解水银密度与温度的关系为p()=[1t2][aoa1a2a],根据实验数据先确定 参数a0,a1,a2,a3.为此,求解下列线性方程组 1101001000a1 13.57 1204008000 13090027000a 对增广矩阵实施初等行变换 100 013.60 100013.60 110100100013.57 0101001000-0.03 20400800013.55 0204008000-0.05 1309002700013.52 03090027000-0.08 100 013.60 013.60 =3:图010100100-0.03 0101001000-0.03 0020060000.01 0020060000.01 00600240000.01 0006000-0.02 B6) 100013.60 R2-(1/2)R 01000-0.25/6 002000 0006000 0.02 解得a0=13.60,a1=-0.025/6,a2=0.00015,a3=-0.0000/3,即 ()=1300~9+0002-32 0.00001 则 P(15)=1360-0.0625+0.03375-0.01125=13.56 (40)=1360-g+0.2416=13.46 (19)证明方程组 2a111+2a12x2+ 2a211+2a2202+…+2a2nn 2an11+2an2x2+ 只有零解,其中a全为整数解: 水银密度与温度的关系为 ρ (t) = h 1 t t2 t 3 i h a0 a1 a2 a3 iT , 根据实验数据先确定 参数 a0, a1, a2, a3. 为此, 求解下列线性方程组      1 0 0 0 1 10 100 1000 1 20 400 8000 1 30 900 27000           a0 a1 a2 a3      =      13.60 13.57 13.55 13.52      对增广矩阵实施初等行变换:      1 0 0 0 13.60 1 10 100 1000 13.57 1 20 400 8000 13.55 1 30 900 27000 13.52      R2 − R1 R3 − R1 R4 − R1 −−−−−−−→      1 0 0 0 13.60 0 10 100 1000 −0.03 0 20 400 8000 −0.05 0 30 900 27000 −0.08      R3 − 2 ∗ R2 R4 − 3 ∗ R1 −−−−−−−−−→      1 0 0 0 13.60 0 10 100 1000 −0.03 0 0 200 6000 0.01 0 0 600 24000 0.01      R4−3∗R3 −−−−−−→      1 0 0 0 13.60 0 10 100 1000 −0.03 0 0 200 6000 0.01 0 0 0 6000 −0.02      R3 − R4 R2 − (1/6) ∗ R4 R2 − (1/2)R3 −−−−−−−−−−−→      1 0 0 0 13.60 0 10 0 0 −0.25/6 0 0 200 0 0.03 0 0 0 6000 −0.02      解得 a0 = 13.60, a1 = −0.025/6, a2 = 0.00015, a3 = −0.00001/3, 即 ρ (t) = 13.60 − 0.025 6 t + 0.00015t 2 − 0.00001 3 t 3 则 ρ (15) = 13.60 − 0.0625 + 0.03375 − 0.01125 = 13.56 ρ (40) = 13.60 − 1 6 + 0.24 − 16 75 = 13.46 (19) 证明方程组    x1 = 2a11x1 + 2a12x2 + · · · + 2a1nxn x2 = 2a21x1 + 2a22x2 + · · · + 2a2nxn · · · · · · · · · · · · · · · xn = 2an1x1 + 2an2x2 + · · · + 2annxn 只有零解, 其中aij全为整数