# How to find B^(-1)?;We know that B^2=B+2I_3

##### 2 Answers
Apr 2, 2017

${B}^{- 1} = \left(\frac{1}{2}\right) B - \left(\frac{1}{2}\right) {I}_{3}$.

#### Explanation:

You have

${B}^{2} = B + 2 {I}_{3}$.

Multiply by ${B}^{- 1}$:

$B = {I}_{3} + 2 B$^{-1}#

And then from simple algebra:

$\left(\frac{1}{2}\right) B - \left(\frac{1}{2}\right) {I}_{3} = {B}^{- 1}$.

Apr 2, 2017

${B}^{-} 1 = \frac{1}{2} \left(B - {I}_{3}\right)$

#### Explanation:

Any matrix $B$ obeys it's characteristic polynomial

Given $B = \left(\begin{matrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{matrix}\right)$ we have

${B}^{3} - 3 B - 2 {I}_{3} = {0}_{3}$ because

$p \left(\lambda\right) = {\lambda}^{3} - 3 \lambda - 2$

Now multiplying by left this relationship we have

${B}^{-} 1 {B}^{3} - 3 {B}^{-} 1 B - 2 {B}^{-} 1 = {0}_{3}$ or

${B}^{2} - 3 {I}_{3} - 2 {B}^{-} 1 = {0}_{3}$ then

$2 {B}^{-} 1 = {B}^{2} - 3 {I}_{3}$ but

${B}^{2} = B + 2 {I}_{3}$ so finally

${B}^{-} 1 = \frac{1}{2} \left(B - {I}_{3}\right)$