已知矩阵a=(0100100000210012),求作可逆矩阵p,使得(ap)^tap是对角矩阵
发布网友
发布时间:2023-09-07 14:14
我来回答
共1个回答
热心网友
时间:2023-09-15 15:56
f = x1^2+x2^2+3x3^2+4x1x2+2x1x3+2x2x3
= (x1+2x2+x3)^2-3x2^2+2x3^2-2x2x3
= (x1+2x2+x3)^2-3(x2^2+(2/3)x2x3)+2x3^2
= (x1+2x2+x3)^2-3(x2+(1/3)x3)^2+(1/3)x3^2+2x3^2
= (x1+2x2+x3)^2-3(x2+(1/3)x3)^2+(7/3)x3^2
y1=x1+2x2+x3
y2=x2+(1/3)x3
y3=x3
C =
1 2 1
0 1 1/3
0 0 1
P=C^-1 即为所求.