复习与习题线性代数问答汇总

1、若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考行列式答 上述行列式中 x 的 3 次方的系数为 -1.分析 根据行列式的定义, 4 阶行列式写成S和式后, 通项为(-1)t ( j1 , j2 , j3 , j4 ) aaaa1 j1 2 j2 3 j3 4 j4其中 j1, j2, j3, j4 为 1, 2, 3, 4 的全排列, t( j1, j2, j3, j4)表示排列 j1, j2, j3, j4 的逆序数. 注意到通项中的 aij 取自不同的行和不同的列, 上述 4 阶行列式的展开式中只有(-1)t( 2134)a12a21a33a44 = (-1)1x1xx

2、= -x3这一项含有 x3, 其系数为-1.(a +1)2(b +1)2(c +1)2(d +1)2(a +1)2(b +1)2(c +1)2(d +1)2(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 + 2)2(b + 2)2(c + 2)2(d + 2)2(a + 3)2(b + 3)2(c + 3)2(d + 3)2a2b2c2d 2(把第一列的-1 倍加到第四列)证明a26a + 96b + 96c + 96d + 9= b2(把第二列的-1 倍加到第三列)c2d 22a + 3 6a + 92b + 3

3、6b + 92c + 3 6c + 92d + 3 6d + 9a2= b2(最后两列成比例) = 0.c2d 2解 因为a, b, g是方程 x3 + px + q = 0 的三个根, 所以1版本号：2011/5/14z990303 ; 272365083.coma b g设a, b, g是方程 x3 + px + q = 0 的三个根, 则行列式 ga b 的值是多少?b g a证明行列式a2 (a +1)2 (a + 2)2 (a + 3)2b2(b +1)2(b + 2)2(b + 3)2c2(c +1)2(c + 2)2(c + 3)2d 2 (d +1)2 (d + 2)2 (d

4、+ 3)2= 0.x x 1 01 x 2 32 3 x 21 1 2 x中, x 的 3 次方的系数是多少?若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考a3 = -pa - q,b3 = -pb - q,g3 = -pg - q,而且x3 + px + q = (x - a)(x - b)(x - g)= x3 + (-a-b-g)x2 + (ab+bg+ga)x + (-abg)比较上式两端各项系数以及常数项得-a-b-g = 0, ab+bg+ga = p, -abg = q.于是利用“对角线法则”可得a bgab ggba= a3 + b3 + g3 -gab -b

5、ga -abg= (-pa - q) + (-pb - q) + (-pg - q) - 3abg= p(-a-b-g) - 3(q + abg)= 0.另解 因为a, b, g是方程 x3 + px + q = 0 的三个根, 所以a+b+g = 0,于是利用行列式的性质可得a bg ab gga + b + g a + b + g a + b + g11ag1bab ´1gbagba= (a+b+g) g= 0.a ´1b矩阵若 n 阶非零实数矩阵的转置和它的伴随相等, 怎么证明它可逆?证明 记 A 的转置为 AT, A 的伴随矩阵为 A*, A 的秩为 r(A).()

6、()()d-a bbTac d(1) 当 n = 2 时, 设 A =, 则 A* =, A =.-cac db故由条件可得 a = d, b = -c, 而且 a, b 不全为 0.于是|A| = ad - bc = a2 + c2 > 0, 因而 A 可逆.(2) 当 n > 2 时, 由条件可知 0 < r(A) = r(AT) = r(A*).又因为r(A) < n-1 时, A 中各元素aij 的式Mij = 0,式Aij = 0,相应的代数.因而 A* = 0, 此时 r(A*) = 0, 这与r(A*) = r(A) > 0若 r(A) = n-1,

7、 则 A*不等于 0, 且 AA* = 0.可见 0 < r(A*) £ 1, 因而 r(A*) = 1 < n-1 = r(A), 这与 r(A*) = r(A).上述表明, r(A) 只能等于 n, 即 A 可逆. 证明 因为 A 不等于零, 所以 AAT 不等于 0.假若 A 不可逆, 则|A| = 0, 于是 AAT = AA* = |A|E = 0,所以 A 可逆.!证明 设 A 中某个元素 aij 不等于零, 则 AAT 的第 i 行第 i 列处的元素为(ai1, ai2, ., ain)(ai1, ai2, ., ain)T = a 2 + ai22 + .

8、 + ain2 > 0.i12版本号：2011/5/14z990303 ; 272365083 .com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考又因为 AAT = AA* = |A|E, 比较等式两端的第 i 行第 i 列处的元素可得2|A| = ai12 + ai22 + . + ain > 0,所以 A 可逆.【注 1】由上面的证明可以看出|A| > 0.【注 2】当 n > 2 时, 由 AAT = AA* = |A|E 可得|A|2 = |A|AT| = |AAT| = |AA*| = |A|n,因而|A|n-2 = 1. 进而由|A|

9、> 0 得|A| = 1.【注 3】当 n > 2 时, 由上面的|A| = 1 可得 AAT = AA* = |A|E = E, 即 A 为正交矩阵.设 A 为 n 阶方阵, 对某整数 k > 1, Ak = 0, 证明(E-A)-1 = E + A + A2 + . + Ak-1.证明 因为 Ak = 0, 所以(E-A)(E + A + A2 + . + Ak-1)= E(E + A + A2 + . + Ak-1) - A(E + A + A2 + . + Ak-1)= E + A + A2 + . + Ak-1 - A - A2 - . - Ak-1 - Ak= E

10、,可见(E-A)-1 = E + A + A2 + . + Ak-1.对称方阵 A 满足 A2 = 0. 证明 A = 0. 设 A 为 m 行 n 列的矩阵, B = ATA, 其中 AT 表示 A 的转置, 则有以下结论:(其中 B = ATA)命题 1 线性方程组 Ax = 0 的解一定是 Bx = 0 的解. 证明 设向量 a 是 Ax = 0 的解, 则 Aa = 0.于是 Ba = ATAa = AT0 = 0.可见 a 是 Bx = 0 的解.命题 2 线性方程组 Bx = 0 的解一定是 Ax = 0 的解. 证明 设向量 b 是 Bx = 0 的解, 则 Bb = 0.于是

11、bTBb = bT0 = 0.(其中 B = ATA)另一方面, bTBb = bTATAb = (Ab)T(Ab) = |Ab|2, 其中|Ab|表示向量 Ab 的长度. 由上述两个方面可得|Ab|2 = 0, 因而|Ab| = 0, 进而得 Ab = 0.可见 b 是 Ax = 0 的解.(其中 B = ATA)命题 3 线性方程组 Ax = 0 与 Bx = 0 同解.证明 综合上面题 1 和命题 2 立得.(注意: 线性方程组 Ax = 0 含有 m 个方程 n 个未知数; Bx个未知数. 根据命题 3, 它们具有相同的基础解系) 命题 4 秩(A) = 秩(B).= 0 含有 n 个

12、方程 n(其中 B = ATA)证明 因为线性方程组 Ax = 0 与 Bx = 0 的基础解系中所含解向量的个数分别为 n-秩(A) 与 n-秩(B),3版本号：2011/5/14z990303 ; 272365083 .com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考故由命题 3 可见 n-秩(A) = n-秩(B).因而秩(A) = 秩(B).推论 若对称方阵 A 满足 A2 = 0, 则 A = 0.证明 设对称方阵 A 满足 A2 = 0, 则 ATA = AA = A2 = 0, 因而秩(ATA) = 0.由命题 4 可知秩(A) = 秩(ATA) = 0,

13、所以 A = 0.【注 1】n 维向量 a = (a1, a2, ., an)的长度|a| =a2 + a2 +"a2 .12n【注 2】若 n 维向量 a = (a1, a2, ., an), 则 aaT = a2 + a2 +"a2 = |a|2.12n【注 3】n 维向量 a = 0 当且仅当 |a| = 0.【注 4】设 A 为 m 行 n 列的矩阵, AT 表示 A 的转置, 则 AT 为 n 行 m 列的矩阵. 记 A 的第一行为 A1, 第二行为 A2, ., 第 m 行为 Am,则 AT 的第一列为 A1T, 第二列为 A2T, ., 第 m 列为 AmT.

14、因而 AAT 的主对角线上的元素依次为A1A1T,A2A2T, ., A AT,m m即|A1|2, |A2|2, ., |Am|2.命题 5 设 A 为 m 行 n 列的矩阵满足 AAT = 0, 则 A = 0. 证明 因为 AAT = 0, 所以|A1|2 = |A2|2 = . = |Am|2 = 0, 因而 A1 = A2 = . = Am = 0, 可见 A = 0.推论 若对称方阵 A 满足 A2 = 0, 则 A = 0. 证明 设对称方阵 A 满足 A2 = 0, 则 AAT = AA = A2 = 0,根据命题 5 可得 A = 0.定理 设 A 为 n 阶实对称矩阵, 则

15、存在正交矩阵 Q 使得æ l1öQ-1AQ = QTAQ = çl2÷ ,ç÷%ç÷lèn ø其中l1, l2, ., ln 为 A 的特征值.æ l1ö【注 4】把上述定理中的矩阵çl2÷ 记为L, 则ç÷%ç÷lèQ-1A2Q = Q-1AAQ = Q-1An ø-1AQ = LLæ lö÷æ lö æ l2ö÷1

16、11ççl÷ çll2=÷ .22ç2ç÷ ç÷%ç÷ ç÷ç÷lllèn ø èn ø2èn ø推论 若对称方阵 A 满足 A2 = 0, 则 A = 0. 证明 设对称方阵 A 满足 A2 = 0, 且 A 的特征值为l1, l2, ., ln.4版本号：2011/5/14z990303 ; 272365083 .com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”

17、，解答参考æ lö÷÷21ç则存在正交矩阵 Q 使得l2-1 2-1= Q A Q = Q 0Q = 0.ç2%ç÷l2èn øæ l1ö÷çl2因而l1 = l2 = . = ln = 0, 即L = 0.ç÷%ç÷lèn ø故由 Q-1AQ = L 得 A = QLQ-1 = Q0Q-1 = 0.()1i, 其中 i2 = -1, 可以验证 A2 = 0 但是 A【注 5】对于 2 阶复对称矩

18、阵 A =i -1不等于 0.【注 6】对于 2 阶矩阵 A =()001, 可以验证 A2 = 0 但是 A 不等于 0.0A, B 为可逆矩阵, 求证 A-1B + B-1A = E.答 该命题不成立.例如: 取 A = B = E, 则 A, B 为可逆矩阵, 但是 A-1B + B-1A = 2E.设 A 是 n 阶矩阵, 证明存在可逆矩阵 B 和等幂阵 C 使 A = BC. ()EO O证明 设 A 的等价标准形为 D =, 则存在可逆矩阵 P, Q 使得 PAQ = D,O且 D2 = D. 于是令 B = P-1Q-1, C = QDQ-1, 则可直接验算 A = BC, 且

19、C2 = C.A, B 为同阶方阵, A2 = E, B2 = E, 则(AB)2 = E 的充要条件是 AB = BA. 证明 若(AB)2 = E, 则BA = E(BA)E = (A2)(BA)(B2) = (AA)(BA)(BB) = A(AB)(AB)B = A(AB)2B = AEB= AB.反之, 若 AB = BA, 则(AB)2 = (AB)(AB) = A(BA)B = A(AB)B = (AA)(BB) = A2B2 = EE = E.设 A 可逆, 求证(A*)-1 = (A-1)*. 证明 因为 A 可逆, 所以|A| ¹ 0, 而且由 AA* = |A|E

20、 得A* = |A|A-1.用 A-1 替换(1)式中的 A 可得(A-1)* = |A-1|(A-1)-1.在(1)式两边分别求逆可得(1)(2)5版本号：2011/5/14z990303 ;272365083.com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考(A*)-1 = (|A|A-1)-1.(3)比较(2)式和(3)式可得(A*)-1 = (A-1)*.设 A 为 n 阶可逆矩阵. 证明: (A*)* = |A|n-2A. 证明 当 A 为 n 阶可逆时, |A| ¹ 0, 并且由 AA* = |A|E 得A* = |A|A-1.|A|A*| = |A

21、|n.由(2)得|A*| = |A|n-1.用 A*替换(1)式中的 A 可得(A*)* = |A*|(A*)-1.进而由(1), (3)式可得(A*)* = |A*|(A*)-1 = |A|n-1(|A|A-1)-1 = |A|n-1|A|-1A = |A|n-2A.(1)(2)(3)(4)【注 1】对于 2 阶矩阵 A =æ a b ö , A* =æ d-b ö , (A*)* =æ a b ö = A.ç c d ÷ç -ca÷ç c d ÷èø&

22、#232;øèø【注 2】当 n > 2 时, 对于 n 阶可逆矩阵 A, 上面已经证明(A*)* = |A|n-2A.【注 3】当 n > 2 时, 对于 n 阶不可逆矩阵 A, 若 A < n-1, 则 A* = 0, (A*)* = 0; 若秩(A) = n-1, 且由 AA* = |A|E = 0 得秩(A) + 秩(A*) £ n, 因而秩(A*) £ 1, 进而(A*)* = 0.设 A, B 均为 n 阶可逆矩阵, 证明: (AB)* = B*A*.证明 因为 A, B 均为 n 阶可逆矩阵, 所以|A|

23、85; 0, |B| ¹ 0, AB 也可逆,AA* = |A|E 得并且由= |A|A-1.A*(1)由 BB* = |B|E 得= |B|B-1.B*(2)用 AB 替换(1)式中的 A 可得(AB)* = |AB|(AB)-1 = |A|B|(B-1A-1) = (|B|B-1)(|A|A-1) = B*A*.Show that if AB is invertible, so is A. Proof If AB is invertible, then we have (AB)D = I for some matrix D. It follows that A(BD) = I a

24、nd, we conclude that A must be invertible. Note that theholds in case A is a square matrix. Show that if AB is invertible, so is B. Proof If AB is invertible, then we have D(AB) = I for some matrix D. It follows that (DA)B = I and, we conclude that B must be invertible. Note that the6版本号：2011/5/14z9

25、90303 ; 272365083 .com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考holds in case B is a square matrix. Show that if A is invertible, then det(A-1) = 1/det(A). Proof We havedet(A)det(A-1) = det(AA-1) = det(I) = 1, where I is the identity matrix of dimension equal to that of A. Therefore det(A-1) = 1/det(A).Let A

26、and P be square matrices, with P invertible. Show det(P-1AP) = det(A). Proof Note that det(P-1), det(A) and det(P) are numbers, so we have det(P-1AP) = det(P-1)det(A)det(P) = det(A)det(P)det(P-1) = det(APP-1) = det(A).Let A be a 3 by 3 matrix. Show that for any real number s, det(sA) = s3det(A). Pro

27、of Lehe 3 by 3 identity matrix, then sI is a 3 by 3 diagonal matrix all ofwhose entries on diagonal are s. So det(sI) = s3 anddet(sA) = det(sIA) = det(sI)det(A) = s3det(A).(a) Prove that if a matrix B can be obtained from a matrix A by a single row operation, then A can be obtained from B by a singl

28、e row operation. (b) Use the previous fact to prove by induction that if B can be obtained from A byk row operations (k ³ 1), then A can be obtained from B by k row operations. (c) Prove that a matrix that contains only zeros can not be row equivalent to a matrix that contains at least one non-

29、zero entry. Proof (a) Assume that B can be obtained from a matrix A by interchanging the ithand the jth rows. Then A can be obtained from B by the same operation, i.e.,interchanging the ith and the jth rows. For instance, let A =æ 12 öand B =æ 34 ö ,ç 3 4 ÷ç 1 2 &#

30、247;èøèøwhere B can be obtained from a matrix A by interchanging the first and the cecond rows. It is easy to see that if we interchange the first and the cecond rows of B then we get A. æ b11"b1n öæ a11"a1n ö= ç÷= ç÷Now, if B#can

31、 be obtained from A#byç b÷ç a÷" b" aè m1mn øè m1mn ømultiplying row i by t ¹ 0. Then the ith row of B is(tai1, tai2, , tain),while the kth row of B is equal to that of A for all 1 £ k ¹ i £ m, i.e.,7版本号：2011/5/14z990303 ; 27236508

32、3 .com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考æ a11a1n öa12#ta"ç÷#B =ç tata ÷ ith row ."çin ÷i1i 2#ç÷ç a÷a"aè m1mn øm2Thus, multiplying row i by t-1 in B, one obtainsæö÷a11a12#t × t -1a"a1n#ç#&

33、#231; t × t -1a÷ = A.t × t -1a"çin ÷÷i1i 2ç#a#ç÷a""aèm1mnøm2æ b11b1n öç÷Finally,supposethatB=#canbeobtainedfromAç b÷" bè m1mn øæ a11a1n ö"=ç÷#by adding t times

34、row j to row i. Then the ith row of B isç a÷" aè m1mn ø(ai1+taj1, ai2+taj2, , ain+tajn),while the kth row of B is equal to that of A for all 1 £ k ¹ i £ m, that is to say,æ#ai 2 + ta j 2#a j 2#öç a + tai1j1ain + ta jn ÷ ith row"B =

35、1;÷÷#aj1#a jn#.ç"÷ jth rowçç÷èøBy adding -t times row j to row i in B, one obtainsæ#ö÷jn ÷ç a+ ta + (-t)aa + ta+ (-t)a+ ta + (-t)a"aççi1j1j1i 2j 2j 2injn÷ = A.÷#ça j1#each typea j 2#of elementary

36、"a jn#ç÷èNote thatøoperationmay be performed by matrixmultiplication, using square matrices called elementary operators. If a matrix Bcan be obtained from a matrix A by a single row operation, then we haveB = EAfor some elementary row operator E. Consequently, A = E-1B, i.e.

37、, A can beE-1obtained from B by a single row operation since operator.is also an elementary row(b) For k = 1, it follows by (a). Now, suppose that we have proven (b) for k = n³ 1 and that B can be obtained from A by n+1 row operations. In view of the8版本号：2011/5/14z990303 ; 272365083 .com若打印/复印本

38、文档，请选择双面格式或用废纸题目摘自“”，解答参考following sequenceA = A0 ® A1 ® A2 ® ® An ® An+1 = B,where Ai is obtained from Ai-1 by a single row operation (1 £ i £ n+1), hence An can be obtained from A by n row operations. By (a) and the hypothesis, A can be obtained from An by n row

39、operations and An can be obtained from B by a single row operation. Consequently, A can be obtained from B by n+1 row operations.(c) In view of (b), it suffices to show that a matrix that contains at least one non-zero entry can not be obtained from a matrix that contains only zeros by a single row

40、operation. But this is obvious since any type of elementary row operation makes a zero matrix into a zero matrix.线性方程组若 AX = 0 的解一定是 BX = 0 的解, 则 B 的行向量都能表示成 A 的行向量的线性组合. ( )A证明 令 C 为分块矩阵(注意 A 与 B 的列数相同, 都等于 X 的行数),B则 AX = 0 与 CX = 0 同解. 因而秩(A) = 秩(C).(注意AX = 0 的基础解系中解向量的个数 = n - 秩(A), 其中n 表示未知数的个数,

41、 即 X 的行数, 亦即 A 的列数;同时 CX = 0 的基础解系中解向量的个数 = n - 秩(C)设 A 中某 r 行的极大无关组.A 的行向量组的极大无关组, 则这 r 行也C 的行向量组(注意, 秩为 r 的向量组中, 任意 r 个线性无关的向量都极大无关组)因而 B 的行向量都能由这 r 行线性表示, 进而能由 A 的行向量线性表示, 即能表示成 A 的行向量的线性组合.Suppose an m´n matrix A has n pivot columns. Explain why for each b Î Rm the equation Ax = b has

42、at most one solution. Answer If both x1 and x2 are solutions of Ax = b, then A(x1 - x2) = Ax1 - Ax2 = b - b = 0.But Ax = 0 has at most one solution according to the hypothesis. It follows that x1 - x2 = 0, i.e. x1 = x2.齐次线性方程组 Ax = 0 有唯一零解是线性方程组 Ax = b 有唯一解的(). (A) 充分必要条件;(B) 充分条件;(C) 必要条件;(D) 无关条件.

43、 答 C.9版本号：2011/5/14z990303 ; 272365083 .com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考设 A, B 都是 n 阶非零矩阵, 且 AB = 0, 则 A 和 B 的秩(). (A) 必有一个等于零;(B) 都小于 n;(C) 必有一个等于 n;(D) 有一个小于 n. 答 B.设 A = 4´6 矩阵, 则齐次线性方程组 Ax = 0 (). (A) 无解;(B) 必有非零解;(C) 只有零解;(D) 不一定. 应该选哪个? 为什么?答 选 B因为 A 是 4´6 距阵, 所以 秩(A) £ 4,而齐次

44、线性方程组 Ax = 0 中未知数的个数为 6.由秩(A) < 6 可知 Ax = 0 必有非零解.设 3 元齐次线性方程组ax1 + x2 + x3 = 0x1 + ax2 + x3 = 0x1 + x2 + ax3 = 0(1) 确定当 a 为何值时, 方程组有非零解;(2) 当方程组有非零解时, 求出它的基础解系和全部解. æ a1 ö1解 (1)该齐次线性方程组的系数矩阵 A =ç 1 a 1 ÷ , |A| = (a+2)(a-1)(a-1).ç 1a ÷1èø该齐次线性方程组有非零解的充分必要条件

45、是: |A| = 0,即 a = -2 或 1.(2) 当 a = -2 时, 对 A 进行初等行变换, 化为如下行最简形:æ 1-1ö0ç 0 1 -1÷ç 00 ÷0èø由此可得x1 - x3 = 0,x2 - x3 = 0,即x1 = x3 ,x2 = x3 ,于是得到基础解系:(1, 1, 1)T,10版本号：2011/5/14z990303 ; 272365083 .com若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考通解(x1, x2, x3)T = k(1, 1, 1)T,其中 k

46、为任意实数.当 a = 1 时, 对 A 进行初等行变换, 化为如下行最简形:æ 11 ö1ç 0 0 0 ÷ç 00 ÷0èø由此可得x1 + x2 + x3 = 0,即x1 = -x2 - x3.于是得到基础解系:(-1, 1, 0)T,(-1, 0, 1)T,通解(x1, x2, x3)T = k1(-1, 1, 0)T + k2(-1, 0, 1)T,其中 k1, k2 为任意实数.Suppose CA = In (the n´n identity matrix). (1) Show that t

47、he equation Ax = 0 has only the trivial solution. (2) Explain why A cannot have more columns than rows. Proof (1) Suppose that Ax = 0 has a nontrivial solution x ¹ 0. Then on one handC(Ax) = C0 = 0,and on the other C(Ax) = In x = x, and we arrive at a contradiction. Therefore Ax =0 cannot have

48、a nontrivial solution.(2) If A had more columns than rows, then A would not have a pivot in every column and the homogeneous equation Ax = 0 would have a nontrivial solution, which would contradict (1).Suppose AD = Im (the m´m identity matrix). Show that for any b Î Rm, the equation Ax = b

49、 has a solution and explain why A cannot have more rows than columns. Proof (1) For any b Î Rm, we have ADb = Imb = b, which means that x = Db is a solution to Ax = b.(2) Suppose that A has n columns. It follows that A has m rows since AD = Im. Then we have m = rank(Im) = rank(AD) £ rank(A

50、) £ n. So A cannot have more rows than columns. Another possible proof: Note that AD = Im implies DTAT = Im. It follows from (2) of the above exercise that AT cannot have more columns than rows, i.e., A cannot have more rows than columns. 11版本号：2011/5/14z990303 ; 272365083 .com若打印/复印本文档，请选择双面格式

51、或用废纸题目摘自“”，解答参考Show that if the columns of B are linearly dependent, then so are the columns ofAB. Proof If the columns of B are linearly dependent, then Bx = 0 for a nonzero vector x. Then (AB)x = A(Bx) = 0 and therefore the columns of AB are linearly dependent.Suppose that an n´n matrix K can

52、not be row reduced to In. Show that the columns of K are linearly dependent. Proof Since the n´n matrix K cannot be row reduced to In, we have rank(K) < n. Therefore the columns of K are linearly dependent.If L is n´n and the equation Lx = 0 has only the trivial solution, show that the

53、columns of L span Rn. Proof If L is n´n and the equation Lx = 0 has only the trivial solution then the columns of L are linearly independent and hence span Rn.Let A be the augmented matrix of a linear system and C its augmented matrix of the corresponding homogeneous linear system We assume tha

54、t the linear system encoded by A is consistent. (a) Prove that if (c1, , cn) and (c1¢, , cn¢) are two solutions (possibly identical) to the linear system encoded by A, then (c1-c1¢, , cn-cn¢) (i.e. their difference) is a solution to the linear system encoded by C. (b) Deduce that

55、 if the linear system encoded by C has only the trivial solution (0, , 0), then the linear system encoded by A has exactly one solution. (c) Prove that if the linear system encoded by C has more than one solution, then the linear system encoded by A has infinitely many solutions.Proof (a) If (c1, ,

56、cn) and (c1¢, , cn¢) are two solutions to the linear system encoded by A, then we have12版本号：2011/5/14z990303 ; 272365083 .comæ a11 " a1n0 öç a" a0 ÷ç 122n÷#ç÷è am1 " amn 0 øæ a11 " a1nb1 öç a" ab 

57、7;ç 122n2 ÷#ç÷è am1 " amn bm ø若打印/复印本文档，请选择双面格式或用废纸题目摘自“”，解答参考a1n öæ c1¢ öæ b1 öæ a11"a1n öæ c1 öæ b1 öæ a11"ç÷ç÷ = ç÷ç÷ç÷ = ç÷ .#

58、and#ç a÷ç c ÷ç b ÷ç a÷ç c¢ ÷ç b ÷"a" amn øè n øè m ømn øè n øè m øèèm1m1Hencea1n öæ c1 - c1¢ öa1n ö éæ c1 öæ c1¢ &#

59、246;ùæ a11æ a11""ç÷ç÷ = ç÷ êç÷ - ç÷ú#a#÷ êç c ÷ç c¢ ÷úç a÷ç c- c¢ ÷ç a""a"è m1mn øè nn øè m1ø ë

60、èn øèn øûmna1n öæ c1¢ öæ b1 ö æ b1 öæ a11a1n öæ c1 öæ a11"æ 0 ö= ç÷ç÷ - ç÷ç÷ =ç÷ - ç# ÷ = ç # ÷ .#ç a÷ç c 

61、7;ç a÷ç c¢ ÷ç b ÷ ç b ÷ ç 0 ÷"a" aè m1mn øè n øè m1mn øè n øè m ø è m ø è øThis means (c1-c1¢, , cn-cn¢) is a solution to the linear system encoded by C.(b) Suppose that (c1, , cn) and (c1¢, , cn¢) are two different solutions to the linear system encoded by A, then (c1-c1¢, , cn-cn¢