线性代数（第六版）同济大学 习题一 （5-6题）个人解答

线性代数（第六版）同济大学 习题一（5-6题）

 

$\begin{array}{rlr}& 5.求解下列方程：& \end{array}$

   ( 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+12−1​2x+11​−11x+1​ ​=0；                    (2)   ​1xx2x3​1aa2a3​1bb2b3​1cc2c3​ ​=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+12−1​2x+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)  ​1xx2x3​1aa2a3​1bb2b3​1cc2c3​ ​= ​1000​1a−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.​​

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$\begin{array}{rlr}& 6.证明：& \end{array}$

   ( 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)   ​a22a1​aba+b1​b22b1​ ​=(a−b)3；  (2)   ​ax+byay+bzaz+bx​ay+bzaz+bxax+by​az+bxax+byay+bz​ ​=(a3+b3) ​xyz​yzx​zxy​ ​；  (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)   ​1aa2a4​1bb2b4​1cc2c4​1dd2d4​ ​=(a−b)(a−c)(a−d)(b−c)(b−d)(c−d)(a+b+c+d)；  (5)   ​x00a0​​−1x0a1​​0−1xa2​​00−1a3​​ ​=a3​x3+a2​x2+a1​x+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)  ​a22a1​aba+b1​b22b1​ ​=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+bx​ay+bzaz+bxax+by​az+bxax+byay+bz​ ​根据性质5 ​axayaz​ayazax​azaxay​ ​+ ​axaybx​ayazby​azaxbz​ ​+ ​axbzaz​aybxax​azbyay​ ​        + ​axbzbx​aybxby​azbybz​ ​+ ​byayaz​bzazax​bxaxay​ ​+ ​byaybx​bzazby​bxaxbz​ ​+ ​bybzaz​bzbxax​bxbyay​ ​+ ​bybzbx​bzbxby​bxbybz​ ​        根据性质3 a3 ​xyz​yzx​zxy​ ​+a2b ​xyx​yzy​zxz​ ​+a2b ​xzz​yxx​zyy​ ​+ab2 ​xzx​yxy​zyz​ ​        +a2b ​yyz​zzx​xxy​ ​+ab2 ​yyx​zzy​xxz​ ​+ab2 ​yzz​zxx​xyy​ ​+b3 ​yzx​zxy​xyz​ ​        根据性质4 a3 ​xyz​yzx​zxy​ ​+b3 ​yzx​zxy​xyz​ ​根据性质2 (a3+b3) ​xyz​yzx​zxy​ ​  (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​ ​a2b2c2d2​2a+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​ ​a2b2c2d2​2a+12b+12c+12d+1​4a+44b+44c+44d+4​(a+3)2(b+3)2(c+3)2(d+3)2​ ​c4​−c1​ ​a2b2c2d2​2a+12b+12c+12d+1​4a+44b+44c+44d+4​6a+96b+96c+96d+9​ ​        c3​−c2​ ​a2b2c2d2​2a+12b+12c+12d+1​2a+32b+32c+32d+3​6a+96b+96c+96d+9​ ​=3 ​a2b2c2d2​2a+12b+12c+12d+1​2a+32b+32c+32d+3​2a+32b+32c+32d+3​ ​c3​=c4​，根据行列式性质2推论 0  (4)  ​1aa2a4​1bb2b4​1cc2c4​1dd2d4​ ​c2​−c1​ ​1aa2a4​0b−a(b−a)(b+a)(b−a)(b+a)(b2+a2)​1cc2c4​1dd2d4​ ​        c3​−c1​ ​1aa2a4​0b−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​ ​1aa2a4​0b−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+ac​1a2+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​​−1x0a1​​0−1xa2​​00−1a3​​ ​c4​+x1​c3​ ​x00a0​​−1x0a1​​0−1xa2​​0−x1​0a3​+xa2​​​ ​=x⋅(−1)3+3 ​x0a0​​−1xa1​​0−x1​a3​+xa2​​​ ​        =x(a3​x2+a2​x+xa0​​+a1​)=a3​x3+a2​x2+a1​x+a0​​​