Question

How to find `a`,such that `BC=I_3`?;`A=((1,3,2),(3,9,6),(2,6,4));B=I_3+A;C=I_3+aA;a inRR`

Answer 1

`a = -1/15`

Explanation:

Matrix `A` obeys it's characteristic polynomial so,
`p(lambda)=lambda^3-14lambda^2 = 0`

then

`p(A) =A^3-14A^2=0_3`

so `BC=a A^2+(a+1)A+I_3 = I_3->a A^2+(a+1)A=0_3`

Taking this last relationship and multiplying it by `A` we have

`aA^3+(a+1)A^2=0_3` or

`A^3+(a+1)/a A^2= 0_3`

now comparing the characteistic polynomial equation with this last equation we conclude

`(a+1)/a=-14->a = -1/15`