AX = b

1 1

A

() ,: ,, ; ,(),. .

Step 1 i mi1 1

Step k n 1

A 0

: 1.2 _

(4)(1)..(4)

1.3 _

Gauss k k k+1,,n xk , . k k.Gauss

Gaussian Elimination10-91.3 _

If ik k then k ik ; If jk k then k jk ; xi

1. (1) (2) (3) (4)

(1) (2)

.

Guass)x1=1.9273, x2=-0.698496, x3=0.9004233

Guass)x1=1.9273, x2=-0.698496, x3=0.9004233

1n A(n,n+1);(1) l (3) 3

4k x11 n 1x1n 1x2 1n 1x2kkn kkxk 1n - 1xk

k = 1, 2, , n

51AX = b1, AX = b2, AX = bmAX = B = (b1, , bm)

X = A-1B

2A = (aij)nnA 0 AA-1 = AX = IA-11m = n, B = I

3 A

2

:3

Step n 1:

A A 0 A LU L AX=b,A=LU, AX=LUX=bUX=Y, LY=b. AX=bUX=YLY=b 1LY=b, :2UX=Y , :

2 , i = 1, 2, , k A

L U 1i=1,2,,n 2 U r , L r r =2,3,n

AX = br = 2, 3, , n2Ur 3Lr (r n)(1)

(4)(5)

4 1LDR3nAAA = LDRLRnDnALDR

2AA=LLT 4

A LU

1ACholeskyA=LLT i = 1, 2,, n

2 3LTX = y

3

LTX = yLDY = bLTX = Y

5 11X R n RnX3XY R n (1) X R nX 0, >0(2) a RX R n

--------(1)--------(2)--------(3)--------(4)x

:

3

3 :1*499/4*4=9

: A B R C1C2 > 0

:

i (i = 1, 2,, n )

2{Xk}X*2

--------(5)--------(6)--------(7) || ||2 Frobenius --------(8)

5A:

5A():

5A():

4

--------(9)

An, || || 3

A A 2 A2 2(A) 4

5.

:

7

6

Wait a minute Who said that ( I + A1 A ) is invertible?

cond ( A )1cond ( kA )= cond ( A ) k cond( A )

, A.A 39206>>1.

cond (H2) =27cond (H3) 748cond (H6) =2.9 106cond (Hn) as n A1

cond (A)