1+ 0 0 0 0 0 点x2)m 4.用行列式定义计算: 010 0 (1) 3 (2) 5 00 0 X1 X2 (3) (4) 0 00 X2 0 00 解:(1) =∑(-1)"a1a2pa31,aa3 5 =(-1)43a13axa3aa3 -1)×1×2×3×4×5=120 002 …|=∑(-1)-aman1…a