All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 212028 by siva12345 last updated on 27/Sep/24
DEFINATIONOFQUADRATICFORM:AQuadraticformisahomogeneouspolynomialofdegreetwoinmultiplevariable.Q=XTAXHereQ=Quadraticform.ax2+by2+cz2+2hxy+2fyz+2gzx=0ByusingtheseQ=XTAX[wecanwritematrixA]A=[xyz][ahghbfgfc][xyz]Byusingthecharacteristicequationofmatrix∣A−λI∣=0Togeteigenequationtocansolveeigenequationstogeteigenvalues.Supposesubstitutetheeigwnvaluein∣A−1λI∣X=0Afterrowtransformationwecangetx1,x2,x3P=[e1e2e3]∴{Herepisamodalmatrix}Sincepisaorthogonalmatrixp−1=pD=PTAPWhereDisdiagonalmatrix.ThequadraticformcanbereduceintonormalformX=YTDY.Here:Index=ThenumberofpositivetermsinitscanonicalformSignature=ThedifferenceofpositiveandnegativeterminthecanonicalformIfallλ>0⇒positivedefinite.Ifallλ<0⇒negativedefinite.Ifallλ≥0atleastoneλ=0⇒positivesemidefinite.Ifallλ≤0atlestoneλ=0⇒negativesemidefinite.Ifsomeλarepositiveandsomeλarenegative⇒indefinite.X=PYisusedtransformationofquadraticformtonormalform(or)canonicalform.Eg:(Q)Reducethequadraticform3x2+2y2+3z2−2xy−2yz=0tocanonicalformbyorthogonaltransformationandalsofindrank,index,natureandsignature.Sol:Thegivenequationis3x2+2y2+3z2−2xy−2yz=03xx+2yy+3zz−xy−xy−yz−yz=0weknowthatQ=XTAX=[XYZ][3−10−12−10−13][XYZ]ThecorrespondingsymmetricmatrixA=[3−10−12−10−13]TofindtheeigenvalueswecanusethechareteristiceauationofmatrixAis∣A−λI∣=0[3−λ−10−12−λ−10−13−λ]=0(3−λ)[(2−λ)(3−λ)−1)]+1[−1(3−λ)]=0(3−λ)(−5λ+λ2+5)+(λ−3)=0λ3−8λ2+19λ−12=0λ=1,3,4Casei:λ=1substitutein[A−λI]X=0[2−10−11−10−12][x1x2x3]=[000]R2⇒R2+2R1[2−1001−20−12][x1x2x3]=[000]R3⇒R3+R2[2−1001−2000][x1x2x3]=[000]2x1−x2=0x2−2x3=0x3=kx2=2kx1=kx1=[121]kCaseii:λ=3Tosubstititeλvaluein[A−λI]X=0[0−10−1−1−10−10][x1x2x3]=[000](R1⇔R2)[−1−1−10−100−10][x1x2x3]=[000]R3⇒R3−R2[−1−1−10−10000][x1x2x3]=[000]x2=0−x1−x2−x3=0x3=kx1=−kx2=[−101]kCaseiii:λ=4Tosubstituteinequation[A−λI]X=0[−1−10−1−2−10−1−1][x1x2x3]=[000]R2⇒R2−R1[−1−100−1−10−1−1][x1x2x3]=[000]R3⇒R3−R2[−1−100−1−1000][x1x2x3]=[000]−x1−x2=0−x3−x3=0x3=kx2=−kx1=kx3=[1−11]kwecanobservethatvectorsareorthogonallymutuallywenormalizethisvectorsandobtain.e1=[162616]e2=[−12012]e3=[13−1313]letP=themodalmatrixinnormalizedformP=[e1e2e3]=[16−1213260−131612−13]SincepisaorthogonalmatrixP−1=PTSoD=PTAPwhereDisdiagnolmatrix=[16−2616−1201213−13−13][3−10−12−10−13][16−1213260−131612−13]D=[100030004]ThequadraticformcanbereducetonormalformYTDYYTDY=[y1y2y3][100030004][y1y2y3]y12+3y22+4y32=0Rank=3Index=3Signature=3Nature=PositivedefiniteByorthogonaltfansformationX=PY[x1x2x3]=[16−1213260−13161213][y1y2y3]x1=y16−y22+y33.x2=2y16−y33.x3=y16+y22+y33.o
Terms of Service
Privacy Policy
Contact: info@tinkutara.com
![DEFINATION OF QUADRATIC FORM: A Quadratic form is a homogeneous polynomial of degree two in multiple variable. Q=X^T AX Here Q=Quadratic form. ax^2 +by^2 +cz^2 +2hxy+2fyz+2gzx=0 By using these Q=X^T AX [we can write matrix A] A= [(x),(y),(z) ] [((a h g)),((h b f)),((g f c)) ] [(x,y,z) ] By using the characteristic equation of matrix ∣A−λI∣ =0 To get eigen equation to can solve eigen equations to get eigen values. Suppose substitute the eigwn value in ∣A−1λI∣X =0 After row transformation we can get x_1 ,x_2 ,x_3 P =[e_(1 ) e_2 e_3 ] ∴ {Here p is a modal matrix} Since p is a orthogonal matrix p^(−1) =p D=P^T AP Where D is diagonal matrix. The quadratic form can be reduce into normal form X = Y^T DY. Here: Index = The number of positive terms in its canonical form Signature = The difference of positive and negative term in the canonical form If all λ>0⇒ positive definite. If all λ<0⇒ negative definite. If all λ≥0 atleast one λ= 0⇒positive semidefinite. If all λ≤ 0 atlest one λ= 0⇒negative semidefinite. If some λ are positive and some λ are negative⇒indefinite. X = PY is used transformation of quadratic form to normal form (or) canonical form. Eg: (Q) Reduce the quadratic form 3x^2 +2y^2 +3z^2 −2xy−2yz = 0 to canonical form by orthogonal transformation and also find rank, index, nature and signature. Sol: The given equation is 3x^2 +2y^2 +3z^2 −2xy−2yz=0 3xx+2yy+3zz−xy−xy−yz−yz=0 we know that Q=X^T AX = [(X),(Y),(Z) ] [(( 3 −1 0)),((−1 2 −1)),(( 0 −1 3)) ] [(X,Y,Z) ] The corresponding symmetric matrix A= [(( 3 −1 0)),((−1 2 −1)),(( 0 −1 3)) ] To find the eigen values we can use the chareteristic eauation of matrix A is ∣A−λI∣= 0 [(( 3−λ −1 0)),((−1 2−λ −1)),(( 0 −1 3−λ)) ]= 0 (3−λ)[(2−λ)(3−λ)−1)]+1[−1(3−λ)]= 0 (3−λ)(−5λ+λ^2 +5)+(λ−3)= 0 λ^3 −8λ^2 +19λ−12= 0 λ=1,3,4 Case i: λ=1 substitute in [A−λI]X = 0 [(( 2 −1 0)),((−1 1 −1)),(( 0 −1 2)) ] [(x_1 ),(x_2 ),(x_3 ) ]= [(0),(0),(0) ] R_2 ⇒R_2 +2R_1 [(( 2 −1 0 )),(( 0 1 −2)),(( 0 −1 2)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ] R_3 ⇒R_3 +R_2 [((2 −1 0 )),((0 1 −2)),((0 0 0)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ] 2 x_(1 ) −x_2 =0 x_2 −2x_3 =0 x_3 = k x_2 = 2k x_1 = k x_1 = [(1),(2),(1) ]k Case ii: λ = 3 To substitite λ value in [A−λI]X = 0 [(( 0 −1 0)),((−1 −1 −1)),(( 0 −1 0)) ] [(x_1 ),(x_2 ),(x_3 ) ]= [(0),(0),(0) ] (R_1 ⇔R_2 ) [(( −1 −1 −1)),(( 0 −1 0)),(( 0 −1 0)) ] [(x_1 ),(x_2 ),(x_3 ) ]= [((0 )),(0),(0) ] R_3 ⇒R_3 −R_2 [(( −1 −1 −1)),(( 0 −1 0)),(( 0 0 0)) ] [(x_1 ),(x_2 ),((x3)) ] = [(0),(0),(0) ] x_2 = 0 −x_1 −x_2 −x_(3 ) =0 x_3 = k x_(1 ) = −k x_2 = [((−1)),(( 0)),(( 1)) ]k Case iii: λ = 4 To substitute in equation [ A−λI ]X = 0 [(( −1 −1 0)),(( −1 −2 −1)),(( 0 −1 −1)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ] R_2 ⇒R_2 −R_1 [(( −1 −1 0)),(( 0 −1 −1)),(( 0 −1 −1)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ] R_3 ⇒R_3 −R_2 [(( −1 −1 0)),(( 0 −1 −1)),(( 0 0 0)) ] [(x_1 ),(x_2 ),(x_3 ) ] = [(0),(0),(0) ] −x_1 −x_(2 ) = 0 −x_3 −x_3 = 0 x_(3 ) = k x_2 = −k x_1 = k x_(3 ) = [(( 1)),((−1)),(( 1 )) ]k we can observe that vectors are orthogonally mutually we normalize this vectors and obtain. e_1 = [((1/( (√6)))),((2/( (√6)))),((1/( (√6)))) ] e_2 = [(((−1)/( (√2)))),(( 0)),(( (1/( (√2))))) ] e_3 = [((1/( (√3)))),(((−1)/( (√3)))),((1/( (√3)))) ] let P = the modal matrix in normalized form P = [e_(1 ) e_2 e_3 ] = [(((1/( (√6))) ((−1)/( (√(2 )))) (1/( (√3))))),(((2/( (√(6 )) )) 0 ((−1)/( (√3))))),(((1/( (√6))) (1/( (√2))) ((−1)/( (√3))))) ] Since p is a orthogonal matrix P^(−1 ) = P^T So D = P^T AP where D is diagnol matrix = [(((1/( (√6))) ((−2)/( (√6))) (1/( (√6))))),((((−1)/( (√2))) 0 (1/( (√2))))),(((1/( (√3))) ((−1)/( (√3))) ((−1)/( (√3))))) ] [(( 3 −1 0)),((−1 2 −1)),(( 0 −1 3)) ] [(((1/( (√(6 )))) −(1/( (√2))) (1/( (√3))))),(((2/( (√6))) 0 ((−1)/( (√3))))),(((1/( (√6))) (1/( (√2))) −(1/( (√3))))) ] D = [((1 0 0 )),((0 3 0)),((0 0 4)) ] The quadratic form can be reduce to normal form Y^T DY Y^( T) DY = [y_1 y_(2 ) y_3 ] [((1 0 0)),((0 3 0)),((0 0 4)) ] [(y_1 ),(y_2 ),(y_3 ) ] y_1 ^2 +3y_2 ^2 +4y_3 ^(2 ) = 0 Rank = 3 Index = 3 Signature= 3 Nature=Positive definite By orthogonal tfansformation X = PY [(x_1 ),(x_2 ),(x_3 ) ] = [(((1/( (√6))) ((−1)/( (√2))) (1/( (√3))))),(((2/( (√6))) 0 ((−1)/( (√3))))),(((1/( (√(6 )))) (1/( (√2))) (1/( (√3))))) ] [(y_1 ),(y_2 ),(y_3 ) ] x_(1 ) = (y_1 /( (√6)))−(y_2 /( (√2)))+(y_3 /( (√3))). x_2 = ((2y_1 )/( (√6)))−(y_3 /( (√(3.)))) x_3 = (y_1 /( (√6)))+(y_2 /( (√2)))+(y_3 /( (√(3.)))) o](Q212028.png)