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let-A-a-b-c-d-use-the-augmented-matrix-A-I-and-elementary-row-operation-to-show-A-1-1-ad-bc-a-b-c-d-and-show-that-det-A-1-1-det-A-

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Question Number 78670 by berket last updated on 19/Jan/20
let A= [((a b)),((c d)) ]use the augmented matrix[A I] and elementary row operation to show A^(−1) = (1/(ad bc)) [((a b)),((c d)) ]and show that det(A^(−1) )=(1/(det(A)))
letA=[abcd]usetheaugmentedmatrix[AI]andelementaryrowoperationtoshowA−1=1adbc[abcd]andshowthatdet(A−1)=1det(A)
Commented by abdomathmax last updated on 20/Jan/20
Pc(A)=det(A−xI) = determinant (((a−x b)),((c d−x))) =(a−x)(d−x)−bc =ad−ax−dx+x^2 −bc =ad−bc −(a+d)x +x^2 cayley hamiton ⇒A^2 −(a+d)A +(ad−bc)I =0 ⇒ A^2 −(a+d)A =(bc−ad)I if bc−ad≠0 we get (1/(bc−ad))A(A−(a+d)I)=I ⇒ A^(−1) =(1/(bc−ad))(A−(a+d)I =(1/(bc−ad))( (((a b)),((c d)) )− (((a+d 0)),((0 a+d)) )) =(1/(bc−ad)) (((−d b)),((c −a)) )
Pc(A)=det(A−xI)=|a−xbcd−x|=(a−x)(d−x)−bc=ad−ax−dx+x2−bc=ad−bc−(a+d)x+x2cayleyhamiton⇒A2−(a+d)A+(ad−bc)I=0⇒A2−(a+d)A=(bc−ad)Iifbc−ad≠0weget1bc−adA(A−(a+d)I)=I⇒A−1=1bc−ad(A−(a+d)I=1bc−ad((abcd)−(a+d00a+d))=1bc−ad(−dbc−a)
Commented by abdomathmax last updated on 20/Jan/20
A^(−1) = ((((d/(ad−bc)) ((−b)/(ad−bc)))),((((−c)/(ad−bc)) (a/(ad−bc)))) )
A−1=(dad−bc−bad−bc−cad−bcaad−bc)

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