aT1+b T2+I3=1 (入+3)x1+x2+2x3=A (2)x1+(A-1)x2+x3=2入 T1+bz2+ar3=1 3(A+1)x1+Ax2+(A+3)x3=5 aT1+bx2+2 3=1 (3)ax1+(2b-1)x2+3r3=1 ax1+bx2+(b+3)x3=2b 解(1)2a1(+2≠0时有解x1=(a-b+2,=(,=m-1a+ 当a=b=-2时,有解x1 当a=b=1时,有解x1=1 其余情形无解; (2)当A≠0.A≠1时有解:x1 入2+4入-15 入+15 42+A+1 工少 当λ=1时有解:x1=2-x3,x2=-7+2x3 当A=0时无解; )≠0.b≠士时有解n1=a+ 当b=1时有解:x2=1-ax1,x3=0; 当a=0b=5时有解n2=-3,n3=3,n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 A1r+ B1y+C12+ D1=0 A2x+B2y+C22+D2=0 直线 13I+ B3y+C32+ D3 =0 Az+ BAy+C42+DA=0 相交,那么 A1 A2 A3 A B1 B2 B3 B C1 C2 C3 CA D D2 D3 D 解:根据例3.3的解,如果L1与L2相交,那么线性方程组 AIr+ Biy+ CI Ar+ Bay+Ca2 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|4=0 A1 B1 c A2 B2 C C 3.求三个平面Ax+B1y+C1z+D1=0(=1,2,3)分别满足下列关系的充要条件 (1)有一个公共点 (2)有一条公共直线(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 ·