线性代数第五版答案(全).pdf

第一章 行列式 1. 利用对角线法则计算下列三阶行列式: 2 0 1 (1) 1 −4 −1; −1 8 3 2 0 1 解 1 −4 −1 −1 8 3 2×(−4)×3+0×(− 1)×(− 1)+ 1×1×8 −0×1×3−2×(− 1)×8− 1×(−4)×(− 1) −24+8+ 16−4 −4. a b c (2) b c a ; c a b a b c 解 b c a c a b acb+bac+cba−bbb−aaa−ccc 3 3 3 3abc−a −b −c . 1 1 1 (3) a b c ; a2 b2 c2 1 1 1 解 a b c a2 b2 c2 2 2 2 2 2 2 bc +ca +ab −ac −ba −cb (a−b)(b−c)(c−a). 1 x y x + y (4) y x + y x . x + y x y x y x + y 解 y x + y x x + y x y 3 3 3 x(x+y )y +yx (x+y )+(x+y )yx −y −(x+y ) −x 3 2 3 3 3 3xy(x+y )−y −3x y −x −y −x 3 3 −2(x +y ). 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)1 2 3 4; 解 逆序数为0 (2)4 1 3 2; 解 逆序数为4: 41, 43, 42, 32. (3)3 4 2 1; 解 逆序数为5: 3 2, 3 1, 4 2, 4 1, 2 1. (4)2 4 1 3; 解 逆序数为3: 2 1, 4 1, 4 3. (5)1 3 ⋅ ⋅ ⋅ (2n− 1) 2 4 ⋅ ⋅ ⋅ (2n); 解 逆序数为n(n −1) : 2 3 2 (1 个) 5 2, 5 4(2 个) 7 2, 7 4, 7 6(3 个) 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ (2n− 1)2, (2n− 1)4, (2n− 1)6, ⋅ ⋅ ⋅, (2n− 1)(2n−2) (n− 1 个) (6)1 3 ⋅ ⋅ ⋅ (2n− 1) (2n) (2n−2) ⋅ ⋅ ⋅ 2. 解 逆序数为n(n− 1) : 3 2(1 个) 5 2, 5 4 (2 个) ⋅ ⋅ ⋅ ⋅