(au1+ba2+x3=1 （(0+3)1+2+2=入 (1) 1+abr2+3=b (2 1+(A-1)z2+3=2 x1+b2+ar3=1 (3(A+1)z1+x2+(A+3)z3=5: ax1+bc2+2r3=1 (3) a1+(2b-1)x2+3rg=1 ax1+bz2+(b+3)xg=2b-1. 解()当a-1(a+2≠0时有解x1=a-+可=驰a子=a-+可 当a=b=-2时，有解x1=x3=-1-2x2 当a=b-1时，有解西1=1-x2-: 其余情形无解 (②倒当入≠0，入≠1时有解五=+公-也，2=+入也，=2入+西 当入=1时有解：1=2-x3,x2=-7+2x3 当入=0时无解： ③当a0,b≠士1时有解=品，=泽，-2 当b=1时有解：2=1-a1,3=0: 当a=0,b=5时有解：2=-了，3=号，1为任意数 其余情形无解 2.利用线性方程组的理论证明：如果直线 L1: Aix+By+C2+D1=0 A2x+B2+Cz+D为=0 与直线 A3+B3!+C3z+D3=0 Aux+Bay+Caz+Da=0 相交，那么 A1 A2 A3 A B1 B2 B3 Ba c-0. 解：根据例3.3的解如果L1与2相交，那么线性方程组 Air+Buy+Cz=-DI =-D为 A3x+B3y+C32=-D3 A+Bay+C:=-Da 有唯一解，从而rank(A)=rank(A)=3,这里A与A分别是上述方程组的系数矩阵与增广矩阵.因此行 列式14=0， -D1 B1 B2 Bs B4 A2 B2 C2 -D2 =0 C1 C2 Cs Ca As B3 C3 -D3 D1 D2 D3 Da A Ba C -Da 3.求三个平面Ax+By+C2+D,=0(位=1,2,3)分别满足下列关系的充要条件. ()有一个公共点： (2②)有一条公共直线： 6(1)    ax1 + bx2 + x3 = 1 x1 + abx2 + x3 = b x1 + bx2 + ax3 = 1; (2)    (λ + 3)x1 + x2 + 2x3 = λ λx1 + (λ − 1)x2 + x3 = 2λ 3(λ + 1)x1 + λx2 + (λ + 3)x3 = 5; (3)    ax1 + bx2 + 2x3 = 1 ax1 + (2b − 1)x2 + 3x3 = 1 ax1 + bx2 + (b + 3)x3 = 2b − 1. : (1) bb(a−1)(a+2) 6= 0RG-: x1= a − b (a − 1)(a + 2) , x2= ab + b − 2 b(a − 1)(a + 2) , x3 = a − b (a − 1)(a + 2) ; b a = b = −2 R, G- x1 = x3 = −1 − 2x2; b a = b = 1 R, G- x1 = 1 − x2 − x3; <$6,-; (2) b λ 6= 0, λ 6= 1 RG-: x1 = λ 2 + 4λ − 15 λ 2 , x2 = λ 2 + λ + 15 λ 2 , x3 = −4λ 2 + λ + 15 λ 2 ; b λ = 1 RG-: x1 = 2 − x3, x2 = −7 + 2x3; b λ = 0 R,-; (3) b a 6= 0, b 6= ±1 RG-: x1 = 5 − b a(b + 1) , x2 = −2 b + 1 , x3 = 2(b − 1) b + 1 ; b b = 1 RG-: x2 = 1 − ax1, x3 = 0; b a = 0, b = 5 RG-: x2 = − 1 3 , x3 = 4 3 , x1 "￾
; <$6,-. 2. 3t&@AB#ST: .t L1 : ( A1x + B1y + C1z + D1 = 0 A2x + B2y + C2z + D2 = 0 B.t L2 : ( A3x + B3y + C3z + D3 = 0 A4x + B4y + C4z + D4 = 0 e, 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ A1 A2 A3 A4 B1 B2 B3 B4 C1 C2 C3 C4 D1 D2 D3 D4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 0. : =>j 3.3 -,  L1 B L2 e, 0t&@AB    A1x + B1y + C1z = −D1 A2x + B2y + C2z = −D2 A3x + B3y + C3z = −D3 A4x + B4y + C4z = −D4 G,H-, C% rank(A) = rank(A˜) = 3, w A B A˜ yS@ABj]^B]^. !O ) |A˜| = 0, ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ A1 A2 A3 A4 B1 B2 B3 B4 C1 C2 C3 C4 D1 D2 D3 D4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = − ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ A1 B1 C1 −D1 A2 B2 C2 −D2 A3 B3 C3 −D3 A4 B4 C4 −D4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 0. 3. s4f Aix + Biy + Ciz + Di = 0 (i = 1, 2, 3) -.*j0&12. (1) GHff(; (2) GH1f(.t; · 6 ·