中南大学：《线性代数》题解案例（双语版）chapter3（3-2）线性方程组的求解

2 -5 02-1 -1 0-2(-1 (-6)(-1) 2 24 -5 当(-1)(2-10)=0时 00(0-2)(x-)/52=1或= →0元-1 -1 方程组有非零解 24 -5 →02-1 -1 00(x-1)(x-10)

            − − − − − − − − → 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=时,A→000→000 2x,+2x 2 同解方程组为x2=x2 x=h11+k20 0 245 (2)当2=10时,A→011|→011→011 000)(000 2~3 同解方程组为x2=-x3∴x 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: -2-222-2-22 A 2a-54 0元-1-1 2 42-52 4元-5 A-2-22 A-2-20 =(元-1)011=(-1)012 4a-5 24a-1 =(-1)(x2-11元+10)=(元-1)2(x-10) 当A=0,即=1或礼=10时,方程组有非0解

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;

12-2 (1)当=时,A=-2-44→000 24-4 2x,+2x 同解方程组为{x2 x=k1+k20 0 8 2 (2)当=10时,4=-254011}011 245)(000 2~3 同解方程组为x2=-x3∴x=k 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