求下列齐次线性方程组的基础解系,并写出其一般解 2x1+x2-3x3+2x4=0 3x1+2x2+x3-2x4=0 x1+x2+4x3-4x4=0

日期:2011-11-27 22:44:56 人气:1

求下列齐次线性方程组的基础解系,并写出其一般解 2x1+x2-3x3+2x4=0 3x1+2x2+x3-2x4=0 x1+x2+4x3-4x4=0

2 1 -3 2 3 2 1 -2 1 1 4 -4 ===== 1 1 4 -4 0 -1 -11 10 0 -1 -11 10 ===== 1 1 4 -4 0 1 11 -10 0 0 0 0 ========== 1 0 -7 6 0 1 11 -10 0 0 0 0 x1=7x3-6x4 x2=-11x3+10x4 取x3=1,x4=0,得 x1=7,x2=-11 ξ1=(
    A+
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