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Question Number 74323 by arthur.kangdani@gmail.com last updated on 22/Nov/19

Commented by arthur.kangdani@gmail.com last updated on 22/Nov/19

Determine the inverse of the matrix of AB+C!

DeterminetheinverseofthematrixofAB+C!

Commented by mathmax by abdo last updated on 22/Nov/19

A.B= (((4 2)),((6 3)) ) (((2 3)),((5 4)) ) = (((18 20)),((27 30)) ) A.B +C = (((18 20)),((27 30)) ) + (((6 4)),((2 3)) ) = (((24 24)),((29 33)) ) =M P_c (x) =det(M−xI)= determinant (((24−x 24)),((29 33−x))) =(24−x)(33−x)−29×24 =24×33−24x−33x+x^2 −29×24 =x^2 −57x +24(33−29) =x^2 −57x +96 cayley hamilton ⇒M^2 −57 M +96I =0 ⇒ 57M−M^2 =96I ⇒ M×((1/(96))( I−M))=I ⇒ M^(−1) =(1/(96))(I−M) =(1/(96))( (((1 0)),((0 1)) ) − (((24 24)),((29 33)) )) =(1/(96)) (((−23 −24)),((−29 −32)) ) = (((−((23)/(96)) −((24)/(96)))),((−((29)/(96)) −((32)/(96)))) )

A.B=(4263)(2354)=(18202730)A.B+C=(18202730)+(6423)=(24242933)=MPc(x)=det(M−xI)=|24−x242933−x|=(24−x)(33−x)−29×24=24×33−24x−33x+x2−29×24=x2−57x+24(33−29)=x2−57x+96cayleyhamilton⇒M2−57M+96I=0⇒57M−M2=96I⇒M×(196(I−M))=I⇒M−1=196(I−M)=196((1001)−(24242933))=196(−23−24−29−32)=(−2396−2496−2996−3296)

Commented by abdomathmax last updated on 22/Nov/19

error at lime 7 M×(57I−M)=96 I ⇒ M×[(1/(96))( 57 I−M)]=I ⇒ M^(−1) =(1/(96)){ (((57 0)),((0 57)) ) − (((24 24)),((29 33)) )} =(1/(96)) (((33 −24)),((−29 24)) )=....

erroratlime7M×(57I−M)=96I⇒M×[196(57I−M)]=I⇒M−1=196{(570057)−(24242933)}=196(33−24−2924)=....

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