www.3112.net > x1+x2+x3+x4+x5=k;3x1+2x2+x3+x4%3x5=0;x2+2x3+2x4+6x...

x1+x2+x3+x4+x5=k;3x1+2x2+x3+x4%3x5=0;x2+2x3+2x4+6x...

解:(1)+(2)+(3)+(4)+(5)得 3(x1+x2 +x3+x4+x5)=15 x1+x2 +x3+x4+x5=5 (6) (1)+(4)-(6)得 x1=0 (2)+(5)-(6)得 x2=2 (1)+(3)-(6)得 x3=-1 (2)+(4)-(6)得 x4=1 (3)+(5)-(6)得 x5=3

x1+x2+x3+x4=0(1)x2+x3+x4+x5=0(2)x1+2 x2+3 x3=2(3)x2+2 x3+3 x4=-2(4)x3+2 x4+3 x5=2(5)(1)-(2):x1=x5(3)-(4):x1+x2+x3-3 x4=4 x1+x2+x3+x4=4+4 x4=0 x4=-1 (4)-(5):x2+x3+x4-3 x5=-4 x2+x3+x4+x5=-4+4 x5=0 x5=1x1=1(2):x2+x3=0(3):1+2 x2+3 x3=2(2)(3):x3=1 x2=-1所以 x1=1,x2=-1,x3=1,x4=-1,x5=1

第一题x=0x(1+2x+x2+x3+x4)=0x=0/(1+2x+x2+x3+x4)这样的话0是不能用来被除的,所以x=0

:x1+x2+x3+x4+x5=a,①3x1+2x2+x3+x4-3x5=0,②x2+2x3+2x4+6x5=b,③5x1+4x2+3x3+3x4-x5=2④先消去x1,①*3-②,x2+2x3+2x4+6x5=3a,①*5-④,x2+2x3+2x4+6x5=5a-2,∴方程组有解时5a-2=3a,a=1.由③,b=3.(2)x3=m,x4=n,x5=p,其中m,n,p是任意数,由③,x2=3-2m-2n-6p,代入①,x1=m+n+5p-2.

方程的一般解为:(-1/3,-2/3,0,0,-1)+k1(4/3,5/3,1,0,1)+k2(1/3,-2/3,0,1,0)

[(1)+(2)+(3)+(4)+(5)]/3得 x1+x2+x3+x4+x5=5 (6) (6)-(1)得x4+x5=4 由(3)得x3+x4+x5=x3+4=3,x3=-1 由(4)得x4+x5+x1=4+x1=4,x1=0 由(1)得x1+x2+x3=0+x2-1=1,x2=2 由(5)得x5+x1+x2=x5+0+2=5,x5=3 x4+x5=x4+3=4,x4=1 则x1=0,x2=2,x3=-1,x4=1,x5=3

增广矩阵=1 1 1 1 51 2 -1 4 -22 -3 -1 -5 -23 1 2 11 0用初等行变换化为1 0 0 0 10 1 0 0 20 0 1 0 30 0 0 1 -1方程组有唯一解:(1,2,3,-1)^T.

之前解答过一个不过第4个方程是 x2+2x3+2x4+6x5=0你参考吧系数矩阵=1 1 1 1 13 2 1 1 -35 4 3 3 -10 1 2 2 6r2-3r1, r3-5r11 1 1 1 10 -1 -2 -2 -60 -1 -2 -3 -30 1 2 2 6r1+r2, r3-r2, r4+r2, r2*(-1)1 0 -1 -1 -50 1 2 2 60 0 0 0 00 0 0 0 0基础解系为 (1,-2,1,0,0)^T, (1,-2,0,1,0)^T, (5,-6,0,0,1)^T

X1+X2+X3+3X4+X5=7 1)3X1+X2+2X3+X4=-2 2) x4=-2-3x1-x2-2x3 2X2+X3+2X4+6X5=23 3)1)+2)+3)4x1+4x2+4x3+6x4+7x5=28 4)1)*4-4) 6x4 -3x5=-7 x4=-2-3x1-x2-2x3 x5=(-12-18x1-6x2-12x3+7)/3=-4-6x1-2x2-4x3+7/3 x1x2x3为任意值

化简是正确的x3 = c1,x4 =c2,x5 = c3代入计算x1,x2用 c1,c2,c3表示就得到了通解{-(9/2),23/2,0,0,0}

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