讨论α,b取何值时,方程组
{αx1+x2+x3=b,
{x1-x2+x3=0,
{x1+2x2+x3=3
有唯一解?无解?无穷多个解?并在有解时求其解.
对方程组的增广矩阵进行初等行变换 (α 1 1 b 1 -1 1 0 1 2 1 3) → (1 -1 1 0 α 1 1 b -1 2 1 3) → (-1 -1 1 0 0 1+α1-α b 0 3 0 3) → (1 -1 1 0 0 1 0 1 0 1+α 1-α b) → (1 -1 1 0 0 1 0 1 0 0 1-α b-α-1) → (1 0 1 1 0 1 0 1 0 0 1-α b-α-1) 当α≠1时原方程组化简为 {x2+x3 =1 {x2 =1 (1-α)x3 =b-α-1 解得 {x1= (2-b)/(1-α) {x2=1 {x3=(b-α-1)/(1-α) 当α=1,b=2时原方程化简为 {x1+x3 =1 {x2 =1 解得 {x1=1-x3 {x2 =1 当α=1,b≠2时,最后一行对应方程 0x1+0x2+0x3=b-2≠0 此时原方程组无解.