+12c+12d+14a+44b+44c+44d+46a+96b+96c+96d+9 c3−c2 a2b2c2d22a+12b+12c+12d+12a+32b+32c+32d+36a+96b+96c+96d+9 =3 a2b2c2d22a+12b+12c+12d+12a+32b+32c+32d+32a+32b+32c+32d+3 c3=c4，根据行列式性质2推论 0 (4) 1aa2a41bb2b41cc2c41dd2d4 c2−c1 1aa2a40b−a(b−a)(b+a)(b−a)(b+a)(b2+a2)1cc2c41dd2d4 c3−c1 1aa2a40b−a(b−a)(b+a)(b−a)(b+a)(b2+a2)0c−a(c−a)(c+a)(c−a)(c+a)(c2+a2)1dd2d4 c4−c1 1aa2a40b−a(b−a)(b+a)(b−a)(b+a)(b2+a2)0c−a(c−a)(c+a)(c−a)(c+a)(c2+a2)0d−a(d−a)(d+a)(d−a)(d+a)(d2+a2) =(b−a)(c−a)(d−a) 1b+a(b+a)(b2+a2)1c+a(c+a)(c2+a2)1d+a(d+a)(d2+a2) c2−c1 (b−a)(c−a)(d−a) 1b+a(b+a)(b2+a2)0c−b(c+a)(c2+a2)−(b+a)(b2+a2)1d+a(d+a)(d2+a2) c3−c1 (b−a)(c−a)(d−a) 1b+a(b+a)(b2+a2)0c−b(c−b)(a2+b2+c2+ab+bc+ac)0d−b(d−b)(a2+b2+d2+ab+bd+ad) =(b−a)(c−a)(d−a) c−b(c−b)(a2+b2+c2+ab+bc+ac)d−b(d−b)(a2+b2+d2+ab+bd+ad) =(b−a)(c−a)(d−a)(c−b)(d−b) 1a2+b2+c2+ab+bc+ac1a2+b2+d2+ab+bd+ad =(b−a)(c−a)(d−a)(c−b)(d−b)(d−c)(a+b+c+d) =(a−b)(a−c)(a−d)(b−c)(b−d)(c−d)(a+b+c+d). (5) x00a0−1x0a10−1xa200−1a3 c4+x1c3 x00a0−1x0a10−1xa20−x10a3+xa2 =x⋅(−1)3+3 x0a0−1xa10−x1a3+xa2 =x(a3x2+a2x+xa0+a1)=a3x3+a2x2+a1x+a0