(④1=1,2=2，g=-2. (⑤)1=,2=合(3-3x4+5s)3,4,x5为自由未知量 2.选择入，使方程组 21-2+3+4=1 1+2红2-x3+4红4=2 x1+7x2-4xg+11x4=入 有解，并求它的一般解 解：仅当入=5时有无穷多解，其一般解为工1=(4-xg-6x4,工2=(3+3-7x4,3,玉4为 自由未知量 3.a,b取何值时，线性方程组 x1+x2+x3+x4+x5=1 31+2r2+x+x4-36=a x2+2r3+2x4+6x5=3 5r1+4x2+3x3+3z4-x5=b 有解在有解的情况下，求一般解 解：仅当a=0,b=2时有解，其一般解为1=-2+x3+x4+5r,2=3-23-2红4-6r5 3,4,5为自由未知量 4.证明方程组 /1-2=a1 工2-工3=02 3-4=03 I4-I5=44 t5-x1=05 有解的充分必要条件是 a1+a2+ag+a4+a5=0. 在有解的情况下，求它的一般解 证明：（→）如线性方程组有解，设(，c2,,4,s)为其一个解，将它代入原方程组并将各式相加 即得a1+2+ag+a4+a5=0. (←)如a1+a2+ag+a4+a5-0,则由最后一个方程得%-x1+a5,依次代入前一个方程得 x4=a4+a5+x1,x3=a+a4+a5+1,2=a2+a3+4+a6+1,将2,3，x4,代入第一个方程 1-(a2+ag+a4+a5+1)=-a2-a3-a4-a5=a 所以原方程组的一般解为 2=a2+a3+a4+as+1 x3=a3+a4+a5+x1 1为自由未知量 r4=a4+5+x1 5=a5+x1 5.求一多项式fc)=a0x3+a1x2+a2x+a4,使f1)=-3,f(-1)=-7,f(2)=-1,f(-2)=-21 解f)-r3-2x2+x-3. .3 (4) x1 = 1, x2 = 2, x3 = −2. (5) x1 = 1 3 x5, x2 = 1 6 (3x3 − 3x4 + 5x5), x3, x4, x5 "gNz . 2. {| λ, '@AB    2x1 − x2 + x3 + x4 = 1 x1 + 2x2 − x3 + 4x4 = 2 x1 + 7x2 − 4x3 + 11x4 = λ G-, Ws8H}-. : cb λ = 5 RG,~ -, <H}-" x1 = 1 5 (4 − x3 − 6x4), x2 = 1 5 (3 + 3x3 − 7x4), x3, x4 " gNz . 3. a, b zR, t&@AB    x1 + x2 + x3 + x4 + x5 = 1 3x1 + 2x2 + x3 + x4 − 3x5 = a x2 + 2x3 + 2x4 + 6x5 = 3 5x1 + 4x2 + 3x3 + 3x4 − x5 = b G-? kG-!, sH}-. : cb a = 0, b = 2 RG-, <H}-" x1 = −2 + x3 + x4 + 5x5, x2 = 3 − 2x3 − 2x4 − 6x5, x3, x4, x5 "gNz . 4. ST@AB    x1 − x2 = a1 x2 − x3 = a2 x3 − x4 = a3 x4 − x5 = a4 x5 − x1 = a5 G-0@&12 a1 + a2 + a3 + a4 + a5 = 0. kG-!, s8H}-. : (⇒) t&@ABG-,  (c1, c2, c3, c4, c5) "<Hf-, v8QRK@AB, Wv()e, P a1 + a2 + a3 + a4 + a5 = 0. (⇐)  a1 + a2 + a3 + a4 + a5 = 0, JNHf@AP x5 = x1 + a5, GHQROHf@A, P x4 = a4 + a5 + x1, x3 = a3 + a4 + a5 + x1, x2 = a2 + a3 + a4 + a5 + x1, v x2, x3, x4, x5 QR=Hf@A, P x1 − (a2 + a3 + a4 + a5 + x1) = −a2 − a3 − a4 − a5 = a1. #$K@ABH}-"    x2 = a2 + a3 + a4 + a5 + x1 x3 = a3 + a4 + a5 + x1 x4 = a4 + a5 + x1 x5 = a5 + x1 x1 "gNz . 5. sH :)f(x) = a0x 3 +a1x 2 +a2x+a3, 'f(1) = −3, f(−1) = −7, f(2) = −1, f(−2) = −21. : f(x) = x 3 − 2x 2 + x − 3. · 3 ·