A,B为n阶矩阵,cond(AB)<=cond(a)cond(b)。()
A,B为n阶矩阵,cond(AB)<=cond(a)cond(b)。()
设Ax=b,其中A∈Rn×n为非奇异阵,证明:
(a)ATA为对称正定矩阵;
(b)cond(ATA)2=[cond(A)2]2.
A.(define (f n) (cond ((n<2 ) 1) ((n>1) (n* f(n-1)) )
B.(define (f n) (cond ((< n 2 ) 1) ((> n 1 ) (* n (f (- n 1)))) ))
C.(define (f n) (cond ((n<2) 1) ((n>1 ) (n* f(n-1) )) ))
D.(define (f n) (cond ((< n 2 ) 1) ((> n 1 ) (* n (f n-1))) ))
A.(define (f n) (cond ((n<2 ) 1) ((n>1) (n* f(n-1)) )
B.(define (f n) (cond ((< n 2 ) 1) ((> n 1 ) (* n (f (- n 1)))) ))
C.(define (f n) (cond ((n<2) 1) ((n>1 ) (n* f(n-1) )) ))
D.(define (f n) (cond ((< n 2 ) 1) ((> n 1 ) (* n (f n-1))) ))
A.Japanese
B.Russian
C.Indian
D.Korean
A.(define (f n) (cond ((n<2 ) 1) ((n>1) (n* f(n-1)) )
B.(define (f n) (cond ((< n 2 ) 1) ((> n 1 ) (* n (f (- n 1)))) ))
C.(define (f n) (cond ((n<2) 1) ((n>1 ) (n* f(n-1) )) ))
D.(define (f n) (cond ((< n 2 ) 1) ((> n 1 ) (* n (f n-1))) ))
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