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

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Answer 1 · 2017-04-04T19:21:43

$a = -1/15$

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

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

How to find aa,such that BC=I_3BC=I_3?;A=((1,3,2),(3,9,6),(2,6,4));B=I_3+A;C=I_3+aA;a inRRA=((1,3,2),(3,9,6),(2,6,4));B=I_3+A;C=I_3+aA;a inRR