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Given-the-matrix-A-1-1-1-0-2-1-2-3-0-and-B-3-3-1-2-2-1-4-5-2-find-the-matrix-product-AB-and-BA-state-the-relationship-between-A-and-B-find-also-the-matri

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Question Number 75099 by Rio Michael last updated on 07/Dec/19
Given the matrix A = ((1,(−1),1),(0,2,( −1)),(2,3,0) ) and B= ((3,3,(−1)),((−2),(−2),1),((−4),(−5),2) ) find the matrix product AB and BA state the relationship between A and B find also the matrix product BM, where M= ((8),((−7)),(1) ) Hence solve the system of equations: x−y + z = 8, 2y −z =−7, 2x + 3y = 1.
GiventhematrixA=(1−1102−1230)andB=(33−1−2−21−4−52)findthematrixproductABandBAstatetherelationshipbetweenAandBfindalsothematrixproductBM,whereM=(8−71)Hencesolvethesystemofequations:x−y+z=8,2y−z=−7,2x+3y=1.
Commented by kaivan.ahmadi last updated on 07/Dec/19
Commented by kaivan.ahmadi last updated on 07/Dec/19
Commented by kaivan.ahmadi last updated on 07/Dec/19
Commented by Rio Michael last updated on 07/Dec/19
nice sir thanks
nicesirthanks
Answered by Kunal12588 last updated on 07/Dec/19
A= [(( 1),(−1),( 1)),(( 0),( 2),(−1)),(( 2),( 3),( 0)) ], B= [(( 3),( 3),(−1)),((−2),(−2),( 1)),((−4),(−5),( 2)) ] AB= [((3+2−4),(3+2−5),(−1−1+2)),((0−4+4),(0−4+5),( 0+2−2)),((6−6+0),(6−6+0),(−2+3+0)) ] ⇒AB= [(1,0,0),(0,1,0),(0,0,1) ]= I BA= [(( 3+0−2),(−3+6−3),( 3−3+0)),((−2+0+2),( 2−4+3),(−2+2+0)),((−4+0+4),( 4−10+6),(−4+5+0)) ] ⇒BA= [(1,0,0),(0,1,0),(0,0,1) ]= I ∵AB=BA=I⇒A^(−1) =B & B^(−1) =A M= [(( 8)),((−7)),(( 1)) ] BM= [(( 3),( 3),(−1)),((−2),(−2),( 1)),((−4),(−5),( 2)) ]_(3×3) [(( 8)),((−7)),(( 1)) ]_(3×1) ⇒BM= [(( 24−21−1)),((−16+14+1)),((−32+35+2)) ]= [(( 2)),((−1)),(( 5)) ] given eq^n can be written as [(( 1),(−1),( 1)),(( 0),( 2),(−1)),(( 2),( 3),( 0)) ] [(x),(y),(z) ]= [(( 8)),((−7)),(( 1)) ] ⇒AX=M ; with X= [(x),(y),(z) ] ⇒A^(−1) AX=A^(−1) M ⇒IX=A^(−1) M ⇒X=BM [∵ A^(−1) =B proved above] ⇒ [(x),(y),(z) ]= [(( 2)),((−1)),(( 5)) ] ⇒x=2,y=−1 & z=5
A=[1−1102−1230],B=[33−1−2−21−4−52]AB=[3+2−43+2−5−1−1+20−4+40−4+50+2−26−6+06−6+0−2+3+0]⇒AB=[100010001]=IBA=[3+0−2−3+6−33−3+0−2+0+22−4+3−2+2+0−4+0+44−10+6−4+5+0]⇒BA=[100010001]=I∵AB=BA=I⇒A−1=B&B−1=AM=[8−71]BM=[33−1−2−21−4−52]3×3[8−71]3×1⇒BM=[24−21−1−16+14+1−32+35+2]=[2−15]giveneqncanbewrittenas[1−1102−1230][xyz]=[8−71]⇒AX=M;withX=[xyz]⇒A−1AX=A−1M⇒IX=A−1M⇒X=BM[∵A−1=Bprovedabove]⇒[xyz]=[2−15]⇒x=2,y=−1&z=5
Commented by Rio Michael last updated on 07/Dec/19
thanks so much sir
thankssomuchsir

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