Question

线性代数大神在哪？第4题

Answer 1

4 a， b， c 均不为0， 则增广矩阵 （A, b） =
[b a 0 c]
[c 0 a b]
[0 c b a]
初等行变换为
[1 a/b 0 c/b]
[0 -ac/b a b-c^2/b]
[0 c b a]
初等行变换为
[1 0 a/c b/c]
[0 1 -b/c (b^2-c^2)/(-ac)]
[0 0 2b (a^2+b^2-c^2)/a]
初等行变换为
[1 0 0 (b^2+c^2-a^2)/(2bc)]
[0 1 0 (a^2-b^2+c^2)/(2ac)]
[0 0 1 (a^2+b^2-c^2)/(2ab)]
r(A, b) = r(A) = 3,
则方程组有唯一解，解是
x = (b^2+c^2-a^2)/(2bc)
y = (a^2-b^2+c^2)/(2ac)
z = (a^2+b^2-c^2)/(2ab)