1.计算下列2阶与3阶行列式 解（四）2 =-5. (3) -1 =1 ④234=2 0 14916 =-22. (6)z工=x3+3+3-3xy2 23-3 u z x 2.利用2阶或3阶行列式解线性方程 2x-3y=5, 2-y+32=9 (1) (2){3z-5y+z=-3 3z+2y=1 -2z=-6 15-3 解()x= 2 =8=1y 1 新---1. 32 -1 1293 12-10 25 3-31 35 12-13 2-1 2-13 3-51 35 3-51 13-2 13 13-2 3.设，时，为基 (1)证明：向量云-可+3-可，万=2石-3-10，元=-+2+6线性无关 (②)求向量d=3-2+在基，，下的坐标 (3)求向量了，使-a+2万-3元+3=0. 解：()设有实数x1,工2，x3满足线性关系式x1云+x2万+3=0，表达成坐标形式就是 (x1+2x2-T3=0 31-3x2+2x3=0 (-工1-10z2+6cg=0, 它的系数行列式是 112-11 3-32=-5≠0， -1-106 因此这个方程组只有零解，即,万，亡线性无关 (2)d=3a-26+亡=3(+3-)-2(2-32-10)+(-+2+6写)= -2+17+23,故的坐标是(-2,17,23)， (3)了-专(d-2元+3)=专(-6+15或+37)=-2+5或+妥可. 4.判断下列每组的三个向量云，了，是否共面？能否将亡表示成其它两个向量的线性组合？若 能，写出具体的表示式子 (1)a(5,2,1),7(-1,4,2),c(-1,-1,5: (②)(3,3,2,(6,6.4,(1,-1,0 14 1. xg 2 yB 3 y ): : (1) ¯ ¯ ¯ ¯ −1 1 2 3 ¯ ¯ ¯ ¯ = −5. (2) ¯ ¯ ¯ ¯ cos θ − sin θ sin θ cos θ ¯ ¯ ¯ ¯ = 1. (3) ¯ ¯ ¯ ¯ ¯ ¯ 1 2 2 0 −1 0 0 0 −1 ¯ ¯ ¯ ¯ ¯ ¯ = 1. (4) ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 2 3 4 4 9 16 ¯ ¯ ¯ ¯ ¯ ¯ = 2. (5) ¯ ¯ ¯ ¯ ¯ ¯ 2 1 1 1 1 5 2 3 −3 ¯ ¯ ¯ ¯ ¯ ¯ = −22. (6) ¯ ¯ ¯ ¯ ¯ ¯ x y z z x y y z x ¯ ¯ ¯ ¯ ¯ ¯ = x 3 + y 3 + z 3 − 3xyz. 2. 3 2 yD 3 y )-t&@AB: (1) ( 2x − 3y = 5, 3x + 2y = 1; (2)    2x − y + 3z = 9, 3x − 5y + z = −3, x + 3y − 2z = −6. : (1) x = ¯ ¯ ¯ ¯ 5 −3 1 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 −3 3 2 ¯ ¯ ¯ ¯ = 13 13 = 1, y = ¯ ¯ ¯ ¯ 2 5 3 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 −3 3 2 ¯ ¯ ¯ ¯ = −13 13 = −1. (2) x = ¯ ¯ ¯ ¯ ¯ ¯ 9 −1 3 −3 −5 1 −6 3 −2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 −1 3 3 −5 1 1 3 −2 ¯ ¯ ¯ ¯ ¯ ¯ = −42 49 = − 6 7 , y = ¯ ¯ ¯ ¯ ¯ ¯ 2 9 3 3 −3 1 1 −6 −2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 −1 3 3 −5 1 1 3 −2 ¯ ¯ ¯ ¯ ¯ ¯ = 42 49 = 6 7 , z = ¯ ¯ ¯ ¯ ¯ ¯ 2 −1 9 3 −5 −3 1 3 −6 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 −1 3 3 −5 1 1 3 −2 ¯ ¯ ¯ ¯ ¯ ¯ = 189 49 = 27 7 . 3.  −→e1 , −→e2 , −→e3 "z. (1) ST:  −→a = −→e1 + 3−→e2 − −→e3 , −→b = 2−→e1 − 3 −→e2 − 10−→e3 , −→c = − −→e1 + 2−→e2 + 6−→e3 t&,*; (2) s −→d = 3−→a − 2 −→b + −→c kz −→e1 , −→e2 , −→e3 WU; (3) s −→f , ' − −→a + 2 −→b − 3 −→c + 3 −→f = 0. : (1) G2 x1, x2, x3 -.t&*j) x1 −→a + x2 −→b + x3 −→c = 0, P*WU6)o    x1 + 2x2 − x3 = 0 3x1 − 3x2 + 2x3 = 0 −x1 − 10x2 + 6x3 = 0, 8j ) ¯ ¯ ¯ ¯ ¯ ¯ 1 2 −1 3 −3 2 −1 −10 6 ¯ ¯ ¯ ¯ ¯ ¯ = −5 6= 0, !Owf@AB{Go-,  −→a , −→b , −→c t&,*. (2) −→d = 3−→a − 2 −→b + −→c = 3(−→e1 + 3−→e2 − −→e3 ) − 2(2−→e1 − 3 −→e2 − 10−→e3 ) + (− −→e1 + 2−→e2 + 6−→e3 ) = −2 −→e1 + 17−→e2 + 23−→e3 , ! −→d WU (−2, 17, 23). (3) −→f = 1 3 ( −→a − 2 −→b + 3−→c ) = 1 3 (−6 −→e1 + 15−→e2 + 37−→e3 ) = −2 −→e1 + 5−→e2 + 37 3 −→e3 . 4. |}sB4f −→a , −→b , −→c )(? c)v −→c *<87f t&BT?  c, ~%)￾. (1) −→a (5, 2, 1), −→b (−1, 4, 2), −→c (−1, −1, 5); (2) −→a (3, 3, 2), −→b (6, 6, 4), −→c (1, −1, 0); · 14 ·