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2024年2月25日 星期日

108年北科大電機碩士班-線性代數詳解

國立台北科技大學108學年度碩士班招生考試

系所組別: 2150電機工程系碩士班戊組
第一節 線性代數

解答: {x1+3x2+10x3=18−2x1+7x2+32x3=29−x1+3x2+14x3=12x1+x2+2x3=8⇒[1310−2732−1314112][x1 x2x3]=[18 29128]⇒augmented matrix [131018−273229−1314121128]2R1+R2→R2,R1+R3→R3,−R1+R4→R4→[13101801352650624300−2−8−10]R3/6→R3,−R4/2→R4→[131018013526501450145]−3R3+R1→R1,−13R3+R2→R2,−R3+R4→R4→[10−23000001450000]R2↔R3→[10−23014500000000]⇒{x1−2x3=3x2+4x3=5⇒(x1,x2,x3,x4)∈{(2k+3,−4k+5,k)∣k∈R}
解答: (a)|11312413a|=0⇒a=5⇒[11321243135b]R2−R1→R2,R3−R1→R3→[11320111022b−2]infinitely many sollutions ⇒12=1b−2⇒b=4⇒{a=5b=4(b) trivially,{a=5b≠4
解答: Let T:E→F and T=[a1a2a3b1b2b3c1c2c3]⇒T(vi)=ui,i=1,2,3⇒{[a1a2a3b1b2b3c1c2c3][111]=[110][a1a2a3b1b2b3c1c2c3][232]=[120][a1a2a3b1b2b3c1c2c3][154]=[121]⇒{[111232154][a1a2a3]=[111][111232154][b1b2b3]=[122][111232154][c1c2c3]=[001][111232154]−1=[2313−13−21073−4313]⇒{[a1a2a3]=[2313−13−21073−4313][111]=[23−143][b1b2b3]=[2313−13−21073−4313][122]=[23013][c1c2c3]=[2313−13−21073−4313][001]=[−13013]⇒T=[23−14323013−13013]x=3v1+2v2−v3=(3,3,3)+(4,6,4)−(1,5,4)=(6,4,3)=au1+bu2+cu3⇒{a+b+c=6a+2b+2c=4c=3⇒(a,b,c)=(8,−5,3)
解答: limn→∞anan−1=α⇒limn→∞anan−1=limn→∞an−1+an−2an−1=1+limn→∞an−2an−1=1+1α⇒α=1+1α⇒α2−α−1=0⇒α=1+√52,Q.E.D.
解答: A=[v1∣v2∣v3]=[125−11−4−14−31−47121]⇒{v1=(1,−1,−1,1,1)Tv2=(1,1,4,−4,2)Tv3=(5,−4,−3,7,1)T by Gram-Schmidt processing⇒{e1=1√5(1,−1,−1,1,1)Te2=12(1,0,1,−1,1)Te3=12(1,0,1,1,−1)T⇒Q=[e1∣e2∣e3]=[√551212−√5500−√551212√55−1212√5512−12]⇒R=QTA=[√5−√54√506−2004]⇒A=QR=[√551212−√5500−√551212√55−1212√5512−12][√5−√54√506−2004]
解答: <1,1>=∫1−11dx=2⇒‖1‖=√2⇒e1=1‖1‖=1√2{<1√2,x>=∫1−11√2xdx=0<x,x>=∫1−1x2dx=23⇒e2=x−<1√2,x>1√2‖x−<1√2,x>1√2‖=x‖x‖=√3√2x{<1√2,x2>=√23<√32x,x2>=0⇒e3=x2−<1√2,x2>1√2−<√32x,x2>√32x‖x2−<1√2,x2>1√2−<√32x,x2>√32x‖=x2−13‖x2−13‖=3√52√2(x2−13)⇒e4=x3−<1√2,x3>1√2−<√32x,x3>√32x−<3√52√2(x2−13),x3>3√52√2(x2−13)‖x3−<1√2,x3>1√2−<√32x,x3>√32x−<3√52√2(x2−13),x3>3√52√2(x2−13)‖=x3−<√32x,x3>√32x‖x3−<√32x,x3>√32x‖=x3−15x‖x3−15x‖=5√52√2(x3−15x)⇒an orthonormal basis: 1√2,√3√2x,3√52√2(x2−13),5√52√2(x3−15x)
解答: A=[1201−121310−10−20100028]⇒rref(A)=[1020−101−10−20001400000][1020−101−10−20001400000][x1x2x3x4x5]=0⇒{x1+2x3−x5=0x2−x3−2x5=0x4+4x5=0⇒x=x3(−21100)+x5(120−41)⇒a basis of  ker(T)={(−21100),(120−41)}
解答: [2−4−236−9−582−7−394−2−2−1−6334]=[10000a11000a2a3100a4a5a610a7a8a9a101][b1b2b3b40b5b6b700b8b9000b100000]=[b1b2b3b4a1b1a1b2+b5a1b3+b6a1b4+b7a2b1a2b2+a3b5a2b3+a3b6+b8a2b4+a3b7+b9a4b1a4b2+a5b5a4b3+a5b6+b8a4b4+a5b7+a6b9+b10a7b1a7b2+a8b5a7b3+a8b6+a9b8a7b4+a8b7+a9b9+a10b10]⇒{b1=2b2=−4b3=−2b4=3⇒{a1b1=6a1b2+b5=−9a1b3+b6=−5a1b4+b7=8⇒{a1=3b5=3b6=1b7=−1⇒⋯B=[10000310001−110022−110−3−3201][2−4−23031−1000500000000]
解答: C=[1−10110]⇒W=CCT=[2−11−110101]det(W−λI)=0⇒eigenvalues: λ=3,1,0⇒ eigenvectors:[2−11],[011],[−1−11]normalization→u1=[√6/3−√6/6√6/6],u2=[0√2/2√2/2],u3=[−√3/3−√3/3√3/3]⇒square roots of the nonzero eigenvalues {σ1=√3σ2=1⇒∑=[√300100]⇒U=[u1∣u2∣u3]=[√630−√33−√66√22−√33√66√22√33]⇒{v1=1σ1CT⋅u1=[√2/2−√2/2]v2=1σ2CT⋅u2=[√2/2√2/2]⇒V=[√2/2√2/2−√2/2√2/2]⇒C=U∑VT=[√630−√33−√66√22−√33√66√22√33][√300100][√2/2√2/2−√2/2√2/2]T
解答: x2−6xy+y2=[x,y][1−3−31][xy]=[x,y][1√2−1√21√21√2][−2004][1√21√2−1√21√2][xy]=[1√2(x+y),1√2(−x+y)][−2004][1√2(x+y)1√2(−x+y)]=[x′,y′][−2004][x′y′]=−2x′2+4y′2=5為一雙曲線雙曲線主軸旋轉角度θ⇒cot2θ=1−16=0⇒θ=45∘,因此圖形如下.
 



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