20.
設\( \displaystyle Z=\Bigg[\; \matrix{cos120^o & -sin120^o \cr sin120^o & cos120^o} \Bigg]\; \)，則\( Z+Z^2+Z^3+...+Z^{100}= \)
①\( \displaystyle \Bigg[\; \matrix{1 & 0 \cr 0 & 1} \Bigg]\; \)　②\( \displaystyle \Bigg[\; \matrix{0 & 0 \cr 0 & 0} \Bigg]\; \)　③\( \displaystyle \Bigg[\; \matrix{\frac{-1}{2} & \frac{-\sqrt{3}}{2} \cr \frac{\sqrt{3}}{2} & \frac{-1}{2}} \Bigg]\; \)　④\( \displaystyle \Bigg[\; \matrix{\frac{-1}{2} & \frac{\sqrt{3}}{2} \cr \frac{-\sqrt{3}}{2} & \frac{-1}{2}} \Bigg]\; \)
[解答]
這題是考旋轉矩陣
\( \displaystyle Z=\Bigg[\; \matrix{cos120^o & -sin120^o \cr sin120^o & cos120^o} \Bigg]\; \)
\( \displaystyle Z^2=\Bigg[\; \matrix{cos240^o & -sin240^o \cr sin240^o & cos240^o} \Bigg]\; \)
\( \displaystyle Z^3=\Bigg[\; \matrix{cos360^o & -sin360^o \cr sin360^o & cos360^o} \Bigg]\;=I_2 \)
\( Z+Z^2+Z^3=\Bigg[\; \matrix{0 & 0 \cr 0 & 0} \Bigg]\; \)
\( (Z+Z^2+Z^3)+(Z^4+Z^5+Z^6)+...+(Z^{97}+Z^{98}+Z^{99})+Z^{100}=Z^1 \)
答案③

[ 本帖最後由 bugmens 於 2011-9-19 07:55 AM 編輯 ]

solver 回覆於: 2011/11/9 下午 10:11:39

因為方程有兩實根，
△>=0
(5-a)^2-4(a^2-7a-14)>=0
25-10a+a^2-4a^2+28a+56>=0
-3a^2+18a+81>=0
a^2-6a-27<=0
(a-9)(a+3)<=0
-3<=a<=9

α+β= a-5
αβ=a^2-7a-14
αβ-α-β+1=a^2-7a-14-(a-5)+1
=a^2-8a-8
=(a-4)^2-24

所求最大值是(-3-4)^2-24=25

昌爸轉貼

49.
已知S(0,1)[f(x)]dx=1

S(0,1)[f(x)]dx=1
S(0,1)[f(2x)/3]dx=1
S(0,1)[f(2x)]dx=3
S(0,2)[f(u)/2]du=3 (u=2x)
S(0,2)[f(u)]du=6
S(0,2)[f(x)]dx=6
S(0,1)[f(x)]dx+S(1,2)[f(x)]dx=6
1+S(1,2)[f(x)]dx=6
S(1,2)[f(x)]dx=5

引用:

原帖由 nanpolend 於 2011-9-19 03:57 AM 發表

第23題錯在矩陣要往前(左邊)乘而非往後(右邊)乘