武汉大学：《线性代数》课程教学资源（习题参考答案）同济第四版《线性代数习题全解》

第一章行列式 8.用克莱姆法则解下列方程组: x1+2x2-x3+4x4=-2 3r1+ 1111 000 1100 0142 5-1-10 5-1-10 -18 D1 0-5-18=-4010 230 00 1000 284 0-1-19 30211 0 1151 DA= 2-3-2-5 2-3-1-2 31011 所以 1 1,x 5x1+6x2 x1+5x2+6x3 x2+5x3+6x4 +5x4+6x5=0 4+5x5=1 解:系数行列式 651 065 1 0065 5 560 5D4x4-6156=5D4×4-6D3×3 015 由递归式D5×5=5D4×4-6D3×3知 D3×3=5D2x2-6D1×1=5(25-6)-6×5=6 5D3×3-6D2×2=5×65-6×19=211

14 第一章 行列式 8 . 用克莱姆法则解下列方程组: (1)    x1 + x2 + x3 + x4 = 5, x1 + 2x2 − x3 + 4x4 = −2, 2x1 − 3x2 − x3 − 5x4 = −2, 3x1 + x2 + 2x3 + 11x4 = 0; 解: D = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 1 1 2 −1 4 2 −3 −1 −5 3 1 2 11 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 1 0 1 −2 3 0 −5 −3 −7 0 −2 −1 8 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 1 0 1 −2 3 0 0 −13 8 0 0 −5 14 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 1 0 1 −2 3 0 0 −1 −54 0 0 0 142 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = −142. D1 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 1 1 1 −2 2 −1 4 −2 −3 −1 −5 0 1 2 11 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 1 −1 −10 0 5 −5 −18 −2 −3 5 28 0 1 0 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 −1 −10 0 −5 −18 −2 5 28 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 −1 −10 −4 0 10 23 0 −22 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = −142; D2 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 5 1 1 1 −2 −1 4 2 −2 −1 −5 3 0 2 11 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 5 1 1 0 −7 −2 3 0 −12 −3 −7 0 −15 −1 8 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 5 1 1 0 −1 3 2 0 0 23 11 0 0 39 31 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 5 1 1 0 −1 3 2 0 0 −1 −19 0 0 0 −284 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = −284; D3 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 5 1 1 2 −2 4 2 −3 −2 −5 3 1 0 11 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = −426; D4 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 5 1 2 −1 −2 2 −3 −1 −2 3 1 2 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 142. 所以, x1 = D1 D = 1 , x2 = D2 D = 2 , x3 = D3 D = 3 , x4 = D4 D = −1. (2)    5x1+ 6x2 = 1, x1+ 5x2+ 6x3 = 0, x2+ 5x3+ 6x4 = 0, x3+ 5x4+ 6x5= 0, x4+ 5x5= 1. 解: 系数行列式 D5×5 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 1 5 6 1 5 6 1 5 6 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c1 ====== 5D4×4 − ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 6 0 0 0 1 5 6 0 0 1 5 6 0 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 5D4×4 − 6 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 0 1 5 6 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 5D4×4 − 6D3×3. 由递归式 D5×5 = 5D4×4 − 6D3×3 知, D3×3 = 5D2×2 − 6D1×1 = 5(25 − 6) − 6 × 5 = 65, D4×4 = 5D3×3 − 6D2×2 = 5 × 65 − 6 × 19 = 211

线性代数(同济四版)习题参考答案 0 65100 06510 0065 0006 0156 =D4×4+64=211+64=1507 51000 10600 D2=00560 展开20560 +(-1)5 1600 0156 0560 0015 D3× 65-1080 500 15000 1+30160 1500 01060 +(-1)5 0005 001 65x00 D4=01500展开11+40150 +(-1)5+/1560 0016 0150 00106 00015 展开-5-6D3×3=-5-6×65=-395 56 15 D5=0 06510 10001 展开 015 0651 6510 56 000 1+D4×4=1+211=212 所以, 工1 6523=665: 9.问入取何值时,齐次线性方程组{x1+m2+x3=0,有非零解?

线性代数 (同济四版) 习题参考答案 15 D5×5 = 5D4×4 − 6D3×3 = 5 × 211 − 6 × 65 = 665. 又 D1 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 6 0 0 0 0 5 6 0 0 0 1 5 6 0 0 0 1 5 6 1 0 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c1 ====== D4×4 + ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 6 0 0 0 5 6 0 0 1 5 6 0 0 1 5 6 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = D4×4 + 64 = 211 + 64 = 1507. D2 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 1 0 0 0 1 0 6 0 0 0 0 5 6 0 0 0 1 5 6 0 1 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c2 ====== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 6 0 0 0 5 6 0 0 1 5 6 0 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + (−1)5+2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 0 0 0 1 6 0 0 0 5 6 0 0 1 5 6 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c2 ====== −D3×3 − 5 × 6 3 = −65 − 1080 = −1145. D3 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 1 0 0 1 5 0 0 0 0 1 0 6 0 0 0 0 5 6 0 0 1 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c3 ====== (−1)1+3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 5 0 0 0 1 6 0 0 0 5 6 0 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + (−1)5+3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 0 0 1 5 0 0 0 1 6 0 0 0 5 6 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c2 ====== D2×2 − 6 × 6 × D2×2 = 703. D4 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 0 1 0 1 5 6 0 0 0 1 5 0 0 0 0 1 0 6 0 0 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c4 ====== (−1)1+4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 5 6 0 0 1 5 0 0 0 1 6 0 0 0 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + (−1)5+4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 0 0 1 5 6 0 0 1 5 0 0 0 1 6 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c4 ====== −5 − 6D3×3 = −5 − 6 × 65 = −395. D5 = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 0 0 1 1 5 6 0 0 0 1 5 6 0 0 0 1 5 0 0 0 0 1 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开c5 ====== (−1)1+5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 5 6 0 0 1 5 6 0 0 1 5 0 0 0 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 5 6 0 0 1 5 6 0 0 1 5 6 0 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1 + D4×4 = 1 + 211 = 212. 所以, x1 = 1507 665 , x2 = − 1145 665 , x3 = 703 665 , x4 = − 395 665 , x5 = 212 665 . 9 . 问 λ, µ 取何值时, 齐次线性方程组    λx1 + x2 + x3 = 0, x1 + µx2 + x3 = 0, x1 + 2µx2 + x3 = 0. 有非零解?

16 第一章行列式 解:系数矩阵 D=1a1-11|=(-13+21 (1-入 要使齐次线性方程组有非零解,则D=0,即 (1-)=0, 得μ=0或A=1 10.问A取何值时,齐次线性方程组 2x1+(3-)) +x3=0,有非零解 +x2+(1-A)x3=0 解:系数矩阵 D 1×(-1y+1|-3+24|+a-x(-1+1-3+ 1-入1 (A-3)-4(1-))+(1-)3-2(1-从)(-3+) 入(A-2)(3- 齐次线性方程组有非零解,则D=0,即 (入-2)(3-入)=0. 得入=0,A=2或A=3

16 第一章 行列式 解: 系数矩阵 D = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ λ 1 1 1 µ 1 1 2µ 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r3−r1 ===== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ λ 1 1 1 µ 1 0 µ 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = µ(−1)3+2 ¯ ¯ ¯ ¯ ¯ λ 1 1 1 ¯ ¯ ¯ ¯ ¯ = µ(1 − λ). 要使齐次线性方程组有非零解, 则 D = 0, 即 µ(1 − λ) = 0, 得 µ = 0 或 λ = 1. 10 . 问 λ 取何值时, 齐次线性方程组    (1 − λ)x1 − 2x2 + 4x3= 0, 2x1+ (3 − λ)x2 + x3= 0, x1 + x2+ (1 − λ)x3= 0. 有非零解? 解: 系数矩阵 D = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 − λ −2 4 2 3 − λ 1 1 1 1 − λ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ c2−c1 ===== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 − λ −3 + λ 4 2 1 − λ 1 1 0 1 − λ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1 × (−1)3+1 ¯ ¯ ¯ ¯ ¯ −3 + λ 4 1 − λ 1 ¯ ¯ ¯ ¯ ¯ + (1 − λ)(−1)3+3 ¯ ¯ ¯ ¯ ¯ 1 − λ −3 + λ 2 1 − λ ¯ ¯ ¯ ¯ ¯ = (λ − 3) − 4(1 − λ) + (1 − λ) 3 − 2(1 − λ)(−3 + λ) = (1 − λ) 3 + 2(1 − λ) 2 + λ − 3 = λ(λ − 2)(3 − λ). 齐次线性方程组有非零解, 则 D = 0, 即 λ(λ − 2)(3 − λ) = 0. 得 λ = 0, λ = 2 或 λ = 3

第二章矩阵及其运算 1.已知线性变换 x2=3y1+y+53 x3=3y1+2y2+3y/3, 求从变量x1,x2,x3到变量弘1,y2,的线性变换 解:方法一.用消元法解方程,得出弘,v,驷.略. 方法二.解矩阵方程.由已知 315 故 221 -7-49 315 3 7x1-4x2+9 y=6x1+3x2-7x3, 方法三.用克拉默法则解方程.系数矩阵 D=315 1-2c3 所以 23 15 同理得y=6x1+3r2-7x3, 3x1+ 2.已知两个线性变换 y1=-321+2, y1+3y2+2y3 r3=4y1+y2+53, 求从1,2,23到x1,x2,x3的线性变换 解:方法一.直接代入.比如 2(-3x1+22)+(-22+323)

第二章 矩阵及其运算 1 . 已知线性变换:    x1 = 2y1 + 2y2 + y3, x2 = 3y1 + y2 + 5y3, x3 = 3y1 + 2y2 + 3y3, 求从变量 x1, x2, x3 到变量 y1, y2, y3 的线性变换. 解: 方法一. 用消元法解方程, 得出 y1, y2, y3. 略. 方法二. 解矩阵方程. 由已知:   x1 x2 x3   =   2 2 1 3 1 5 3 2 3     y1 y2 y2   , 故   y1 y2 y2   =   2 2 1 3 1 5 3 2 3   −1   x1 x2 x3   =   −7 −4 9 6 3 −7 3 2 −4     y1 y2 y3   . 即    y1 = −7x1 − 4x2 + 9x3, y2 = 6x1 + 3x2 − 7x3, y3 = 3x1 + 2x2 − 4x3. 方法三. 用克拉默法则解方程. 系数矩阵 D = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 2 1 3 1 5 3 2 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ c1−2c3 ====== c2−2c3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 0 0 1 −7 −9 5 −3 −4 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1. 所以, y1 = 1 D ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x1 2 1 x2 1 5 x3 2 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = x1 ¯ ¯ ¯ ¯ ¯ 1 5 2 3 ¯ ¯ ¯ ¯ ¯ − x2 ¯ ¯ ¯ ¯ ¯ 2 1 2 3 ¯ ¯ ¯ ¯ ¯ + x3 ¯ ¯ ¯ ¯ ¯ 2 1 1 5 ¯ ¯ ¯ ¯ ¯ = −7x1 − 4x2 + 9x3 ; 同理得 y2 = 6x1 + 3x2 − 7x3, y3 = 3x1 + 2x2 − 4x3. 2 . 已知两个线性变换    x1 = 2y1 + y3, x2 = −2y1 + 3y2 + 2y3, x3 = 4y1 + y2 + 5y3,    y1 = −3z1 + z2, y2 = 2z1 + z3, y3 = −z2 + 3z3, 求从 z1, z2, z3 到 x1, x2, x3 的线性变换. 解: 方法一. 直接代入. 比如: x1 = 2y1 + y3 = 2(−3z1 + z2) + (−z2 + 3z3) = −6z1 + z2 + 3z3. 17

第二章矩阵及其运算 方法简单,但我们应尽可能使用本章学习的矩阵知识 方法二由已知 201 所以 415 415 -613 12-49 r1=-621+z2+323, r2=1221-42+923, 1021-2+162 111 123 051 求3AB-2A及A 解 058 111 30-56 A 4.计算下列乘积: (2)(1,2,3)

18 第二章 矩阵及其运算 方法简单, 但我们应尽可能使用本章学习的矩阵知识. 方法二. 由已知   x1 x2 x3   =   2 0 1 −2 3 2 4 1 5     y1 y2 y2   ,   y1 y2 y2   =   −3 1 0 2 0 1 0 −1 3     z1 z2 z3   . 所以,   x1 x2 x3   =   2 0 1 −2 3 2 4 1 5     y1 y2 y2   =   2 0 1 −2 3 2 4 1 5     −3 1 0 2 0 1 0 −1 3     z1 z2 z3   =   −6 1 3 12 −4 9 −10 −1 16     z1 z2 z3   . 即    x1 = −6z1 + z2 + 3z3, x2 = 12z1 − 4z2 + 9z3, x3 = −10z1 − z2 + 16z3. 3 . 设 A =   1 1 1 1 1 −1 1 −1 1  , B =   1 2 3 −1 −2 4 0 5 1   , 求 3AB − 2A 及 ATB. 解: 3AB − 2A = 3   1 1 1 1 1 −1 1 −1 1     1 2 3 −1 −2 4 0 5 1   − 2   1 1 1 1 1 −1 1 −1 1   = 3   0 5 8 0 −5 6 2 9 0   − 2   1 1 1 1 1 −1 1 −1 1   =   −2 13 22 −2 −17 20 4 29 −2   . ATB =   1 1 1 1 1 −1 1 −1 1     1 2 3 −1 −2 4 0 5 1   =   0 5 8 0 −5 6 2 9 0   . 4 . 计算下列乘积: (1)   4 3 1 1 −2 3 5 7 0     7 2 1   ; (2) (1, 2, 3)   3 2 1   ; (3)   2 1 3   (−1, 2);