《线性代数》Chapter 3（2）线性方程组的求解

2 4 -5 →0x-1 -1 0-2(-1) (-6(九-1) 24 -5 当(-1(2-10)=0时, 00(10-0-1)/即x=1或入=时 0n -1 2 方程组有非零解 24 -5 →0x-1 -1 00(x-1)(-10) K图心

            − − − − − − − − → 2 ( 6)( 1) 0 2( 1) 0 1 1 2 4 5                   − − − − − → 2 (10 )( 1) 0 0 0 1 1 2 4 5                − − − − − → 0 0 ( 1)( 10) 0 1 1 2 4 5      . 1 10 , ( 1)( 10) 0 , 方程组有非零解 即 或 时 当 时 = =  − − =    

24-4 12-2 (1)当花=时,A→000→000 2x 2 +2x 同解方程组为{x2=x2 x=A1|+k2|0 3-3 2 5(201 01 (2)当=10时,4→011→011→011 2-3 同解方程组为x2=-x3∴:x=k1 3 3

(1) 当 = 1时,           − → 0 0 0 0 0 0 2 4 4 A           − → 0 0 0 0 0 0 1 2 2 同解方程组为      = = = − + 3 3 2 2 1 2 2 2 3 x x x x x x x           +          −  = 1 0 2 0 1 2 x k1 k2 (2) 当 = 10时,           → 0 0 0 0 1 1 2 4 5 A           → 0 0 0 0 1 1 2 0 1           → 0 0 0 0 1 1 1 0 2 1 同解方程组为      = = − = − 3 3 2 3 2 3 1 1 x x x x x x           − −  = 1 1 2 1 x k           − −  =  2 2 1 x k

方法2: -2-22-2-22 A 22-54 0-1a 44-5244-5 A-2-22 A-2-20 =(-1)011=(-1)0 2 24-5 24a =(-1)(x2-113+10)=(x-1)2(4-10) 当4=0,即=1或λ=10时,方程组有非解; K图心

方法2： 2 4 5 2 5 4 2 2 2 − − − − − =    A 2 4 5 0 1 1 2 2 2 − − − − − =     2 4 5 0 1 1 2 2 2 ( 1) − − − = −    2 4 1 0 1 2 2 2 0 ( 1) − − − = −    ( 1)( 11 10) 2 =  −  −  + ( 1) ( 10) 2 =  −  − 当A = 0,即 = 1或 = 10时,方程组有非0解;

1-22 12-2 (1)当花=时,A=-2-44→000 24-4 2x 2 +2x 同解方程组为{x2=x2 x=A1|+k2|0 3-3 8-22)(245(10 (2)当λ=10时, 4 254})|0 245)(000)(000 2-3 同解方程组为x2=-x3∴x=k12 3 3

(1) 当 = 1时,           − − − − − = 2 4 4 2 4 4 1 2 2 A           − → 0 0 0 0 0 0 1 2 2 同解方程组为      = = = − + 3 3 2 2 1 2 2 2 3 x x x x x x x           +          −  = 1 0 2 0 1 2 x k1 k2 (2) 当 = 10时,           − − = 2 4 5 2 5 4 8 2 2 A           → 0 0 0 0 1 1 2 4 5           → 0 0 0 0 1 1 1 0 2 1 同解方程组为      = = − = − 3 3 2 3 2 3 1 1 x x x x x x           − −  = 1 1 2 1 x k           − −  =  2 2 1 x k