å. 解: ç³»æ°è¡åå¼ |A|=
2-λ 2 -2
2 5-λ -4
-2 -4 5-λ
r3+r2
2-λ 2 -2
2 5-λ -4
0 1-λ 1-λ
c2-c3
2-λ 4 -2
2 9-λ -4
0 0 1-λ
= (1-λ)[(2-λ)(9-λ)-8]
= (1-λ)(λ^2-11λ+10)
= -(λ-1)^2(λ-10).
æ以λâ 1ä¸Î»â 10æ¶,æ¹ç¨ç»æå¯ä¸è§£.
å½Î»=1æ¶, å¢å¹¿ç©éµ(A,b)=
1 2 -2 1
2 4 -4 2
-2 -4 4 -2
r2-2r1,r3+2r1
1 2 -2 1
0 0 0 0
0 0 0 0
æ
æ¤æ¶æ¹ç¨ç»ææ ç©·å¤è§£, é解为: (1,0,0)^T+c1(-2,1,0)^T+c2(2,0,1)^T.
å½Î»=10æ¶, å¢å¹¿ç©éµ(A,b)=
-8 2 -2 1
2 -5 -4 2
-2 -4 -5 -2
r1+4r2,r3+r2
0 -18 -18 9
2 -5 -4 2
0 -9 -9 0
r1-2r3
0 0 0 9
2 -5 -4 2
0 -9 -9 0
r(A)=2, r(A,b)=3,æ¤æ¶æ¹ç¨ç»æ 解.
äº. 解:
(α1,α2,α3,α4)=
1 1 -2 -3
-2 -1 4 8
1 1 -3 -6
1 1/2 -2 -4
r4+(1/2)r2, r3-r1, r2+2r1
1 1 -2 -3
0 1 0 2
0 0 -1 -3
0 0 0 0
r3*(-1),r1+2r3-r2
1 0 0 1
0 1 0 2
0 0 1 3
0 0 0 0
æ以α1,α2,α3æ¯ä¸ä¸ªæ大æ å
³ç»
ä¸ Î±4 = α1+2α2+3α3æ¥èªï¼æ±å©å¾å°çåç
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