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solve the given matrix equation for X. Simplify your answers as much as possible. (In the words of Albert Einstein, "Everything should be made as simple as possible, but not simpler") Assume that all matrices are invertible. a.XA-1 = A3b. AXB = (BA)2 c.(A-1 X)-1 = (AB-1)-1(AB2) d.ABXA-1 B-1 =1 + A

Respuesta :

LammettHash

a.

[tex]XA^{-1}=A^3[/tex]

[tex](XA^{-1})A=A^3A[/tex]

[tex]X(A^{-1}A)=A^4[/tex]

[tex]X=A^4[/tex]

b.

[tex]AXB=(BA)^2[/tex]

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

[tex](A^{-1}A)X(BB^{-1})=A^{-1}(BA)^2B^{-1}[/tex]

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

c.

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

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

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

[tex]X^{-1}A=B^3[/tex]

[tex](XX^{-1})A=XB^3[/tex]

[tex]XB^3=A[/tex]

[tex]X(B^3(B^3)^{-1})=A(B^3)^{-1}[/tex]

[tex]X=A(B^3)^{-1}[/tex]

d. Not totally sure what the equation is supposed to be, but I guess it's

[tex]ABXA^{-1}B^{-1}=A[/tex]

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

[tex]((AB)^{-1}(AB))X((BA)^{-1}(BA))=(AB)^{-1}A(BA)[/tex]

[tex]X=(AB)^{-1}A(BA)[/tex]

[tex]X=(B^{-1}A^{-1})A(BA)[/tex]

[tex]X=B^{-1}(A^{-1}A)(BA)[/tex]

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

[tex]X=A[/tex]

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