# 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

Apr 4, 2017

$a = - \frac{1}{15}$

#### Explanation:

Matrix $A$ obeys it's characteristic polynomial so,
$p \left(\lambda\right) = {\lambda}^{3} - 14 {\lambda}^{2} = 0$

then

$p \left(A\right) = {A}^{3} - 14 {A}^{2} = {0}_{3}$

so $B C = a {A}^{2} + \left(a + 1\right) A + {I}_{3} = {I}_{3} \to a {A}^{2} + \left(a + 1\right) A = {0}_{3}$

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

$a {A}^{3} + \left(a + 1\right) {A}^{2} = {0}_{3}$ or

${A}^{3} + \frac{a + 1}{a} {A}^{2} = {0}_{3}$

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

$\frac{a + 1}{a} = - 14 \to a = - \frac{1}{15}$