线性代数第五版答案(全)

第一章 行列式 1( 利用对角线法则计算下列三阶行列式( (1) ( 解 (2(((4)(3(0(((1)(((1)(1(1(8 (0(1(3(2(((1)(8(1(((4)(((1) ((24(8(16(4((4( (2) ( 解 (acb(bac(cba(bbb(aaa(ccc (3abc(a3(b3(c3( (3) ( 解 (bc2(ca2(ab2(ac2(ba2(cb2 ((a(b)(b(c)(c(a)( (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)( 2( 按自然数从小到大为 标准 excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载 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 ( 3 2 (1个) 5 2( 5 4(2个) 7 2( 7 4( 7 6(3个) ( ( ( ( ( ( (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) ( ( ( ( ( ( (2n(1)2( (2n(1)4( (2n(1)6( ( ( (( (2n(1)(2n(2) (n(1个) 4 2(1个) 6 2( 6 4(2个) ( ( ( ( ( ( (2n)2( (2n)4( (2n)6( ( ( (( (2n)(2n(2) (n(1个) 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 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 ( (2) ( 解 EMBED Equation.3 EMBED Equation.3 ( (3) ( 解 EMBED Equation.3 ( (4) ( 解 EMBED Equation.3 EMBED Equation.3 (abcd(ab(cd(ad(1( 5( 证明: (1) ((a(b)3; 证明 EMBED Equation.3 EMBED Equation.3 ((a(b)3 ( (2) ; 证明 ( (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 数学高考答题卡模板高考数学答题卡模板三年级数学混合运算测试卷数学作业设计案例新人教版八年级上数学教学计划 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( 6( 设n阶行列式D(det(aij), 把D上下翻转、或逆时针旋转90(、或依副对角线翻转( 依次得 ( ( ( 证明 ( D3(D ( 证明　因为D(det(aij)( 所以 ( 同理可证 EMBED Equation.3 ( ( 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( (3) ; 解 根据第6题结果( 有 此行列式为范德蒙德行列式( ( (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 ( (5) D(det(aij)( 其中aij(|i(j|; 解 aij(|i(j|( (((1)n(1(n(1)2n(2( (6) , 其中a1a2 ( ( ( an(0( 解 ( 8( 用克莱姆法则解下列方程组( (1) ( 解 因为 ( ( ( ( ( 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( 第二章　矩阵及其运算 1( 已知线性变换( ( 求从变量x1( x2( x3到变量y1( y2( y3的线性变换( 解 由已知( ( 故 EMBED Equation.3 ( ( 2( 已知两个线性变换 ( ( 求从z1( z2( z3到x1( x2( x3的线性变换( 解 由已知 EMBED Equation.3 ( 所以有 ( 3( 设 ( ( 求3AB(2A及ATB( 解 ( ( 4( 计算下列乘积( (1) ( 解 EMBED Equation.3 EMBED Equation.3 ( (2) ( 解 ((1(3(2(2(3(1)((10)( (3) ( 解 EMBED Equation.3 EMBED Equation.3 ( (4) ( 解 EMBED Equation.3 ( (5) ( 解 ((a11x1(a12x2(a13x3 a12x1(a22x2(a23x3 a13x1(a23x2(a33x3) ( 5( 设 ( ( 问( (1)AB(BA吗? 解 AB(BA( 因为 ( ( 所以AB(BA( (2)(A(B)2(A2(2AB(B2吗? 解 (A(B)2(A2(2AB(B2( 因为 ( EMBED Equation.3 ( 但 EMBED Equation.3 ( 所以(A(B)2(A2(2AB(B2( (3)(A(B)(A(B)(A2(B2吗? 解 (A(B)(A(B)(A2(B2( 因为 ( ( ( 而 ( 故(A(B)(A(B)(A2(B2( 6( 举反列说明下列命题是错误的( (1)若A2(0( 则A(0( 解 取 ( 则A2(0( 但A(0( (2)若A2(A( 则A(0或A(E( 解 取 ( 则A2(A( 但A(0且A(E( (3)若AX(AY( 且A(0( 则X(Y ( 解 取 ( ( ( 则AX(AY( 且A(0( 但X(Y ( 7( 设 ( 求A2( A3( ( ( (( Ak( 解 ( ( ( ( ( ( ( (( ( 8( 设 ( 求Ak ( 解 首先观察 EMBED Equation.3 ( ( ( ( ( ( ( ( ( (( EMBED Equation.3 EMBED Equation.3 ( 用数学归纳法证明( 当k(2时( 显然成立( 假设k时成立，则k(1时， ( 由数学归纳法原理知( ( 9( 设A( B为n阶矩阵，且A为对称矩阵，证明BTAB也是对称矩阵( 证明 因为AT(A( 所以 (BTAB)T(BT(BTA)T(BTATB(BTAB( 从而BTAB是对称矩阵( 10( 设A( B都是n阶对称矩阵，证明AB是对称矩阵的充分必要条件是AB(BA( 证明 充分性( 因为AT(A( BT(B( 且AB(BA( 所以 (AB)T((BA)T(ATBT(AB( 即AB是对称矩阵( 必要性( 因为AT(A( BT(B( 且(AB)T(AB( 所以 AB((AB)T(BTAT(BA( 11( 求下列矩阵的逆矩阵( (1) ( 解 ( |A|(1( 故A(1存在( 因为 ( 故 EMBED Equation.3 ( (2) ( 解 ( |A|(1(0( 故A(1存在( 因为 ( 所以 EMBED Equation.3 ( (3) ( 解 ( |A|(2(0( 故A(1存在( 因为 ( 所以 EMBED Equation.3 ( (4) (a1a2( ( (an (0) ( 解 ( 由对角矩阵的性质知 ( 12( 解下列矩阵方程( (1) ( 解 EMBED Equation.3 EMBED Equation.3 ( (2) ( 解 ( (3) ( 解 EMBED Equation.3 ( (4) ( 解 EMBED Equation.3 ( 13( 利用逆矩阵解下列线性方程组( (1) ( 解 方程组可 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 示为 ( 故 ( 从而有 ( (2) ( 解 方程组可表示为 ( 故 ( 故有 ( 14( 设Ak(O (k为正整数)( 证明(E(A)(1(E(A(A2(( ( ((Ak(1( 证明 因为Ak(O ( 所以E(Ak(E( 又因为 E(Ak((E(A)(E(A(A2(( ( ((Ak(1)( 所以 (E(A)(E(A(A2(( ( ((Ak(1)(E( 由定理2推论知(E(A)可逆( 且 (E(A)(1(E(A(A2(( ( ((Ak(1( 证明 一方面( 有E((E(A)(1(E(A)( 另一方面( 由Ak(O( 有 E((E(A)((A(A2)(A2(( ( ((Ak(1((Ak(1(Ak) ((E(A(A2(( ( ((A k(1)(E(A)( 故 (E(A)(1(E(A)((E(A(A2(( ( ((Ak(1)(E(A)( 两端同时右乘(E(A)(1( 就有 (E(A)(1(E(A)(E(A(A2(( ( ((Ak(1( 15( 设方阵A满足A2(A(2E(O( 证明A及A(2E都可逆( 并求A(1及(A(2E)(1( 证明 由A2(A(2E(O得 A2(A(2E( 即A(A(E)(2E( 或 ( 由定理2推论知A可逆( 且 ( 由A2(A(2E(O得 A2(A(6E((4E( 即(A(2E)(A(3E)((4E( 或 由定理2推论知(A(2E)可逆( 且 ( 证明 由A2(A(2E(O得A2(A(2E( 两端同时取行列式得 |A2(A|(2( 即 |A||A(E|(2( 故 |A|(0( 所以A可逆( 而A(2E(A2( |A(2E|(|A2|(|A|2(0( 故A(2E也可逆( 由 A2(A(2E(O (A(A(E)(2E (A(1A(A(E)(2A(1E( ( 又由 A2(A(2E(O((A(2E)A(3(A(2E)((4E ( (A(2E)(A(3E)((4 E( 所以 (A(2E)(1(A(2E)(A(3E)((4(A(2 E)(1( ( 16( 设A为3阶矩阵( ( 求|(2A)(1(5A*|( 解 因为 ( 所以 EMBED Equation.3 (|(2A(1|(((2)3|A(1|((8|A|(1((8(2((16( 17( 设矩阵A可逆( 证明其伴随阵A*也可逆( 且(A*)(1((A(1)*( 证明 由 ( 得A*(|A|A(1( 所以当A可逆时( 有 |A*|(|A|n|A(1|(|A|n(1(0( 从而A*也可逆( 因为A*(|A|A(1( 所以 (A*)(1(|A|(1A( 又 ( 所以 (A*)(1(|A|(1A(|A|(1|A|(A(1)*((A(1)*( 18( 设n阶矩阵A的伴随矩阵为A*( 证明( (1)若|A|(0( 则|A*|(0( (2)|A*|(|A|n(1( 证明 (1)用反证法证明( 假设|A*|(0( 则有A*(A*)(1(E( 由此得 A(A A*(A*)(1(|A|E(A*)(1(O ( 所以A*(O( 这与|A*|(0矛盾，故当|A|(0时( 有|A*|(0( (2)由于 ( 则AA*(|A|E( 取行列式得到 |A||A*|(|A|n( 若|A|(0( 则|A*|(|A|n(1( 若|A|(0( 由(1)知|A*|(0( 此时命题也成立( 因此|A*|(|A|n(1( 19( 设 ( AB(A(2B( 求B( 解 由AB(A(2E可得(A(2E)B(A( 故 EMBED Equation.3 ( 20( 设 ( 且AB(E(A2(B( 求B( 解 由AB(E(A2(B得 (A(E)B(A2(E( 即 (A(E)B((A(E)(A(E)( 因为 ( 所以(A(E)可逆( 从而 ( 21( 设A(diag(1( (2( 1)( A*BA(2BA(8E( 求B( 解 由A*BA(2BA(8E得 (A*(2E)BA((8E( B((8(A*(2E)(1A(1 ((8[A(A*(2E)](1 ((8(AA*(2A)(1 ((8(|A|E(2A)(1 ((8((2E(2A)(1 (4(E(A)(1 (4[diag(2( (1( 2)](1 (2diag(1( (2( 1)( 22( 已知矩阵A的伴随阵 ( 且ABA(1(BA(1(3E( 求B( 解 由|A*|(|A|3(8( 得|A|(2( 由ABA(1(BA(1(3E得 AB(B(3A( B(3(A(E)(1A(3[A(E(A(1)](1A ( 23( 设P(1AP((( 其中 ( ( 求A11( 解 由P(1AP((( 得A(P(P(1( 所以A11( A=P(11P(1. |P|(3( ( ( 而 ( 故 EMBED Equation.3 ( 24( 设AP(P(( 其中 ( ( 求((A)(A8(5E(6A(A2)( 解 ((()((8(5E(6(((2) (diag(1(1(58)[diag(5(5(5)(diag((6(6(30)(diag(1(1(25)] (diag(1(1(58)diag(12(0(0)(12diag(1(0(0)( ((A)(P((()P(1 ( 25( 设矩阵A、B及A(B都可逆( 证明A(1(B(1也可逆( 并求其逆阵( 证明 因为 A(1(A(B)B(1(B(1(A(1(A(1(B(1( 而A(1(A(B)B(1是三个可逆矩阵的乘积( 所以A(1(A(B)B(1可逆( 即A(1(B(1可逆( (A(1(B(1)(1([A(1(A(B)B(1](1(B(A(B)(1A( 26( 计算 ( 解 设 ( ( ( ( 则 EMBED Equation.3 ( 而 ( ( 所以 EMBED Equation.3 EMBED Equation.3 ( 即 EMBED Equation.3 ( 27( 取 ( 验证 ( 解 ( 而 ( 故 ( 28( 设 ( 求|A8|及A4( 解　令 ( ( 则 ( 故 EMBED Equation.3 ( ( ( 29( 设n阶矩阵A及s阶矩阵B都可逆( 求 (1) ( 解 设 ( 则 EMBED Equation.3 EMBED Equation.3 ( 由此得 ( ( 所以 ( (2) ( 解 设 ( 则 ( 由此得 ( ( 所以 ( 30( 求下列矩阵的逆阵( (1) ( 解 设 ( ( 则 ( ( 于是 ( (2) ( 解 设 ( ( ( 则 ( 第三章　矩阵的初等变换与线性方程组 1( 把下列矩阵化为行最简形矩阵( (1) ( 解 (下一步( r2(((2)r1( r3(((3)r1( ) ~ (下一步( r2(((1)( r3(((2)( ) ~ (下一步( r3(r2( ) ~ (下一步( r3(3( ) ~ (下一步( r2(3r3( ) ~ (下一步( r1(((2)r2( r1(r3( ) ~ ( (2) ( 解 (下一步( r2(2(((3)r1( r3(((2)r1( ) ~ (下一步( r3(r2( r1(3r2( ) ~ (下一步( r1(2( ) ~ ( (3) ( 解 (下一步( r2(3r1( r3(2r1( r4(3r1( ) ~ (下一步( r2(((4)( r3(((3) ( r4(((5)( ) ~ (下一步( r1(3r2( r3(r2( r4(r2( ) ~ ( (4) ( 解 (下一步( r1(2r2( r3(3r2( r4(2r2( ) ~ (下一步( r2(2r1( r3(8r1( r4(7r1( ) ~ (下一步( r1(r2( r2(((1)( r4(r3( ) ~ (下一步( r2(r3( ) ~ ( 2( 设 ( 求A( 解 是初等矩阵E(1( 2)( 其逆矩阵就是其本身( 是初等矩阵E(1( 2(1))( 其逆矩阵是 E(1( 2((1)) ( ( 3( 试利用矩阵的初等变换( 求下列方阵的逆矩阵( (1) ( 解 ~ ~ ~ ~ 故逆矩阵为 ( (2) ( 解 ~ ~ ~ ~ ~ 故逆矩阵为 ( 4( (1)设 ( ( 求X使AX(B( 解 因为 EMBED Equation.3 ( 所以 ( (2)设 ( ( 求X使XA(B( 解 考虑ATXT(BT( 因为 EMBED Equation.3 ( 所以 ( 从而 ( 5( 设 ( AX (2X(A( 求X( 解 原方程化为(A(2E)X (A( 因为 ( 所以 ( 6( 在秩是r 的矩阵中,有没有等于0的r(1阶子式? 有没有等于0的r阶子式? 解 在秩是r的矩阵中( 可能存在等于0的r(1阶子式( 也可能存在等于0的r阶子式( 例如( ( R(A)(3( 是等于0的2阶子式( 是等于0的3阶子式( 7( 从矩阵A中划去一行得到矩阵B( 问A( B的秩的关系怎样? 解 R(A)(R(B)( 这是因为B的非零子式必是A的非零子式( 故A的秩不会小于B的秩( 8( 求作一个秩是4的方阵( 它的两个行向量是 (1( 0( 1( 0( 0)( (1( (1( 0( 0( 0)( 解 用已知向量容易构成一个有4个非零行的5阶下三角矩阵( ( 此矩阵的秩为4( 其第2行和第3行是已知向量( 9( 求下列矩阵的秩( 并求一个最高阶非零子式( (1) ; 解 (下一步( r1(r2( ) ~ (下一步( r2(3r1( r3(r1( ) ~ (下一步( r3(r2( ) ~ ( 矩阵的 ( 是一个最高阶非零子式( (2) ( 解 (下一步( r1(r2( r2(2r1( r3(7r1( ) ~ (下一步( r3(3r2( ) ~ ( 矩阵的秩是2( 是一个最高阶非零子式( (3) ( 解 (下一步( r1(2r4( r2(2r4( r3(3r4( ) ~ (下一步( r2(3r1( r3(2r1( ) ~ (下一步( r2(16r4( r3(16r2( ) ~ ~ ( 矩阵的秩为3( 是一个最高阶非零子式( 10( 设A、B都是m(n矩阵( 证明A~B的充分必要条件是R(A)(R(B)( 证明 根据定理3( 必要性是成立的( 充分性( 设R(A)(R(B)( 则A与B的标准形是相同的( 设A与B的标准形为D( 则有 A~D( D~B( 由等价关系的传递性( 有A~B( 11( 设 ( 问k为何值( 可使 (1)R(A)(1( (2)R(A)(2( (3)R(A)(3( 解 EMBED Equation.3 ( (1)当k(1时( R(A)(1( (2)当k((2且k(1时( R(A)(2( (3)当k(1且k((2时( R(A)(3( 12( 求解下列齐次线性方程组: (1) ( 解　对系数矩阵A进行初等行变换( 有 A( ~ ( 于是 ( 故方程组的解为 (k为任意常数)( (2) ( 解 对系数矩阵A进行初等行变换( 有 A( ~ ( 于是 ( 故方程组的解为 (k1( k2为任意常数)( (3) ( 解 对系数矩阵A进行初等行变换( 有 A( ~ ( 于是 ( 故方程组的解为 ( (4) ( 解 对系数矩阵A进行初等行变换( 有 A( ~ ( 于是 ( 故方程组的解为 (k1( k2为任意常数)( 13( 求解下列非齐次线性方程组: (1) ( 解 对增广矩阵B进行初等行变换( 有 B( ~ ( 于是R(A)(2( 而R(B)(3( 故方程组无解( (2) ( 解 对增广矩阵B进行初等行变换( 有 B( ~ ( 于是 ( 即 (k为任意常数)( (3) ( 解 对增广矩阵B进行初等行变换( 有 B( ~ ( 于是 ( 即 (k1( k2为任意常数)( (4) ( 解 对增广矩阵B进行初等行变换( 有 B( ~ ( 于是 ( 即 (k1( k2为任意常数)( 14( 写出一个以 为通解的齐次线性方程组( 解 根据已知( 可得 ( 与此等价地可以写成 ( 或 ( 或 ( 这就是一个满足题目要求的齐次线性方程组( 15( (取何值时( 非齐次线性方程组 ( (1)有唯一解( (2)无解( (3)有无穷多个解? 解 ( (1)要使方程组有唯一解( 必须R(A)(3( 因此当((1且(((2时方程组有唯一解. (2)要使方程组无解( 必须R(A)(R(B)( 故 (1(()(2(()(0( (1(()(((1)2(0( 因此(((2时( 方程组无解( (3)要使方程组有有无穷多个解( 必须R(A)(R(B)(3( 故 (1(()(2(()(0( (1(()(((1)2(0( 因此当((1时( 方程组有无穷多个解. 16( 非齐次线性方程组 当(取何值时有解？并求出它的解( 解　 ~ ( 要使方程组有解( 必须(1(()(((2)(0( 即((1( (((2( 当((1时( ~ ( 方程组解为 或 ( 即 (k为任意常数)( 当(((2时( ~ ( 方程组解为 或 ( 即 (k为任意常数)( 17( 设 ( 问(为何值时( 此方程组有唯一解、无解或有无穷多解? 并在有无穷多解时求解( 解 B( ~ ( 要使方程组有唯一解( 必须R(A)(R(B)(3( 即必须 (1(()(10(()(0( 所以当((1且((10时( 方程组有唯一解. 要使方程组无解( 必须R(A)(R(B)( 即必须 (1(()(10(()(0且(1(()(4(()(0( 所以当((10时( 方程组无解. 要使方程组有无穷多解( 必须R(A)(R(B)(3( 即必须 (1(()(10(()(0且(1(()(4(()(0( 所以当((1时( 方程组有无穷多解(此时，增广矩阵为 B~ ( 方程组的解为 ( 或 (k1( k2为任意常数)( 18( 证明R(A)(1的充分必要条件是存在非零列向量a及非零行向量bT( 使A(abT( 证明 必要性( 由R(A)(1知A的标准形为 ( 即存在可逆矩阵P和Q( 使 ( 或 ( 令 ( bT((1( 0( (((( 0)Q(1( 则a是非零列向量( bT是非零行向量( 且A(abT( 充分性( 因为a与bT是都是非零向量( 所以A是非零矩阵( 从而R(A)(1( 因为 1(R(A)(R(abT)(min{R(a)( R(bT)}(min{1( 1}(1( 所以R(A)(1( 19( 设A为m(n矩阵( 证明 (1)方程AX(Em有解的充分必要条件是R(A)(m( 证明 由定理7( 方程AX(Em有解的充分必要条件是 R(A)(R(A( Em)( 而| Em|是矩阵(A( Em)的最高阶非零子式( 故R(A)(R(A( Em)(m( 因此( 方程AX(Em有解的充分必要条件是R(A)(m( (2)方程YA(En有解的充分必要条件是R(A)(n( 证明 注意( 方程YA(En有解的充分必要条件是ATYT(En有解( 由(1) ATYT(En有解的充分必要条件是R(AT)(n( 因此，方程YA(En有解的充分必要条件是R(A)(R(AT)(n( 20( 设A为m(n矩阵( 证明( 若AX(AY( 且R(A)(n( 则X(Y( 证明 由AX(AY( 得A(X(Y)(O( 因为R(A)(n( 由定理9( 方程A(X(Y)(O只有零解( 即X(Y(O( 也就是X(Y( 第四章　向量组的线性相关性 1( 设v1((1( 1( 0)T( v2((0( 1( 1)T( v3((3( 4( 0)T( 求v1(v2及3v1(2v2(v3( 解 v1(v2((1( 1( 0)T((0( 1( 1)T ((1(0( 1(1( 0(1)T ((1( 0( (1)T( 3v1(2v2(v3(3(1( 1( 0)T (2(0( 1( 1)T ((3( 4( 0)T ((3(1(2(0(3( 3(1(2(1(4( 3(0(2(1(0)T ((0( 1( 2)T( 2( 设3(a1(a)(2(a2(a)(5(a3(a)( 求a( 其中a1((2( 5( 1( 3)T( a2((10( 1( 5( 10)T( a3((4( 1( (1( 1)T( 解 由3(a1(a)(2(a2(a)(5(a3(a)整理得 ((1( 2( 3( 4)T( 3( 已知向量组 A( a1((0( 1( 2( 3)T( a2((3( 0( 1( 2)T( a3((2( 3( 0( 1)T( B( b1((2( 1( 1( 2)T( b2((0( (2( 1( 1)T( b3((4( 4( 1( 3)T( 证明B组能由A组线性表示( 但A组不能由B组线性表示( 证明 由 EMBED Equation.3 EMBED Equation.3 知R(A)(R(A( B)(3( 所以B组能由A组线性表示( 由 知R(B)(2( 因为R(B)(R(B( A)( 所以A组不能由B组线性表示( 4( 已知向量组 A( a1((0( 1( 1)T( a2((1( 1( 0)T( B( b1(((1( 0( 1)T( b2((1( 2( 1)T( b3((3( 2( (1)T( 证明A组与B组等价( 证明 由 ( 知R(B)(R(B( A)(2( 显然在A中有二阶非零子式( 故R(A)(2( 又R(A)(R(B( A)(2( 所以R(A)(2( 从而R(A)(R(B)(R(A( B)( 因此A组与B组等价( 5( 已知R(a1( a2( a3)(2( R(a2( a3( a4)(3( 证明 (1) a1能由a2( a3线性表示( (2) a4不能由a1( a2( a3线性表示( 证明 (1)由R(a2( a3( a4)(3知a2( a3( a4线性无关( 故a2( a3也线性无关( 又由R(a1( a2( a3)(2知a1( a2( a3线性相关( 故a1能由a2( a3线性表示( (2)假如a4能由a1( a2( a3线性表示( 则因为a1能由a2( a3线性表示( 故a4能由a2( a3线性表示( 从而a2( a3( a4线性相关( 矛盾( 因此a4不能由a1( a2( a3线性表示( 6( 判定下列向量组是线性相关还是线性无关( (1) ((1( 3( 1)T( (2( 1( 0)T( (1( 4( 1)T( (2) (2( 3( 0)T( ((1( 4( 0)T( (0( 0( 2)T( 解 (1)以所给向量为列向量的矩阵记为A( 因为 ( 所以R(A)(2小于向量的个数( 从而所给向量组线性相关( (2)以所给向量为列向量的矩阵记为B( 因为 ( 所以R(B)(3等于向量的个数( 从而所给向量组线性相无关( 7( 问a取什么值时下列向量组线性相关？ a1((a( 1( 1)T( a2((1( a( (1)T( a3((1( (1( a)T( 解 以所给向量为列向量的矩阵记为A( 由 知( 当a((1、0、1时( R(A)(3( 此时向量组线性相关( 8( 设a1( a2线性无关( a1(b( a2(b线性相关( 求向量b用a1( a2线性表示的表示式( 解 因为a1(b( a2(b线性相关( 故存在不全为零的数(1( (2使 (1(a1(b)((2(a2(b)(0( 由此得 ( 设 ( 则 b(ca1((1(c)a2( c(R( 9( 设a1( a2线性相关( b1( b2也线性相关( 问a1(b1( a2(b2是否一定线性相关？试举例说明之( 解 不一定( 例如( 当a1((1( 2)T, a2((2( 4)T, b1(((1( (1)T, b2((0( 0)T时( 有 a1(b1((1( 2)T(b1((0( 1)T, a2(b2((2( 4)T((0( 0)T((2( 4)T( 而a1(b1( a2(b2的对应分量不成比例( 是线性无关的( 10( 举例说明下列各命题是错误的( (1)若向量组a1( a2( ( ( (( am是线性相关的( 则a1可由a2( ( ( (( am线性表示( 解 设a1(e1((1( 0( 0( ( ( (( 0)( a2(a3( ( ( ( (am(0( 则a1( a2( ( ( (( am线性相关( 但a1不能由a2( ( ( (( am线性表示( (2)若有不全为0的数(1( (2( ( ( (( (m使 (1a1( ( ( ( ((mam((1b1( ( ( ( ((mbm(0 成立( 则a1( a2( ( ( (( am线性相关, b1( b2( ( ( (( bm亦线性相关( 解 有不全为零的数(1( (2( ( ( (( (m使 (1a1( ( ( ( ((mam ((1b1( ( ( ( ((mbm (0( 原式可化为 (1(a1(b1)( ( ( ( ((m(am(bm)(0( 取a1(e1((b1( a2(e2((b2( ( ( (( am(em((bm( 其中e1( e2( ( ( (( em为单位坐标向量( 则上式成立( 而a1( a2( ( ( (( am和b1( b2( ( ( (( bm均线性无关( (3)若只有当(1( (2( ( ( (( (m全为0时( 等式 (1a1( ( ( ( ((mam((1b1( ( ( ( ((mbm(0 才能成立( 则a1( a2( ( ( (( am线性无关, b1( b2( ( ( (( bm亦线性无关( 解 由于只有当(1( (2( ( ( (( (m全为0时( 等式 由(1a1( ( ( ( ((mam((1b1( ( ( ( ((mbm (0 成立( 所以只有当(1( (2( ( ( (( (m全为0时( 等式 (1(a1(b1)((2(a2(b2)( ( ( ( ((m(am(bm)(0 成立( 因此a1(b1( a2(b2( ( ( (( am(bm线性无关( 取a1(a2( ( ( ( (am(0( 取b1( ( ( (( bm为线性无关组( 则它们满足以上条件( 但a1( a2( ( ( (( am线性相关( (4)若a1( a2( ( ( (( am线性相关, b1( b2( ( ( (( bm亦线性相关( 则有不全为0的数( (1( (2( ( ( (( (m使 (1a1( ( ( ( ((mam(0( (1b1( ( ( ( ((mbm(0 同时成立( 解 a1((1( 0)T( a2((2( 0)T( b1((0( 3)T( b2((0( 4)T( (1a1((2a2 (0((1((2(2( (1b1((2b2 (0((1(((3/4)(2( ((1((2(0( 与题设矛盾( 11( 设b1(a1(a2( b2(a2(a3( b3(a3(a4( b4(a4(a1( 证明向量组b1( b2( b3( b4线性相关( 证明 由已知条件得 a1(b1(a2( a2(b2(a3( a3(b3(a4( a4(b4(a1( 于是 a1 (b1(b2(a3 (b1(b2(b3(a4 (b1(b2(b3(b4(a1( 从而 b1(b2(b3(b4(0( 这说明向量组b1( b2( b3( b4线性相关( 12( 设b1(a1( b2(a1(a2( ( ( (( br (a1(a2( ( ( ( (ar( 且向量组a1( a2( ( ( ( ( ar线性无关( 证明向量组b1( b2( ( ( ( ( br线性无关( 证明 已知的r个等式可以写成 ( 上式记为B(AK( 因为|K|(1(0( K可逆( 所以R(B)(R(A)(r( 从而向量组b1( b2( ( ( ( ( br线性无关( 13( 求下列向量组的秩, 并求一个最大无关组( (1)a1((1( 2( (1( 4)T( a2((9( 100( 10( 4)T( a3(((2( (4( 2( (8)T( 解　由 ( 知R(a1( a2( a3)(2( 因为向量a1与a2的分量不成比例( 故a1( a2线性无关( 所以a1( a2是一个最大无关组( (2)a1T((1( 2( 1( 3)( a2T((4( (1( (5( (6)( a3T((1( (3( (4( (7)( 解 由 ( 知R(a1T( a2T( a3T)(R(a1( a2( a3)(2( 因为向量a1T与a2T的分量不成比例( 故a1T( a2T线性无关( 所以a1T( a2T是一个最大无关组( 14( 利用初等行变换求下列矩阵的列向量组的一个最大无关组( (1) ( 解 因为 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 ( 所以第1、2、3列构成一个最大无关组. (2) ( 解 因为 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 ( 所以第1、2、3列构成一个最大无关组( 15( 设向量组 (a( 3( 1)T( (2( b( 3)T( (1( 2( 1)T( (2( 3( 1)T 的秩为2( 求a( b( 解 设a1((a( 3( 1)T( a2((2( b( 3)T( a3((1( 2( 1)T( a4((2