线性代数第(3)问
1个回答
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使用初孝腔等行歼慎岩变换,
写出增广矩阵为
1 -3 -2 -1 6
3 -8 1 5 0
-2 1 -4 1 -12
-1 4 -1 -3 2 r2-3r1,r3+2r1,r4+r1
~氏御
1 -3 -2 -1 6
0 1 7 8 -18
0 -5 -8 -1 0
0 1 -3 -4 8 r1+3r4,r3+5r4,r4-r2
~
1 0 -11 -13 30
0 1 7 8 -18
0 0 -23 -21 40
0 0 -10 -12 26 r3+r4,r3/(-33)
~
1 0 -11 -13 30
0 1 7 8 -18
0 0 1 1 -2
0 0 -10 -12 26 r1+11r3,r2-7r3,r4+10r3
~
1 0 0 -2 8
0 1 0 1 -4
0 0 1 1 -2
0 0 0 -2 6 r1-r4,r4/(-2) ,r2-r4,r3-r4
~
1 0 0 0 2
0 1 0 0 -1
0 0 1 0 1
0 0 0 1 -3
解得x1=2,x2=-1,x3=1,x4=-3
写出增广矩阵为
1 -3 -2 -1 6
3 -8 1 5 0
-2 1 -4 1 -12
-1 4 -1 -3 2 r2-3r1,r3+2r1,r4+r1
~氏御
1 -3 -2 -1 6
0 1 7 8 -18
0 -5 -8 -1 0
0 1 -3 -4 8 r1+3r4,r3+5r4,r4-r2
~
1 0 -11 -13 30
0 1 7 8 -18
0 0 -23 -21 40
0 0 -10 -12 26 r3+r4,r3/(-33)
~
1 0 -11 -13 30
0 1 7 8 -18
0 0 1 1 -2
0 0 -10 -12 26 r1+11r3,r2-7r3,r4+10r3
~
1 0 0 -2 8
0 1 0 1 -4
0 0 1 1 -2
0 0 0 -2 6 r1-r4,r4/(-2) ,r2-r4,r3-r4
~
1 0 0 0 2
0 1 0 0 -1
0 0 1 0 1
0 0 0 1 -3
解得x1=2,x2=-1,x3=1,x4=-3
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