Question

请教大佬们一道线性代数题？

谢谢大佬

Answer 1

lAl=0,a11的余子式M11不等于0，所以代数余子式A11不等于0,所以r(A)=n-1,
AA*=|A|E=0
这说明A*的列向量都是AX=O的解，
又A11不等于0，
β=（A11,A12.A1n）^T构成AX=O的基础解系，
AX=O的通解为x=kβ

Answer 2

f = (x1)^2+(x2)^2+(x3)^2+4x1x2+4x1x3+4x2x3
= (x1+2x2+2x3)^2 - 3(x2)^2 - 3(x3)^2 - 4x2x3
= (x1+2x2+2x3)^2 - 3[x2+(2/3)x3])^2 - (5/3)(x3)^2
= (y1)^2 - 3(y2)^2 - (5/3)(y3)^2
所用可逆性变换是 y1 = x1+2x2+2x3, y2 = x2+(2/3)x3, y3 = x3.

Answer 3

配方法，步骤为先归并所有含x₁的项，再归并x₂项，依次类推，因归并步骤差异而解得并不唯一，需要检查可逆性：
f=x₁²+x₂²+x₃²+4x₁x₃+4x₂x₃+4x₁x₂
=(x₁+2x₂+2x₃)²-4x₂²-4x₃²-8x₂x₃+x₂²+x₃²+4x₂x₃
=(x₁+2x₂+2x₃)²-3x₂²-3x₃²-4x₂x₃
=(x₁+2x₂+2x₃)²-3(x₂+2/3x₃)²+4/3x₃²-3x₃²
=(x₁+2x₂+2x₃)²-3(x₂+2/3x₃)²-5/3x₃²
令y₁=x₁+2x₂+2x₃,y₂=x₂+2/3x₃,y₃=x₃即x₁=y₁-2y₂-4/3y₃,x₂=y₂-2/3y₃,x₃=y₃
得变换x=Cy，C=
[1 -2 -4/3;
0 1 -2;
0 0 1]
显然是可逆变换，det(C)=1≠0
f的标准型为f=y₁²-3y₂-5/3y₃²