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Find-the-2-2-matrix-A-such-that-4-0-0-4-A-A-1-3-0-0-3-

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Question Number 125768 by Don08q last updated on 13/Dec/20
Find the 2×2 matrix A such that (((−4),( 0)),(( 0),(−4)) ) − A = A^(−1) ((3,0),(0,3) )
Findthe2×2matrixAsuchthat(−400−4)−A=A−1(3003)
Answered by 676597498 last updated on 13/Dec/20
A= ((a,b),(c,d) ) det(A)=ad−cd A^(−1) =(1/(ad−cd)) ((d,(−b)),((−c ),( a)) ) (((−4),( 0)),(0,(−4)) ) − ((d,(−b)),((−c),( d)) ) =(1/(ad−cd)) ((d,(−b)),((−c),( a)) ) ((3,0),(0,3) ) ⇒ (((−4−d),b),(c,(−4−d)) ) =(3/(ad−cd)) ((d,(−b)),((−c),( a)) ) equate enteries and solve
A=(abcd)det(A)=ad−cdA−1=1ad−cd(d−b−ca)(−400−4)−(d−b−cd)=1ad−cd(d−b−ca)(3003)⇒(−4−dbc−4−d)=3ad−cd(d−b−ca)equateenteriesandsolve
Answered by Olaf last updated on 13/Dec/20
−4I−A = A^(−1) ×3I = 3A^(−1) A^2 +4A+3I = 0^∼ (matrice null) (A+I)(A+3I) = 0^∼ A = −I or A = −3I
−4I−A=A−1×3I=3A−1A2+4A+3I=0∼(matricenull)(A+I)(A+3I)=0∼A=−IorA=−3I
Commented by mindispower last updated on 13/Dec/20
AM=0⇏A=0 or M=0 (((0 0)),((1 0)) ). (((0 0)),((1 0 )) )=0
AM=0⇏A=0orM=0(0010).(0010)=0
Commented by Olaf last updated on 14/Dec/20
Sure sir but in your example the determinant of A and M are 0. It′s not the case in the exercise and we verify that the two solutions found are really solutions... of course.
SuresirbutinyourexamplethedeterminantofAandMare0.It′snotthecaseintheexerciseandweverifythatthetwosolutionsfoundarereallysolutions…ofcourse.
Answered by liberty last updated on 13/Dec/20
(((a b)),((c d)) ) [ (((−4 0)),(( 0 −4)) )− (((a b)),((c d)) ) ] = (((3 0)),((0 3)) ) (((a b)),((c d)) ) (((−4−a − b)),(( −c −4−d)) ) = (((3 0)),((0 3)) ) (((−a^2 −4a−bc −ab−4b−bd)),((−ac−4c−cd −bc−4d−d^2 )) ) = (((3 0)),((0 3)) ) (•) −ab−4b−bd=0 ; b(a+4+d)=0 → { ((b=0)),((a+d=−4)) :} (••)−ac−4c−cd=0 ; c(a+d+4)=0→ { ((c=0)),((a+d=−4)) :} for b=0 ⇒−a^2 −4a=3 ; a^2 +4a+3=0 (a+1)(a+3)=0→ { ((a=−1 ∧d=−3)),((a=−3 ∧d=−1)) :} A= (((−1 0)),(( 0 −3)) ) or A= (((−3 0)),(( 0 −1)) )
(abcd)[(−400−4)−(abcd)]=(3003)(abcd)(−4−a−b−c−4−d)=(3003)(−a2−4a−bc−ab−4b−bd−ac−4c−cd−bc−4d−d2)=(3003)(∙)−ab−4b−bd=0;b(a+4+d)=0→{b=0a+d=−4(∙∙)−ac−4c−cd=0;c(a+d+4)=0→{c=0a+d=−4forb=0⇒−a2−4a=3;a2+4a+3=0(a+1)(a+3)=0→{a=−1∧d=−3a=−3∧d=−1A=(−100−3)orA=(−300−1)
Commented by liberty last updated on 13/Dec/20
checking A= (((−1 0)),(( 0 −3)) ) ⇒A^(−1) =(1/3) (((−3 0)),(( 0 −1)) ) (•) (((−4 0 )),(( 0 −4)) ) − (((−1 0)),(( 0 −3)) ) = (1/3) (((−3 0)),(( 0 −1)) ) (((3 0)),((0 3)) ) ⇔ (((−3 0)),(( 0 −1)) ) = (((−3 0)),(( 0 −1)) ) I (true)
checkingA=(−100−3)⇒A−1=13(−300−1)(∙)(−400−4)−(−100−3)=13(−300−1)(3003)⇔(−300−1)=(−300−1)I(true)

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