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Question Number 4812 by sanusihammed last updated on 15/Mar/16

Commented by prakash jain last updated on 16/Mar/16

Isn′t (A−B)^T =A^T −B^T ?

Isn′t(A−B)T=AT−BT?

Commented by prakash jain last updated on 15/Mar/16

(A+3 [(1,(−1),0),(1,2,4) ])^T = [(2,1),(0,5),(3,8) ] (A+3 [(1,(−1),0),(1,2,4) ])^T = [(2,1),(0,5),(3,8) ] (A+3 [(1,(−1),0),(1,2,4) ])= [(2,1),(0,5),(3,8) ]^T = [(2,0,3),(1,5,8) ] A= [(2,0,3),(1,5,8) ]−3 [(1,(−1),0),(1,2,4) ]= [((−1),3,3),((−2),(−1),(−4)) ]

(A+3[1−10124])T=[210538](A+3[1−10124])T=[210538](A+3[1−10124])=[210538]T=[203158]A=[203158]−3[1−10124]=[−133−2−1−4]

Commented by Yozzii last updated on 16/Mar/16

(3A^T +2 ((1,0),(0,2) ))^T = ((8,0),(3,1) ) Matrix addition is only possible between matrices of equal dimensions so that addition of corresponding elements occurs. So in (ii) we could let A be a 2×2 matrix of the form A= ((a,b),(c,d) ) ⇒A^T = ((a,c),(b,d) )⇒3A^T = (((3a),(3c)),((3b),(3d)) ) ⇒3A^T + ((2,0),(0,4) )= (((3a+2),(3c)),((3b),(3d+4)) ) ∴(3A^T +2 ((1,0),(0,2) ))^T = (((3a+2),(3b)),((3c),(3d+4)) )= ((8,0),(3,1) ) Equating elements we get a=2,b=0,c=1 and d=−1. Hence A is given by A= ((2,0),(1,(−1)) ). Checking: 3A^T = ((6,3),(0,(−3)) ) 3A^T + ((2,0),(0,4) )= ((8,3),(0,1) )⇒(3A^T +2 ((1,0),(0,2) ))^T = ((8,0),(3,1) ). Alternatively you could utilise the idea of the trasposition of matrices to find A by a series of operations on the original equation. Mr. Jain has done this in his answer for (i). (iii) can be done simply by such a manipulation.

(3AT+2(1002))T=(8031)Matrixadditionisonlypossiblebetweenmatricesofequaldimensionssothatadditionofcorrespondingelementsoccurs.Soin(ii)wecouldletAbea2×2matrixoftheformA=(abcd)⇒AT=(acbd)⇒3AT=(3a3c3b3d)⇒3AT+(2004)=(3a+23c3b3d+4)∴(3AT+2(1002))T=(3a+23b3c3d+4)=(8031)Equatingelementswegeta=2,b=0,c=1andd=−1.HenceAisgivenbyA=(201−1).Checking:3AT=(630−3)3AT+(2004)=(8301)⇒(3AT+2(1002))T=(8031).AlternativelyyoucouldutilisetheideaofthetraspositionofmatricestofindAbyaseriesofoperationsontheoriginalequation.Mr.Jainhasdonethisinhisanswerfor(i).(iii)canbedonesimplybysuchamanipulation.

Commented by prakash jain last updated on 16/Mar/16

(2A^T −5 [(1,0),((−1),2) ])^T =4A−9 [(1,1),((−2),0) ] 2A−5 [(1,(−1)),(0,2) ]=4A−9 [(1,1),((−2),0) ] 2A=9 [(1,1),((−2),0) ]−5 [(1,(−1)),(0,2) ] 2A= [(4,(14)),((−18),(−10)) ] A= [(2,7),((−9),(−5)) ]

(2AT−5[10−12])T=4A−9[11−20]2A−5[1−102]=4A−9[11−20]2A=9[11−20]−5[1−102]2A=[414−18−10]A=[27−9−5]

Commented by Yozzii last updated on 16/Mar/16

(2A^T −5 ((1,0),((−1),2) ))^T =4A−9 ((1,1),((−2),0) ) lhs=rhs Let A be a 2×2 matrix of the form A= ((a,b),(c,d) ) . ∴ 2A^T = (((2a),(2c)),((2b),(2d)) )⇒2A^T −5 ((1,0),((−1),2) )= (((2a−5),(2c)),((2b+5),(2d−10)) ) ∴ lhs= (((2a−5),(2b+5)),((2c),(2d−10)) ) rhs= (((4a−9),(4b−9)),((4c+18),(4d)) ) Since lhs=rhs ⇒2a−5=4a−9⇒a=2 ⇒4c+18=2c⇒c=−9 ⇒4b−9=2b+5⇒b=7 ⇒4d=2d−10⇒d=−5 ∴ A= ((2,7),((−9),(−5)) ) Checking: 2A^T = ((4,(−18)),((14),(−10)) ) (2A^T −5 ((1,0),((−1),2) ))^T = (((−1),(−18)),((19),(−20)) )^T = (((−1),(19)),((−18),(−20)) ) =lhs. rhs= ((8,(28)),((−36),(−20)) )+ (((−9),(−9)),((18),0) )= (((−1),(19)),((−18),(−20)) )=rhs

(2AT−5(10−12))T=4A−9(11−20)lhs=rhsLetAbea2×2matrixoftheformA=(abcd).∴2AT=(2a2c2b2d)⇒2AT−5(10−12)=(2a−52c2b+52d−10)∴lhs=(2a−52b+52c2d−10)rhs=(4a−94b−94c+184d)Sincelhs=rhs⇒2a−5=4a−9⇒a=2⇒4c+18=2c⇒c=−9⇒4b−9=2b+5⇒b=7⇒4d=2d−10⇒d=−5∴A=(27−9−5)Checking:2AT=(4−1814−10)(2AT−5(10−12))T=(−1−1819−20)T=(−119−18−20)=lhs.rhs=(828−36−20)+(−9−9180)=(−119−18−20)=rhs

Commented by Yozzii last updated on 16/Mar/16

Didn′t know that (though simple). Thanks!

Didn′tknowthat(thoughsimple).Thanks!

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