因此当k=1或2时这3个向量共面：要使云，乙共线必须使它的的相应坐不成比例，即十=二产=会， 满得k=2.因此当k=2时云，共线. 6.设，，为取问向量可能式表为向量a,,的线要与明？如能，则写出表得式 ()=+22+4，万=-或+，=++3，元=6+3+15 (②云=-2+3，元=+22+2，元=-+6+5，司=2+ 解设元=玉1+2万+x3c,问题归为为满线要必程与. ()必程与为 (x1+x2+x3=6 21-+=3 （4z1+x2+3cg-15, 11111 系数行列式2-11 A1 Q不能断下必程与是式有浅用加减游去安清得=3-子 2=3-号3 =3头可以得到线要表示式22)+仔-+己，于中k为任意数 1+F2 (2)必程与为 -21+2x2+6r3=1系数行列式 31+22+5g=0, 111-1 -226-36, 325 必程与有满 10251」 11 2=30516 x3= 线要表示式为可=-装云+元-品元. 7.当a为何即时，下列四点共面： M(1.a,a2),M2(1,-1,1).M2,1,-2),M(-1,2,2). 解：根可推论4.5，此4点共面的么分必要条件是 11-12-1-1-1 -1-a1-a2-a=-7a2-5a+2=0, 1-a2-2-a22-a2 满得a=-1或， 习题1-5 1.根可n维向量的下义证明：对任意n维向量a,有 (1)0a=0: (2)(-1)a--a (3)k0=0(任意数k) (4从ka=0推出k=0或a=0. 证明：对任意的a=(a1,…,an,则 .16 !Obk = 1D2Rw3f (. &'−→a , −→c (t@A'8e,WU*ej,  1 1 = 1 − k −1 = k 2 , -P k = 2. !Ob k = 2 R −→a , −→c (t. 6.  −→e1 , −→e2 , −→e3 "z.
 −→v c)" −→a , −→b , −→c t&BT? c, J~%P). (1) −→a = −→e1 + 2−→e2 + 4−→e3 , −→b = −→e1 − −→e2 + −→e3 , −→c = −→e1 + −→e2 + 3−→e3 , −→v = 6−→e1 + 3−→e2 + 15−→e3 ; (2) −→a = −→e1 − 2 −→e2 + 3−→e3 , −→b = −→e1 + 2−→e2 + 2−→e3 , −→c = − −→e1 + 6−→e2 + 5−→e3 , −→v = 2−→e1 + −→e2 . :  −→v = x1 −→a + x2 −→b + x3 −→c ,
a""-t&@AB. (1) @AB"    x1 + x2 + x3 = 6 2x1 − x2 + x3 = 3 4x1 + x2 + 3x3 = 15, j ) ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 2 −1 1 4 1 3 ¯ ¯ ¯ ¯ ¯ ¯ = 0, Uc}@AB)G-. D-P ( x1 = 3 − 2 3 x3 x2 = 3 − 1 3 x3, I x3 = 3k >$Pt&) −→v = (3 − 2k) −→a + (3 − k) −→b + 3k −→c , < k "￾
. (2) @AB"    x1 + x2 − x3 = 2 −2x1 + 2x2 + 6x3 = 1 3x1 + 2x2 + 5x3 = 0, j ) ¯ ¯ ¯ ¯ ¯ ¯ 1 1 −1 −2 2 6 3 2 5 ¯ ¯ ¯ ¯ ¯ ¯ = 36, @ABG-: x1 = ¯ ¯ ¯ ¯ ¯ ¯ 2 1 −1 1 2 6 0 2 5 ¯ ¯ ¯ ¯ ¯ ¯ 36 = − 11 36 , x2 = ¯ ¯ ¯ ¯ ¯ ¯ 1 2 −1 −2 1 6 3 0 5 ¯ ¯ ¯ ¯ ¯ ¯ 36 = 16 9 , x3 = ¯ ¯ ¯ ¯ ¯ ¯ 1 1 2 −2 2 1 3 2 0 ¯ ¯ ¯ ¯ ¯ ¯ 36 = − 19 36 . t&)" −→v = − 11 36 −→a + 16 9 −→b − 19 36 −→c . 7. b a "R, l(: M1(1, a, a2 ), M2(1, −1, 1), M3(2, 1, −2), M4(−1, 2, 2). : =>^# 4.5, O 4 (0@&12 ¯ ¯ ¯ ¯ ¯ ¯ 1 − 1 2 − 1 −1 − 1 −1 − a 1 − a 2 − a 1 − a 2 −2 − a 2 2 − a 2 ¯ ¯ ¯ ¯ ¯ ¯ = −7a 2 − 5a + 2 = 0, -P a = −1 D 2 7 .
1–5 1. => n F MST: ￾
n F α, G (1) 0α = 0; (2) (−1)α = −α; (3) k0 = 0 (￾
 k); (4) C kα = 0 ^% k = 0 D α = 0. : ￾
 α = (a1, · · · , an), J: · 16 ·