来自李孝明的问题
请问方程组x1+x2-3x4-x5=0,x1-x2+2x3-x4+x5=0,4x1-2x2+6x3-5x4+x5=0的基础解系与通解怎么求
请问方程组x1+x2-3x4-x5=0,x1-x2+2x3-x4+x5=0,4x1-2x2+6x3-5x4+x5=0的基础解系与通解怎么求
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2020-11-12 01:29
请问方程组x1+x2-3x4-x5=0,x1-x2+2x3-x4+x5=0,4x1-2x2+6x3-5x4+x5=0的基础解系与通解怎么求
请问方程组x1+x2-3x4-x5=0,x1-x2+2x3-x4+x5=0,4x1-2x2+6x3-5x4+x5=0的基础解系与通解怎么求
系数矩阵A=110-3-11-12-114-26-51r2-r1,r3-4r1110-3-10-22220-6675r2*(-1/2),r1-r2,r3+6r2101-2001-1-1-10001-1r1+2r3,r2+r21010-201-10-20001-1所以方程组的基...