线性代数_赵树源第4版文档.doc

=赵1A2(1)这是让用对角线法则计算行列式1 2 33 1 22 3 1= 1*1*1 + 2*2*2 + 3*3*3 - 3*1*2*3= 1+8+27 - 18= 18=赵1A12(3)1 1 1 00 1 0 10 1 1 10 0 1 0解: 根据行列式的定义,每行每列恰取一个元素的乘积构成一个和项且只需考虑非零的和项.第1列非零元只有a11, 第4行非零元只有a43所以行列式= (-1)t(1243)a11a22a34a43 + (-1)t(1423)a11a24a32a43= -1 + 1 = 0.=1A18(1)1 1 2 31 2 3 -13 -1 -1 -22 3 -1 -1第1步: r2-r1, r3-3r1, r4-2r1, 得1 1 2 30 1 1 -40 -4 -7 -110 1 -5 -7第2步: r3 + 4r2, r4 - r2, 得1 1 2 30 1 1 -40 0 -3 -270 0 -6 -3第3步: r4 - 2r3 得1 1 2 30 1 1 -40 0 -3 -270 0 0 51所以 行列式 = -153=1A18(2) 2 -5 3 1 1 3 -1 3 0 1 1 -5-1 -4 2 -3=1A18(3)-2 2 -4 0 4 -1 3 5 3 1 -2 -3 2 0 5 1解:r2+2r1,r4+r1, r1*(1/2) 第1行提出2, r3+3r1-1 1 -2 0 0 3 -5 5 0 4 -8 -3 0 2 1 1r2-r4,r3-2r4-1 1 -2 0 0 1 -6 4 0 0 -10 -5 0 2 1 1r4-2r2-1 1 -2 0 0 1 -6 4 0 0 -10 -5 0 0 13 -7= 2*(-1)*1*(10*7+5*13)= -2*135= -270.=1A210 x x . xx 0 x . xx x 0 . x . .x x x . 0解: c1+c2+.+cn (所有列加到第1列)(n-1)x x x . x(n-1)x 0 x . x(n-1)x x 0 . x . .(n-1)x x x . 0ri-r1,i=2,3,.,n (所有行减第1行)(n-1)x x x . x 0 -x 0 . 0 0 0 -x . 0 . . 0 0 0 . -x行列式 = (-x)(n-1) (n-1)x = (-1)(n-1) (n-1)xn=1A25(2)x 1 1 11 x 1 11 1 x 11 1 1 x第1步: c1+c2+c3+c4 (即2,3,4列都加到第1列), 提出第1列公因子 (3+x), 得1 1 1 11 x 1 11 1 x 11 1 1 x第2步: 第1行乘 -1 加到 2,3,4行, 得1 1 1 10 x-1 0 00 0 x-1 00 0 0 x-1所以行列式 = (3+x)(x-1)3=1A26解: 2的代数余子式A31 = (-1)(3+1)*0 40 3= 0-2的代数余子式A32 = (-1)(3+2)*-3 4 5 3= -(-9-20) = 29.=1A27解:第3列的余子式分别为: 5,3,-7,4所以第3列的代数余子式分别为: (-1)(1+3)*5,(-1)(1+3)*3,(-1)(1+3)*(-7),(-1)(1+3)*4即 5,-3,-7,-4而第3列元素分别为-1,2,0,1所以 D = (-1)*5+2*(-3)+0*(-7)+1*(-4) = -15.=1A28解: A41+A42+A43+A44 =1 0 4 02 -1 -1 20 -6 0 01 1 1 1按第3行展开= (-1)(3+2)*(-6)*1 4 02 -1 21 1 1= 6*(-3) = -18.=1A29解: A11+A12+A13+A14 =1 1 1 1d c b bb b b bc d a d=0. (1,3行成比例)=1A30(1) 1 0 a 1 0 -1 b -1-1 -1 c -1-1 1 d 0 0 -1 -1 1 0 1= (-1)(1+3)a* -1 -1 -1 + (-1)(2+3)b* -1 -1 -1 -1 1 0 -1 1 0 1 0 1 1 0 1+ (-1)(3+3)c* 0 -1 -1 + (-1)(4+3)d* 0 -1 -1 -1 1 0 -1 -1 -1= a + b + d.=1A30(2)与(1)类似, 略=1A321 2 3 4 . n-1 n1 1 2 3 . n-2 n-11 x 1 2 . n-3 n-21 x x 1 . n-4 n-3 . . . .1 x x x . 1 21 x x x . x 1ri - r(i+1), i=1,2,.,n-10 1 1 1 . 1 10 1-x 1 1 . 1 10 0 1-x 1 . 1 10 0 0 1-x. 1 1 . . . . .0 0 0 0 .1-x 11 x x x . x 1按第1列展开= (-1)(1+n)* 1 1 1 . 1 11-x 1 1 . 1 1 0 1-x 1 . 1 1 0 0 1-x. 1 1 . . . . 0 0 0 .1-x 1ci-c(n-1), i=1,2,.,n-2 0 0 0 . 0 1-x 0 0 . 0 1-1 -x 0 . 0 1-1 -1 -x . 0 1-1. . . .-1 -1 -1 . -x 1按第1行展开=(-1)(1+n)*(-1)(1+n-1)*-x 0 0 . 0-1 -x 0 . 0-1 -1 -x . 0-1. . . .-1 -1 -1 . -x行列式 = - (-x)(n-2) = (-1)(n-1)x(n-2)=1A33a b 0 . 0 00 a b . 0 00 0 a . 0 0 . .0 0 0 . a bb 0 0 . 0 a解: 按第1列展开得a*(-1)(1+1)*a(n-1)+b*(-1)(1+n) * b(n-1)=an + (-1)(1+n) * bn.注: 按展开定理, 是第1列的每个数乘其代数余子式之和.代数余子式 Aij = (-1)(i+j)Mij第1列非零元只有 a (a11) 和 b(a1n)b 位于第n行第1列, 所以其代数余子式要乘 (-1)(n+1)=1A34用特殊分块矩阵的行列式的结果原行列式 = |1 2| * |x 2| = (x-2)(x2-4) = (x-2)2(x+2). |1 x| |2 x|所以 x=2.1 2 1 11 x 2 30 0 x 20 0 2 x第1步: r2-r1, 再交换 第3,4列, (注意这里行列式变符号) 得1 2 1 10 x-2 2 10 0 2 x0 0 x 2第2步: r4 - (x/2)r31 2 1 10 x-2 2 10 0 2 x0 0 0 2- (1/2)x2所以行列式 = - (x-2)*2* 2 - (1/2)x2 = (x-2)(x2 - 4) 或写成 (x+2)(x-2)2=1A351+x 1 1 1 1 1-x 1 1 1 1 1+y 1 1 1 1 1-y x=0 或 y=0 时, 行列式有两行相等, 行列式为0当 xy0 时ri-r1 (i=2,3,4)1+x 1 1 1 -x -x 0 0 -x 0 y 0 -x 0 0 -yc1-c2+(x/y)c3-(x/y)c4x 1 1 10 -x 0 00 0 y 00 0 0 -y= x2y2所以 x=0 或 y=0.=1A36此为Vandermonde行列式a1=-1,a2=2,a3=1,a4=3行列式 = 3*2*4 * (-1)*1 * 2 = -48.=1A37a+b x+b x+a x a bx2 a2 b2r1+r2 提出第1行公因子 (x+a+b) 1 1 1 x a bx2 a2 b2此为Vandermonde行列式行列式 = (x+a+b)(a-x)(b-x)(b-a)=1A38按第3,4行展开得2 0 * (-1)(3+4+1+2) 1 1 = 2*2 = 40 1 0 22 1 * (-1)(3+4+1+4) 2 1 = 4*(-4) = -160 2 4 00 1 * (-1)(3+4+2+3) 3 1 = (-1)*6 = -61 0 0 20 1 * (-1)(3+4+2+4) 3 1 = 01 2 0 01 1 * (-1)(3+4+3+4) 3 2 = 2*12 = 240 2 0 4行列式=4-16-6+24 = 6=赵1A39=赵1A40. 疯了, 非让用Crammer法则解线性方程组!(1)D = |2 5| = -1 |3 7|D1 = |2 1| = 1. D2 = |1 5| = -3 |3 2| |2 7|x1=D1/D = -1, x2=D2/D = 3.(2)D = 4 5 = -43 0 3 -7齐次线性方程组只有零解. 故 x1=x2=0=赵1A41(1) 验证代入即可(略)系数行列式= 2 -3 4 -3 3 -1 11 -13 4 5 -7 -213 -25 1 11所有列加到第1列, 第1列全是0, 故行列式=0=赵1A41(2)系数行列式=1 2 3 -13 2 1 15 5 2 02 3 1 -1=赵1A42.判断齐次线性方程组是否只有零解2x1+2x2-x3=0x1-2x2+4x3=05x1+8x2-2x3=0分析: n元齐次线性方程组只有零解的充分必要条件是 系数行列式不等于0.解: 系数行列式=|2 2 -1| |0 0 -1|1 -2 4| c1+2c3,c2+2c3 |9 6 4| = 36-6 = 30|5 8 -2| |1 4 -2|所以方程组只有零解.=赵1A43.如果齐次线性方程组有非零解, k应取什么值?kx+y+z=0x+ky-z=02x-y+z=0分析: n元齐次线性方程组有非零解的充分必要条件是 系数行列式等于0.解: 系数行列式=|k 1 1| |k-2 2 0|1 k -1| r1-r3,r2+r3 | 3 k-1 0| =(k-1)(k-2)-6 = (k+1)(k-4)|2 -1 1| | 2 -1 1|所以 k=-1 或 k=4.=赵1A44.k取什么值时,齐次线性方程组仅有零解kx+y-z=0x+ky-z=02x-y+z=0分析: n元齐次线性方程组只有零解的充分必要条件是 系数行列式不等于0.解: 系数行列式=|k 1 -1| |k+2 0 0|1 k -1| r1+r3,r2+r3 | 3 k-1 0| = (k+2)(k-1)|2 -1 1| | 2 -1 1|所以 k1 且 k-2 时仅有零解.=赵1B1.行列式 = (k-1)2-22 = (k+1)(k-3)选(C).=赵1B2.行列式=k2-2-k-4 = k2-k-6 = (k+2)(k-3)选(D)=赵1B3.行列式=1-2. 选(C).=赵1B4.(A)(12435)=0+0+1+0 = 1(B)(54321)=4+3+2+1 = 10(C)(32514)=2+1+2+0 = 5(D)(54231)=4+3+1+1 = 9(B)正确.=赵1B5.列标按自然序, 计算行标的逆序数(ik42)i=1,k=3时, (ik42)=(1342)=0+1+1=2 带正号所以 i=3,k=1时, 带负号故(2)(3)正确.=赵1B6.(A)不是 行标有2个3(B)(42135)=3+1+0+0 = 4 不是(C)不是 列标2个1(D)(21435)=1+0+1+0 =2 是选(D).=赵1B7.(A)(13254)=0+1+0+1 = 2 属于(B)(51432)=4+0+2+1 = 7 属于(C)(31452)=2+0+1+1 = 4 不属于(D)(54321)=4+3+2+1 =10 属于选(D).=赵1B8.D1: c1提4,c2-2c1, c2提(-3) 化成D所以 D1 = 4*(-3)D = -12.(A)正确.=赵1B9.c2-3c1,c3+c1, r2提(-3), c2提(-2), c3提4化成原行列式所以行列式 = (-3)*(-2)*4 = 24.(B)正确.=赵1B10.(A)=赵1B11.D1=D2=D, D3 = -D, D4=2D故(B)正确.=赵1B12.行列式 = (-1)(4321)*2a = 2a所以 a = -1/2故(A)正确 #=赵1B13.ri-r1, i=2.3,.,n 化为上三角行列式行列式 = (-1)(n-1).故(C)正确 #=赵1B14.A41 = (-1)(1+4) * (-1)(1+3)*a2 *(a3a6-a4a5) = -a2a3a6 + a2a4a5故(B)正确 #=赵1B15.行列式 = (-a1)(-a2).(-an)(-1)(n(n-1).21) = (-1)(n-1+n-2+.+1) * (-1)n * a1a2.an = (-1)n(n-1)/2 + na1a2.an = (-1)n(n+1)/2a1a2.an故(D)正确.=赵1B16.(B)行列式 = (-1)(n(n-1).21) = (-1)(n-1+n-2+.+1) = (-1)n(n-1)/2其余行列式都是由单位矩阵的行列式交换两行得到的, 值都是-1.=赵1B17.选1,3行,用Laplace展开得行列式 = (-1)(1+3+1+3) |a b| |x y| |c d| |u v| = (ad-bc)(xv-yu).(C)正确 #=赵1B18f(x) =c1+c2+c3+c4x -1 1 x-1x -1 x+1 -1x x-1 1 -1x -1 1 -1r1-r4,r2-r4,r3-r40 0 0 x0 0 x 00 x 0 0x -1 1 -1= - x4(D)正确 #=赵1B19.(1)行列式 = (-1)(4123) = (-1)3 = -1.(2)行列式 = (-1)(3241) = (-1)4 = 1.(3)因为行和都是0, 故行列式 = 0.(4)行列式 =| 1 2 2 200+1| | 1 2 2 200| | 1 2 2 1|-1 3 3 300-1| = |-1 3 3 300| + |-1 3 3 -1| = 0+0 =0|-2 2 1 100-2| |-2 2 1 100| |-2 2 1 -2| 3 5 1 100+3| | 3 5 1 100| | 3 5 1 3|所以(A)正确 #=赵1B20.方程组有唯一解, 故系数行列式等于0|2 k|k 2| = 4-k2 = (2+k)(2-k).所以 k=-2 或 k=2.选(B) #=赵1B21.(A) 充分条件=赵1B22.系数行列式a11 -a12a21 -a22= -1x1 = (-1)* |b1 -a12| = |b1 a12| |b2 -a22| |b2 a22|x2 = (-1)* |a11 b1| |a21 b2|(C) 正确 #=赵1B23.因为方程组有非零解, 所以系数行列式等于02 -1 11 k -1k 1 1= 2k + k + 1 - k2 + 2 +1= -k2 +3k + 4= -(k-4)(k+1)所以 k=-1 或 k=4.选(D) #=赵1B24.齐次线性方程组只有零解, 故系数行列式不等于0所以 k-1 且 k4选(C) #=第2章=2A08A=50 30 25 10 530 60 25 20 1050 60 0 25 5B = (0.95,1.2,2.35,3,5.2)该厂各月份的总产值为AB = (198.25, 271.25, 220.5)=2A09设 A = 0.8 0.1 0.1 B = (30,20) 0.4 0.3 0.3BA = ( 32, 8, 9)三种金属的数量分别为 32, 8, 9 吨.=2A21解: |2|A|AT| = 2n |A|n |AT| = 2n |A|(n+1) = 2n m(n+1).=2A41解矩阵方程 AX+B=X解: 由AX+B=X得 (A-E)X=-B(A-E,-B)=-1 1 0 -1 1-1 0 1 -2 0-1 0 -2 5 -3r1-r2,r3-r2 0 1 -1 1 1-1 0 1 -2 0 0 0 -3 7 -3r2*(-1),r3*(-1/3)0 1 -1 1 11 0 -1 2 00 0 1 -7/3 1r1+r3,r2+r30 1 0 -4/3 21 0 0 -1/3 10 0 1 -7/3 1r1<->r21 0 0 -1/3 10 1 0 -4/3 20 0 1 -7/3 1所以 X =-1/3 1-4/3 2-7/3 1=2A51设A为三阶矩阵，且|A|=1/2，求|(3A)-1 - 2A*|的值解: (3A)(-1) = (1/3) A(-1)A* = |A|A(-1) = (1/2) A(-1)所以 |(3A)-1 - 2A*|= | (1/3) A(-1) - (1/2) A(-1) |= | (-2/3) A(-1) |= (-2/3)3 | A(-1) |= (-2/3)3 * 2= - 16/27.=2A52设A,B为三阶矩阵, 且|A|=2,|B|=3, 求 |-2(ATB-1)-1|解: |-2(ATB-1)-1| = (-2)3 |ATB-1|-1 = -8 |AT|-1 |B-1|-1 = -8 |A|-1 |B| = -8*(1/2)*3 = -12.=2A54(1) 2 2 3 1 0 0 1 -1 0 0 1 0-1 2 1 0 0 1r1-2r2, r3+r2 0 4 3 1 -2 0 1 -1 0 0 1 0 0 1 1 0 1 1r1-4r3, r2+r2 0 0 -1 1 -6 -4 1 0 1 0 2 1 0 1 1 0 1 1r2+r1,r3+r1,r1*(-1) 0 0 1 -1 6 4 1 0 0 1 -4 -3 0 1 0 1 -5 -3交换行得 1 0 0 1 -4 -3 0 1 0 1 -5 -3 0 0 1 -1 6 4=2A54(3) 1 1 1 1 1 0 0 0-1 1 1 1 0 1 0 0-1 -1 1 1 0 0 1 0-1 -1 -1 1 0 0 0 1r2+r1,r3+r1,r4+r1 1 1 1 1 1 0 0 0 0 2 2 2 1 1 0 0 0 0 2 2 1 0 1 0 0 0 0 2 1 0 0 1r2*(1/2),r3*(1/2),r4*(1/2), 1 1 1 1 1 0 0 0 0 1 1 1 1/2 1/2 0 0 0 0 1 1 1/2 0 1/2 0 0 0 0 1 1/2 0 0 1/2r1-r2,r2-r3,r3-r4 1 0 0 0 1/2 -1/2 0 0 0 1 0 0 0 1/2 -1/2 0 0 0 1 0 0 0 1/2 -1/2 0 0