Respuesta :

LammettHash

[tex]A^2=2I[/tex]

Recall that [tex]I=AA^{-1}=A^{-1}A[/tex], so

[tex]A^2=2(A^{-1}A)[/tex]

Multiply on the right by [tex]A^{-1}[/tex] and regroup terms:

[tex]A^2A^{-1}=2(A^{-1}A)A^{-1}[/tex]

[tex]A(AA^{-1})=2A^{-1}(AA^{-1})[/tex]

[tex]A=2A^{-1}[/tex]

and finally solve for [tex]A^{-1}[/tex]:

[tex]\boxed{A^{-1}=\dfrac12A}[/tex]

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