同济大学4版线性代数答案.docVIP

第一章 行列式 1( 利用对角线法则计算下列三阶行列式( (4)( 解 (x(x(y)y(yx(x(y)((x(y)yx(y3((x(y)3(x3 (3xy(x(y)(y3(3x2 y(x3(y3(x3 ((2(x3(y3)( 3( 写出四阶行列式中含有因子a11a23的项( 解 含因子a11a23的项的一般形式为 ((1)ta11a23a3ra4s( 其中rs是2和4构成的排列( 这种排列共有两个( 即24和42( 所以含因子a11a23的项分别是 ((1)ta11a23a32a44(((1)1a11a23a32a44((a11a23a32a44( ((1)ta11a23a34a42(((1)2a11a23a34a42(a11a23a34a42( 4( 计算下列各行列式( (1)( 解 ( (2)( 解 ( 5( 证明: (1)((a(b)3; 证明 ((a(b)3 ( (2); 证明 ( 1 (3); 证明 (c4(c3( c3(c2( c2(c1得) (c4(c3( c3(c2得) ( (4) ((a(b)(a(c)(a(d)(b(c)(b(d)(c(d)(a(b(c(d); 证明 =(a(b)(a(c)(a(d)(b(c)(b(d)(c(d)(a(b(c(d)( (5)(xn(a1xn(1( ( ( ( (an(1x(an ( 证明 用数学归纳法证明( 当n(2时( ( 命题成立( 假设对于(n(1)阶行列式命题成立( 即 Dn(1(xn(1(a1 xn(2( ( ( ( (an(2x(an(1( 则Dn按第一列展开( 有 (xD n(1(an(xn(a1xn(1( ( ( ( (an(1x(an ( 因此( 对于n阶行列式命题成立( 7( 计算下列各行列式(Dk为k阶行列式)( (1), 其中对角线上元素都是a( 未写出的元素都是0( 解 (按第n行展开) (an(an(2(an(2(a2(1)( (2); 解 将第一行乘((1)分别加到其余各行( 得 ( 再将各列都加到第一列上( 得 ([x((n(1)a](x(a)n(1( (4); 解 (按第1行展开) ( 再按最后一行展开得递推公式 D2n(andnD2n(2(bncnD2n(2( 即D2n((andn(bncn)D2n(2( 于是 ( 而 ( 所以 ( (2)( 解 因为 ( ( ( ( ( ( 所以 ( ( ( ( ( 9( 问(( (取何值时( 齐次线性方程组有非零解？ 解 系数行列式为 ( 令D(0( 得 ((0或((1( 于是( 当((0或((1时该齐次线性方程组有非零解( 10( 问(取何值时( 齐次线性方程组有非零解？ 解 系数行列式为 ((1(()3((((3)(4(1(()(2(1(()((3(() ((1(()3(2(1(()2(((3( 令D(0( 得 ((0( ((2或((3( 于是( 当((0( ((2或((3时( 该齐次线性方程组有非零解( 大学生校园网—VvSchool.CN努力打造大学生最实用的网络平台！ 4