åÆÍÆ¡Ú âY 2019.12.30 ò! {âK(zK7©,4K,O28©)
1. α1, α2, α3 ̨ènëï ̨ßPn › A = (α1, α2, α3)ß
B = (α1 + α2 + α3, α1 + 2α2 + 4α3, α1 + 3α2 + 9α3)ße |A| = −3ßOé |B|.
1 1 1 1 1 1 )μdB=(α1,α2,α3)1 2 3ߧ±ß|B|=|A|1 2 3=(−3)×2=−6.
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1 4 9 1 4 9
){μ|B|=|α1 +α2 +α3,α2 +3α3,2α2 +8α3|=|α1 +α2 +α3,α2 +3α3,2α3|=2|α1,α2,α3|=−6.
2.n ê A=2 1 2 ßnëï ̨α=(λ,1,1)TßeAαÜαÇ5ɒ߶~Íλ.
1 2 −2λ λ
)μAα=2 1 2 1=2λ+3ßdAαÜαÇ5É’ßλ = 2λ+3 = 3λ+4,):λ=−1. 304 1 3λ+4 λ 1 1
λ λ 1 1
){μ(Aα,α)=2λ+3 1→0 λ+1ßdAαÜαÇ5É’ßλ+1=0ß=λ=−1.
3. λ¤äûߢg.f(x1,x2,x3)=x21 +x2 +5×23 +2λx1x2 −2x1x3 +4x2x3 ¥1⁄2. 1 λ −1
)μg. f ÈA› è A = λ 1 2 ßg.1⁄2øá^ᥠA §k^SÃfTMμ −1 2 5
1 λ 1 λ −1
∆1 =det(1)=1>0,∆2 = =1−λ2 >0,∆3 = λ 1 2 =−5λ2 −4λ>0ß
)μ−54 < λ < 0.
λ 1 −1 2 5 1001 0 0
){μ‹”CÜμA→ 0 1−λ2 λ+2 →0 . 0 −(5λ2 + 4λ)/4 0
A 1⁄2ßK‹”È È ̨èμ−(5λ2 + 4λ)/4 > 0ß)μ− 4 < λ < 0.
4. A è ê ßα1, α2 ¥Ç5Ã'ëï ̨ßÖ Aα1 = 0, Aα2 = 2α1 + α2߶ A §kAä. )μdα1,α2 Ç5Ã'ߧ±α1 ̸=0,2α1 +α2 ̸=0. dAα1 =0=0α1ßKλ1 =0¥òáAä¶ qkA(2α1 +α2)=0+Aα2 =2α1 +α2ßλ2 =1¥,òáAäßKA§kAäè0,1.
){μ¥ A(α1, α2) = (α1, α2) 0 1 = (α1, α2)Bßd α1, α2 Ç5Ã'ß
P =(α1,α2)å_ßu¥P AP =Bß=A∼BßAÜB= 0 1 kÉ”Aäμ0,1.
! ( K12©) ÆÇ5êß|
x1 +λx2 2x1 +x2 3x1 +(2+λ)x2
+μx3 +x4 +x3 +2x4 +(4+μ)x3 +4x4
e (1, −1, 1, −1)T ¥Têß|òá)߶μ(1) êß|œ)¶(2) êß| ̃v x2 = x3 ‹). )μÚ (1, −1, 1, −1)T ì\êß|ß λ = μ.
Æ γ = (1, −1, 1, −1)T ¥êß|òáA). ȇgêß| Ax = 0 ?11– CÜμ 1 λ λ 1 1 0 −2λ 1−λ
A=2 1 1 2→0 1 3 1 . 3 2+λ 4+λ 4 0 0 2(2λ−1) 2λ−1
(1)λ̸=1/2ûßA→0 1 0 −1/2ßAx=0ƒ:)Xùè1ß
å¶êß|œ)èμξ = (1, −1, 1, −1)T + k(−1, 1/2, −1/2, 1)T, k ∈ R; 1 0 −1 1/2
λ=1/2ûßA→0 1 3 1 ßAx=0ƒ:)Xùè2ß 0000
å¶êß|œ)èμξ = (1, −1, 1, −1)T + k1(1, −3, 1, 0)T + k2(−1/2, −1, 0, 1)T, k1, k2 ∈ R. (2)ex2 =x3ßλ̸=1/2ûßd−1+k/2=1−k/2ß)k=2ß
dûêß|œ)èμξ = (1, −1, 1, −1)T + 2(−1, 1/2, −1/2, 1)T = (−1, 0, 0, 1)T; λ=1/2ûßd−1−3k1 −k2 =1+k1ß)k2 =−2−4k1ß
dûêß|œ)èμξ = (2, 1, 1, −3)T + k(3, 1, 1, −4)T. ){μÚ (1, −1, 1, −1)T ì\êß|ß λ = μ.
Èêß| Ax = b O2› ?11– CÜμ
1 λ λ 10 10 −2λ 1−λ −λ (A,b)=2 1 1 2 0→0 1 3 1 1 .
3 2+λ 4+λ 4 1 0 0 2(2λ−1) 2λ−1 2λ−1 1001 0
(1)λ̸=1/2ûß(A,b)→0 1 0 −1/2 −1/2ß 0 0 1 1/2 1/2
r(A, b) = r(A) = 3 < 4, 4 − 3 = 1ßêß|dðı|)ßÖŸƒ:)Xùè 1ß å¶êß|œ)èμξ = (0, −1/2, 1/2, 0)T + k(−2, 1, −1, 2)T, k ∈ R;
1 0 −1 1/2 −1/2 λ = 1/2 ûß(A, b) → 0 1 3 1 1 ß
r(A, b) = r(A) = 2 < 4, 4 − 2 = 2ßêß|dðı|)ßÖŸƒ:)Xùè 2ß
å¶êß|œ)èμξ = (−1/2, 1, 0, 0)T + k1(1, −3, 1, 0)T + k2(−1, −2, 0, 2)T, k1, k2 ∈ R. (2)ex2 =x3ßλ̸=1/2ûßd−1/2+k=1/2−kß)k=1/2ß dûêß|œ)èμξ = (0, −1/2, 1/2, 0)T + 0.5(−2, 1, −1, 2)T = (−1, 0, 0, 1)T; λ=1/2ûßd1−3k1 −2k2 =k1ß)k1 =1/4−k2/2ß
dûêß|œ)èμξ = (−1/4, 1/4, 1/4, 0)T + k(−3/2, −1/2, −1/2, 2)T.
n. ( K12©) (1⁄2~Í k߶ï ̨| α1 = (1, 1, k)T, α2 = (1, k, 1)T, α3 = (k, 1, 1)T ådï ̨| β1 = (1, 1, k)T, β2 = (−2, k, 4)T, β3 = (−2, k, k)T Ç5L—ßï ̨| β1, β2, β3 ÿUdï ̨| α1,α2,α3 Ç5L—.
)μ- A = (α1, α2, α3), B = (β1, β2, β3)ße r(A) = 3ßKï ̨| α1, α2, α3 “¥òá4åÃ'|ß 1 1 k
ÜKøÿ‹ßr(A)<3ßK1 k 1=0⇒k=11⁄2k=−2. k 1 1
1 1 1 k=1ûßA=(α1,α2,α3)=1 1 1ß=r(A)=1ß
1 −2 −2 1 0 0
B=(β1,β2,β3)=1 1 1→0 1 0ß=r(B)=3ß 141001
β1, β2, β3 ÿUdï ̨| α1, α2, α3 Ç5L—.
1 1 −2 1 1 −2
k=−2ûßA=(α1,α2,α3)= 1 −2 1 →0 −3 3 ß=r(A)=2ß −2 1 1 0 3 −3
B = (β1,β2,β3) = 1 −2 −2ß= r(B) = 2ßdûßβ1 = α1,β2 = −2α1ß
1 1 −2
Èu C = (β1,α2,β3)ßdu |C| = 1 −2 −2 = 18 ̸= 0ß= α2 ÿU β1,β2,β3 Ç5L—ß −2 1 −2
Œ‹Kø)¥ k = 1. ){μ1– CÜμ
1 1 k 1 −2 −2 A = (α1,α2,α3,β1,β2,β3) = 1 k 1 1 k k
k11k4k ß 1 0 0 0 0 0 1 0 0 0 0 0 →1 k−1 1−k 0 k+2 k+2→0 1 0 0 0 0
k 1−k 1−k2 0 2k+4 3k 0 0 1 0 0 0 ùè 3ß {α1, α2, α3, β1, β2, β3} 4åÃ'|ï ̨áÍè 3.
α1, α2, α3 åd β1, β2, β3 Ç5L—ßK β1, β2, β3 è4åÃ'|ß
1 −2 −2
u¥|β1,β2,β3|=1 k k =(k+2)(k−4)̸=0.
β1, β2, β3 ÿUd α1, α2, α3 Ç5L—ßK α1, α2, α3 ÿ¥4åÃ'|ßÇ5É'ß
u¥|α1,α2,α3|=1 k 1=−(k+2)(k−1)2 =0ß)k=1.
o. ( K12©) ï ̨ α1, α2, · · · , αn ¥Ç5êß| Ax = 0 òáƒ:)Xßï ̨ β ÿ¥)ß= Aβ ̸= 0ß y2μï ̨| β, β + α1, β + α2, · · · , β + αn Ç5Ã'.
y2μkò|Í k, k1, k2, · · · , kn߶ kβ + k1(β + α1) + k2(β + α2) + · · · + kn(β + αn) = 0ß nn
=(k+ki)β=−kiαiß ̧>”¶› AßdAαi =0,(i=1,2,···,n) i=1 i=1
μ(k + ki)Aβ = − kiAαi = 0ßœ Aβ ̸= 0ßk + ki = 0ßduα1,α2,··· ,αn ¥ƒ:)Xß
i=1 i=1 i=1
Ç5Ã’ß= k1 = k2 = · · · = kn = 0ßKÌ k = 0ߧ±ï ̨| β, β + α1, · · · , β + αn Ç5Ã’.
.( K12©)A,Cèn 1⁄2› ßeB¥’uZ› êßAZ+ZA=Cçò)ß y2μB ¥1⁄2› .
y2μdKøßAB+BA=CßœA,Cè1⁄2› ßÈ°ßK ̧>=òß
C =CT =(AB+BA)T =(AB)T +(BA)T =BTAT +ATBT =BTA+ABT =ABT +BTAß =BT 襛 êßAZ+ZA=C)ßdK()çò5)ßBT =BßB¥È°› . λ ¥ B ?òAäßx ¥·u λ òáAï ̨ßKμ
xTCx = xT(AB + BA)x = xTABx + xTBAx = xTAλx + (Bx)TAx = λxTAx + λxTAx = 2λxTAx. du A,C —è1⁄2› ß xTCx > 0,xTAx > 0ßu¥ λ > 0ß= B ?øAä—åu 0ß §± B è1⁄2› .
15 −11 5 8. ( K12©) Ç5CÜ T 3ƒ α1, α2, α3 e› èA = 20 −15 8.
(1) ¶ T 3ƒ β1 = 2α1 + 3α2 + α3, β2 = 3α1 + 4α2 + α3, β3 = α1 + 2α2 + 2α3 e› ¶
(2)eï ̨x=α1 +6α2 −α3,y=β1 −β2 +β3߶Tx,Ty3α1,α2,α3 eãI. )μlƒ α1, α2, α3 ƒ β1, β2, β3 eLfi› è PßdKøß
231 −6 5 −2 P=3 4 2ßå¶P−1=4 −3 1.
1 1 2 1 −1 1 (1) T 3ƒ β1,β2,β3 e› è BßK
−6 5 −215−115231 100 B=P−1AP=4 −3 120 −15 83 4 2=0 2 0¶
1 −1 1 8 −7 6 1 1 2 0 0 3 x1 y1
(2)Tx,Ty3ƒα1,α2,α3 eãI©Oèx2,y2ßœx3ƒα1,α2,α3 eãIè 6 ß x3 y3 −1
x1 15 −11 51 −56
x2 = 20 −15 8 6 = −78ßdu (β1,β2,β3) = (α1,α2,α3)3 4 2ß
x3 8−76−1−40
2 3 1 112
1 2 3 11 0
§± y = (β1,β2,β3)−1 = (α1,α2,α3)3 4 2−1 = (α1,α2,α3)1ß 111212
y1 15 −11 5 0 −1 u¥y2=20 −15 81= 1 .
y3 8−762 5
(2)){μ(2) T x, T y 3ƒ α1, α2, α3 eãI©Oè x2 , y2ß
1 x1 15 −11 œx3ƒα1,α2,α3 eãIè 6 ßx2=20 −15
8 6 =−78.
x1 y1 x3 y3
−56 −1 x3 8 −7 6 −1 −40
1 1 1 œy3ƒβ1,β2,β3 eãIè−1ßTy3ƒβ1,β2,β3 eãIèB−1=−2ß
113 y1 1 2 3 1 1 −1
Ty3ƒα1,α2,α3 eãIèy2=P−2=3 4 2−2= 1 . y3 311235
‘. ( K12©) A èn ê ßα1, α2, α3 ¥Ç5Ã’nëï ̨ß ÖAα1 =α1 +α2 +α3,Aα2 =2α2 +α3,Aα3 =2α2 +3α3.
(1) ¶› B߶ A(α1, α2, α3) = (α1, α2, α3)B¶
(2) ¶› A Aä¶
(3) ¶å_› P߶ P−1AP èÈ› . )μ(1) dKß
A(α1,α2,α3) = (α1,α2,α3)1 2 2ß= B = 1 2 2¶
(2) œ α1,α2,α3 Ç5Ã’ß› C = (α1,α2,α3) å_ßK C−1AC = Bß= A Ü B Éqß
λ−1 0 0 u¥ßÇkÉ”Aä.|λE−B|= −1 λ−2 −2 =(λ−1)2(λ−4)=0.
−1 −1 λ−3 )μB (=A)Aäèλ1 =λ2 =1,λ3 =4¶
(3)λ1 =λ2 =1ûßd(E−B)x=0)ƒ:)Xèμξ1 =(−1,1,0)T,ξ2 =(−2,0,1)Tß λ3 =4ûßd(4E−B)x=0)ƒ:)Xèμξ3 =(0,1,1)T.
−1 −2 0 1 0 0
– Q = (ξ1,ξ2,ξ3) = 1 0 1ß= Q−1BQ = 0 1 0ßd Q−1BQ = Q−1C−1ACQß
011 001 −1 −2 0
P› P = CQ = (α1,α2,α3) 1 0 1 = (−α1 +α2,−2α1 +α3,α2 +α3),P=觶å_› . 011
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