解數學題目

Question

行列式&方成祖

題目
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Answer 1

3.
xy-x-3y+1=0 =>(x-3)(y-1)=2 --(A)
zx+2x-3z-9=0=>(x-3)(z+2)=3 --(B)
yz+2y-z-8=0 =>(y-1)(z+2)=6 --(C)
以上三式相乘=>[(x-3)(y-1)(z+2)]²=36
=> (x-3)(y-1)(z+2)=±6
與(A)(B)(C)式比較得
(x,y,z)=(4,3,1) or (2, -1, -5)

6.高斯消去法
1, -1, 2, 1-a --(A)
1, 3, -3, 1+a --(B)
3, 1, 1, a --(C)
-(A)+(B), -3(A)+(C),則
1, -1, 2, 1-a
0, 4, -5, 2a --(B')
0, 4, -5, 4a-3 --(C')
-(B')+(C')=> 0, 0, 0, 2a-3
有解=> 2a-3=0
故a=3/2

8.
設a=18度,
行列式=0=>x²-2cosa*x+1=0
(公式)x=cos a±i sin a
(棣美佛)=>x^10=cos(10a)±i sin(10a)=-1

9.造一個方程式使x,y,z為3根
xy+yz+zx=[(x+y+z)²-(x²+y²+z²)]/2=-4
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
=>xyz=-12
(根與係數)x,y,z為方程式t³-3t²-4t+12=0之3根
=>t²(t-3)-4(t-3)=0
=>(t+2)(t-2)(t-3)=0
故(x,y,z)=(-2, 2, 3)