设n（n>1）阶方阵A为正对角线为1,其余为x的方阵.求A的秩

题目

设n（n>1）阶方阵A为正对角线为1,其余为x的方阵.求A的秩
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答案

A =
1 x x ...x
x 1 x ...x
x x 1 ...x
......
x x x ...1
A -->
c1+c2+...+cn --所有列加到第1列
1+(n-1)x x x ...x
1+(n-1)x 1 x ...x
1+(n-1)x x 1 ...x
.........
1+(n-1)x x x ...1
(1) 若 1+(n-1)x ≠ 0,第1列乘 1/[1+(n-1)x]
A-->
1 x x ...x
1 1 x ...x
1 x 1 ...x
......
1 x x ...1
ri-r1,i=2,3,...,n
1 x x ...x
0 1-x 0 ...0
0 0 1-x...0
......
0 0 0 ...1-x
若 x = 1,则 r(A) = 1
若 x ≠ 1,则 r(A) = n
(2) 若 1+(n-1)x = 0
A -->
0 x x ...x
0 1 x ...x
0 x 1 ...x
......
0 x x ...1
ri-r1,i=2,3,...,n
0 x x ...x
0 1-x 0 ...0
0 0 1-x...0
......
0 0 0 ...1-x
若 x = 1,则 r(A) = 1
若 x ≠ 1,则 r(A) = n-1.
综上有
若 x = 1,则 r(A) = 1
若 x ≠ 1,且 x ≠ -1/(n-1) 则 r(A) = n.
若 x ≠ 1,且 x = -1/(n-1) 则 r(A) = n-1.