第三章向量与线性方程组习题3.11判断下列方程组是否有解，若有解，用高斯消元法求出一般解（1）解：A=(A|b)=(21−11142−21221−1−11)→r3−r1r2−r1(21−111000−10000−20)→r3−2r2¿r1+r2¿r2×(−1)¿¿¿(21−1010001000000)因为r(A)=r(A)=2<4，所以方程组有无穷多个解，矩阵所对应的方程组为{2x1+x2−x3=1x4=0取x2=t1,x3=t2,则(x1x2x3x4)=(12000)+t1(−12100)+t2(12010)，t1,t2为任意实数。（2）解：A=(A|b)=(42−123−121011308)⃗r1−r2(13−3−83−121011308)→r3−11r1r2−3r1(13−3−80−1011340−303396)→r3−3r2(13−3−80−101134000−6)r(A)=2≠r(A)=3所以此方程组无解。（3）解：A=(A|b)=(23141−24−538−2134−19−6)→r1↔r2(1−24−5231438−2134−19−6)→r4−4r1¿r2−2r1¿r3−3r1¿¿(1−24−507−714014−142807−714)→r2÷7¿r3−2r2¿r4−r2¿¿(1−24−501−1200000000)→r1+2r2(102−101−1200000000)¿r(A)=r(A)=2<3，所以此方程组有无穷多个解，矩阵所对应的方程组为{x1+2x3=−1x2−x3=2令x3=t，则{x1=−1−2tx2=2+tx3=t(x1x2x3)=(−120)+t(−211)，t为任意实数。（4）解：A=(A|b)=(1111012−13125)→r3−3r1(1111012−10−2−12)→r3+2r2r1−r2(10−12012−10030)→r3÷3(10−12012−10010)→r2−2r3r1+r3(1002010−10010)r(A)=r(A)=3=n，所以方程组有唯一解{x1=2x2=−1x3=02求下列齐次线性方程组的通解（1）解：A=(2−4533−6424−81711)→r1−r2r3−2r2(−12113−6420075)→r2+3r1(−121100750075)→r3−r2(−121100750000)矩阵所对应的方程组为：{−x1+2x2+x3+x4=07x3+5x4=0令x2=t1,x4=t2，则{x1=2t1+2t2x2=t1x3=−5t2x4=7t2⇒(x1x2x3x4)=t1(2100)+t2(20−57)，t1,t2为任意实数。（2）解：A=(3−51−223−51−17−43415−79)→r1↔r3(−17−4323−513−51−2415−79)→r4+4r1¿r2+2r1¿r3+3r1¿¿¿(−17−43017−137016−117043−2321)→r2−r3(−17−4301−20016−117043−2321)→r4−43r2r3−16r2(−17−4301−2000217006321)→r3÷7¿r1−7r2¿r4−3r3¿¿(−1010301−2000310000)¿r(A)=3
