Question

问k取何值时 kx1+x2+x3=1 x1+kx2+x3=k x1+x2+kx3=k^2 有无穷多个解?

Answer 1

题目三式分别为①②③

把②带入①得

kx1+x2+k-x1-kx2=1

(k-1)x1=1-k-(1-k)x2

x1=x2-1④

把④分别带入①、③得

k(x2-1)+x2+x3=1

x2-1+x2+kx3=k²

化简得

x3=(k+1)(1-x2)⑤

2x2+kx3=k²+1⑥

把⑤带入⑥得

2x2+k(k+1)(1-x2)=k²+1

2x2+k²-k²x2+k-kx2=k²+1

(2-k-k²)x2=1-k

(1-k)(2+k)x2=1-k

x2=1/k+2

因为分母为0时无解,得k=-2时无解

又由题意得当K=1时①②③式相等,则当K=1时有无限解

所以当k≠1且k≠-2时有唯一解