ξ1 = (1,0,0,...,0,(3-70n)/67)^T
ξ2 = (0,1,0,...,0,(3-70(n-1))/67)^T
.
ξn-1 = (0,0,0,...,1,(3-70*2)/67)^T
方程组的通解为 c1ξ1+c2ξ2+...+cn-1ξn-1
矩阵形式你会的
ξ1 = (1,0,0,...,0,(3-70n)/67)^T
ξ2 = (0,1,0,...,0,(3-70(n-1))/67)^T
.
ξn-1 = (0,0,0,...,1,(3-70*2)/67)^T
方程组的通解为 c1ξ1+c2ξ2+...+cn-1ξn-1
矩阵形式你会的