求下列方程组的通解（a+3）x1+x2+2x3=aax1+（a-1）x2+x3=2a3（a+1）x1+ax2+（a+3）x3=3

Question

求下列方程组的通解
（a+3）x1+x2+2x3=a
ax1+（a-1）x2+x3=2a
3（a+1）x1+ax2+（a+3）x3=3

Answer 1

解方程组：
(a+3)x₁+x₂+2x₃=a.(1)
ax₁+(a-1)x₂+x₃=2a.(2)
3(a+1)x₁+ax₂+(a+3)x₃=3.(3)
∣a+3 1 2∣
Δ= ∣ a a-1 1∣=(a+3)[(a-1)(a+3)-a]-[a(a+3)-3(a+1)]+2[a²-3(a+1)(a-1)]=a²(a-1)；
∣3(a+1) a a+3∣
∣a 1 2 ∣
Δ₁=∣2a a-1 1 ∣=a[(a-1)(a+3)-a]-[2a(a+3)-3]+2[2a²-3(a-1)]=a³+3a²-15a+9；
∣3 a a+3∣
∣a+3 a 2 ∣
Δ₂=∣a 2a 1 ∣=(a+3)[2a(a+3)-3]-a[a(a+3)-3(a+1)]+2[3a-6a(a+1)]=a³+12a-9；
∣3(a+1) 3 a+3∣
∣a+3 1 a∣
Δ₃=∣ a a-1 2a∣=(a+3)[3(a-1)-2a²]-[3a-6a(a+1)]+a[a²-3(a+1)(a-1)]=-4a³+3a²+12a-9
∣3(a+1) a 3∣
故x₁=(a³+3a²-15a+9)/[a²(a-1)]；
x₂=(a³+12a-9)/[a²(a-1)]；
x₃=(-4a³+3a²+12a-9)/[a²(a-1)]；