解用加减消元法,可得 11022 a12a2)x1=ba2-a2b2, 1122 当a1(2a121≠0时,求得方程组(1)的解为 b 22 122 xX l122 120121 b2-b1 (2) 21 1022 120121 − = − − = − ( ) . ( ) , 1 1 2 2 1 2 2 1 2 1 1 2 1 2 1 1 1 2 2 1 2 2 1 1 1 2 2 1 2 2 a a a a x a b b a a a a a x b a a b 解 用加减消元法,可得 当 a11a22 - a12a21 0 时,求得方程组(1)的解为 − − = − − = . , 11 22 12 21 11 2 1 21 2 11 22 12 21 1 22 12 2 1 a a a a a b b a x a a a a b a a b x (2)