举报
vipfwl
你如果知道合同变换(行列同时相应变换),这题就解了
x1x2 或 x1x3无所谓
写出二次型的矩阵, 用合同变换化对角形
用4阶举例如下
2 -1 0 0
-1 2 -1 0
0 -1 2 -1
0 0 -1 2
r2 + (1/2)r1, c2+ (1/2)c1
2 0 0 0
0 3/2 -1 0
0 -1 2 -1
0 0 -1 2
r3 + (2/3)r2, c3+ (2/3)c2
2 0 0 0
0 3/2 0 0
0 0 4/3 -1
0 0 -1 2
r4 + (3/4)r3, c4+ (3/4)c3
2 0 0 0
0 3/2 0 0
0 0 4/3 0
0 0 0 5/4
看出规律了吧