【课后习题】线性代数第六版第一章行列式习题一

习题一

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 ∣ \left|\begin{array}{rrr}2 & 0 & 1 \\ 1 & -4 & -1 \\ -1 & 8 & 3\end{array}\right| ∣∣∣∣∣∣21−10−481−13∣∣∣∣∣∣;

(2) ∣ a b c b c a c a b ∣ \left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right| ∣∣∣∣∣∣abcbcacab∣∣∣∣∣∣;

(3) ∣ 1 1 1 a b c a 2 b 2 c 2 ∣ \left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{array}\right| ∣∣∣∣∣∣1aa21bb21cc2∣∣∣∣∣∣;

(4) ∣ x y x + y y x + y x x + y x y ∣ \left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right| ∣∣∣∣∣∣xyx+yyx+yxx+yxy∣∣∣∣∣∣.

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 ) 13 \cdots(2 n-1) 24 \cdots(2 n) 13⋯(2n−1)24⋯(2n);

(6) 13 ⋯ ( 2 n − 1 ) ( 2 n ) ( 2 n − 2 ) ⋯ 2 13 \cdots(2 n-1)(2 n)(2 n-2) \cdots 2 13⋯(2n−1)(2n)(2n−2)⋯2.

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

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 ∣ \left|\begin{array}{rrrr}4 & 1 & 2 & 4 \\ 1 & 2 & 0 & 2 \\ 10 & 5 & 2 & 0 \\ 0 & 1 & 1 & 7\end{array}\right| ∣∣∣∣∣∣∣∣41100125120214207∣∣∣∣∣∣∣∣;

(2) ∣ 2 1 4 1 3 − 1 2 1 1 2 3 2 5 0 6 2 ∣ \left|\begin{array}{rrrr}2 & 1 & 4 & 1 \\ 3 & -1 & 2 & 1 \\ 1 & 2 & 3 & 2 \\ 5 & 0 & 6 & 2\end{array}\right| ∣∣∣∣∣∣∣∣23151−12042361122∣∣∣∣∣∣∣∣;

(3) ∣ − a b a c a e b d − c d d e b f c f − e f ∣ \left|\begin{array}{rrr}-a b & a c & a e \\ b d & -c d & d e \\ b f & c f & -e f\end{array}\right| ∣∣∣∣∣∣−abbdbfac−cdcfaede−ef∣∣∣∣∣∣;

(4) ∣ 1 1 1 a b c b + c c + a a + b ∣ \left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ b+c & c+a & a+b\end{array}\right| ∣∣∣∣∣∣1ab+c1bc+a1ca+b∣∣∣∣∣∣;

(5) ∣ a 1 0 0 − 1 b 1 0 0 − 1 c 1 0 0 − 1 d ∣ \left|\begin{array}{rrrr}a & 1 & 0 & 0 \\ -1 & b & 1 & 0 \\ 0 & -1 & c & 1 \\ 0 & 0 & -1 & d\end{array}\right| ∣∣∣∣∣∣∣∣a−1001b−1001c−1001d∣∣∣∣∣∣∣∣;

(6) ∣ 1 2 3 4 1 3 4 1 1 4 1 2 1 1 2 3 ∣ \left|\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 3 & 4 & 1 \\ 1 & 4 & 1 & 2 \\ 1 & 1 & 2 & 3\end{array}\right| ∣∣∣∣∣∣∣∣1111234134124123∣∣∣∣∣∣∣∣.

5. 求解下列方程:

(1) ∣ x + 1 2 − 1 2 x + 1 1 − 1 1 x + 1 ∣ = 0 \left|\begin{array}{rcr}x+1 & 2 & -1 \\ 2 & x+1 & 1 \\ -1 & 1 & x+1\end{array}\right|=0 ∣∣∣∣∣∣x+12−12x+11−11x+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 \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 ∣∣∣∣∣∣∣∣1xx2x31aa2a31bb2b31cc2c3∣∣∣∣∣∣∣∣=0

其中 a , b , c a, b, c a,b,c 互不相等.

6. 证明 :

(1) ∣ a 2 a b b 2 2 a a + b 2 b 1 1 1 ∣ = ( a − b ) 3 \left|\begin{array}{ccc}a^2 & a b & b^2 \\ 2 a & a+b & 2 b \\ 1 & 1 & 1\end{array}\right|=(a-b)^3 ∣∣∣∣∣∣a22a1aba+b1b22b1∣∣∣∣∣∣=(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 ∣ \left|\begin{array}{lll}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\end{array}\right|=\left(a^3+b^3\right)\left|\begin{array}{lll}x & y & z \\ y & z & x \\ z & x & y\end{array}\right| ∣∣∣∣∣∣ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz∣∣∣∣∣∣=(a3+b3)∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣;

(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 \left|\begin{array}{llll}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 ∣∣∣∣∣∣∣∣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) ∣ 1 1 1 1 a b c d a 2 b 2 c 2 d 2 a 4 b 4 c 4 d 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| ∣∣∣∣∣∣∣∣1aa2a41bb2b41cc2c41dd2d4∣∣∣∣∣∣∣∣ = ( a − b ) ( a − c ) ( a − d ) ( b − c ) ( b − d ) ( c − d ) ( a + b + c + d ) =(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(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 ∣ = a 3 x 3 + a 2 x 2 + a 1 x + a 0 \left|\begin{array}{rrrr} x & -1 & 0 & 0 \\ 0 & x & -1 & 0 \\ 0 & 0 & x & -1 \\ a_0 & a_1 & a_2 & a_3 \end{array}\right|=a_3 x^3+a_2 x^2+a_1 x+a_0 ∣∣∣∣∣∣∣∣x00a0−1x0a10−1xa200−1a3∣∣∣∣∣∣∣∣=a3x3+a2x2+a1x+a0

7. 设 n n n 阶行列式 D = det ⁡ ( a i j ) D=\operatorname{det}\left(a_{i j}\right) D=det(aij), 把 D D D 上下翻转、或逆时针旋转 9 0 ∘ 90^{\circ} 90∘ 、或依副对角线翻转, 依次得

D 1 = ∣ a n 1 ⋯ a n n ⋮ ⋮ a 11 ⋯ a 1 n ∣ , D 2 = ∣ a 1 n ⋯ a n n ⋮ ⋮ a 11 ⋯ a n 1 ∣ , D 3 = ∣ a a n ⋯ a 1 n ⋮ ⋮ a n 1 ⋯ a 11 ∣ , D_1=\left|\begin{array}{ccc} a_{n 1} & \cdots & a_{n n} \\ \vdots & & \vdots \\ a_{11} & \cdots & a_{1 n} \end{array}\right|, D_2=\left|\begin{array}{ccc} a_{1 n} & \cdots & a_{n n} \\ \vdots & & \vdots \\ a_{11} & \cdots & a_{n 1} \end{array}\right|, D_3=\left|\begin{array}{ccc} a_{\mathrm{an}} & \cdots & a_{1 n} \\ \vdots & & \vdots \\ a_{n 1} & \cdots & a_{11} \end{array}\right|, D1=∣∣∣∣∣∣∣an1⋮a11⋯⋯ann⋮a1n∣∣∣∣∣∣∣,D2=∣∣∣∣∣∣∣a1n⋮a11⋯⋯ann⋮an1∣∣∣∣∣∣∣,D3=∣∣∣∣∣∣∣aan⋮an1⋯⋯a1n⋮a11∣∣∣∣∣∣∣,
证明 D 1 = D 2 = ( − 1 ) n ( n − 1 ) 2 D , D 3 = D D_1=D_2=(-1)^{\frac{n(n-1)}{2}} D, D_3=D D1=D2=(−1)2n(n−1)D,D3=D.

8. 计算下列各行列式 ( D k D_k Dk 为 k k k 阶行列式):

(1) D n = ∣ a 1 ⋱ 1 a ∣ D_n=\left|\begin{array}{lll}a & & 1 \\ & \ddots & \\ 1 & & a\end{array}\right| Dn=∣∣∣∣∣∣a1⋱1a∣∣∣∣∣∣, 其中对角线上元素都是 a a a, 末写出的元素都是 0 ;

(2) D n = ∣ x a ⋯ a a x ⋯ a ⋮ ⋮ ⋮ a a ⋯ x ∣ D_n=\left|\begin{array}{cccc}x & a & \cdots & a \\ a & x & \cdots & a \\ \vdots & \vdots & & \vdots \\ a & a & \cdots & x\end{array}\right| Dn=∣∣∣∣∣∣∣∣∣xa⋮aax⋮a⋯⋯⋯aa⋮x∣∣∣∣∣∣∣∣∣

(3) D n + 1 = ∣ a n ( a − 1 ) n ⋯ ( a − n ) n a n − 1 ( a − 1 ) n − 1 ⋯ ( a − n ) n − 1 ⋮ ⋮ ⋮ a a − 1 ⋯ a − n 1 1 ⋯ 1 ∣ D_{n+1}=\left|\begin{array}{cccc}a^n & (a-1)^n & \cdots & (a-n)^n \\ a^{n-1} & (a-1)^{n-1} & \cdots & (a-n)^{n-1} \\ \vdots & \vdots & & \vdots \\ a & a-1 & \cdots & a-n \\ 1 & 1 & \cdots & 1\end{array}\right| Dn+1=∣∣∣∣∣∣∣∣∣∣∣anan−1⋮a1(a−1)n(a−1)n−1⋮a−11⋯⋯⋯⋯(a−n)n(a−n)n−1⋮a−n1∣∣∣∣∣∣∣∣∣∣∣,提示:利用范德蒙德行列式的结果;

(5) D n = ∣ 1 + a 1 a 1 ⋯ a 1 a 2 1 + a 2 ⋯ a 2 ⋮ ⋮ ⋮ a n a n ⋯ 1 + a n ∣ D_n=\left|\begin{array}{cccc}1+a_1 & a_1 & \cdots & a_1 \\ a_2 & 1+a_2 & \cdots & a_2 \\ \vdots & \vdots & & \vdots \\ a_n & a_n & \cdots & 1+a_n\end{array}\right| Dn=∣∣∣∣∣∣∣∣∣1+a1a2⋮ana11+a2⋮an⋯⋯⋯a1a2⋮1+an∣∣∣∣∣∣∣∣∣;

(6) D n = det ⁡ ( a i j ) D_n=\operatorname{det}\left(a_{i j}\right) Dn=det(aij), 其中 a i j = ∣ i − j ∣ a_{i j}=|i-j| aij=∣i−j∣;

(7) D n = ∣ 1 + a 1 1 ⋯ 1 1 1 + a 2 ⋯ 1 ⋮ ⋮ ⋮ 1 1 ⋯ 1 + a n ∣ D_n=\left|\begin{array}{cccc}1+a_1 & 1 & \cdots & 1 \\ 1 & 1+a_2 & \cdots & 1 \\ \vdots & \vdots & & \vdots \\ 1 & 1 & \cdots & 1+a_n\end{array}\right| Dn=∣∣∣∣∣∣∣∣∣1+a11⋮111+a2⋮1⋯⋯⋯11⋮1+an∣∣∣∣∣∣∣∣∣, 其中 a 1 a 2 ⋯ a n ≠ 0 a_1 a_2 \cdots a_n \neq 0 a1a2⋯an=0.

9. 设 D = ∣ 3 1 − 1 2 − 5 1 3 − 4 2 0 1 − 1 1 − 5 3 − 3 ∣ , D D=\left|\begin{array}{rrrr}3 & 1 & -1 & 2 \\ -5 & 1 & 3 & -4 \\ 2 & 0 & 1 & -1 \\ 1 & -5 & 3 & -3\end{array}\right|, D D=∣∣∣∣∣∣∣∣3−521110−5−13132−4−1−3∣∣∣∣∣∣∣∣,D 的 ( i , j ) (i, j) (i,j) 元的代数余子式记作 A i j A_{i j} Aij, 求

A 31 + 3 A 32 − 2 A 33 + 2 A 34 . A_{31}+3 A_{32}-2 A_{33}+2 A_{34} \text {. } A31+3A32−2A33+2A34.