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#