Question

已知矩阵n=107014110122001,A=(aij)n*n,aij=i^2+j^2 ,求A的行列式和秩?

Answer 1

由已知,A =

1^2+1^2 1^2+2^2 1^2+3^2 ...1^2+n^2

2^2+1^2 2^2+2^2 2^2+3^2 ...2^2+n^2

3^2+1^2 3^2+2^2 3^2+3^2 ...3^2+n^2

..

n^2+1^2 n^2+2^2 n^2+3^2 ...n^2+n^2

ri-r(i-1),i=n,n-1,...,2 (即从最后一行开始,每行减上一行)

1^2+1^2 1^2+2^2 1^2+3^2 ...1^2+n^2

2^2-1^2 2^2-1^2 2^2-1^2 ...2^2-1^2

3^2-2^2 3^2-2^2 3^2-2^2 ...3^2-2^2

..

n^2-(n-1)^2 n^2-(n-1)^2 n^2-(n-1)^2 ...n^2-(n-1)^2

当n>2时,第2,3,...,n行成比例,故 |A|=0,r(A)=2.