解线性方程组(2-k)x1+2x2-2x3=12x1+(5-k)x2-4x3=2-2x1-4x2+(5-k)x3=-k-1此方程组有唯一解时,k不等于1且k不等于10,

Question

解线性方程组
(2-k)x1+2x2-2x3=1 2x1+(5-k)x2-4x3=2 -2x1-4x2+(5-k)x3=-k-1
此方程组有唯一解时,k不等于1且k不等于10,

Answer 1

(2-k)x1+2x2-2x3=1
2x1+(5-k)x2-4x3=2
-2x1-4x2+(5-k)x3=-k-1有唯一解,
<==>系数行列式≠0,
2-k 2 -2
2 5-k -4
-2 -4 5-k,按对角线法则展开,得
（2-k)(5-k)(5-k)+16+16-[4(5-k)+4(5-k)+16(2-k)]
=(2-k)(25-10k+k^2)+32-(40-8k+32-16k)
=50-20k+2k^2
32 -25k+10k^2-k^3
-72+24k
=10-21k+12k^2-k^3
=(1-k)(10-11k+k^2)
=(1-k)^2(10-k)≠0,
∴k≠1,且k≠10.