If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 2}\\ 2&5&{ - 4}\\ 3&7&{ - 5} \end{array}} \right]\) then A^-1 is

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 2}\\ 2&5&{ - 4}\\ 3&7&{ - 5} \end{array}} \right]\) 

[A - λI] = 0

\(\left[ {\begin{array}{*{20}{c}} {1 - \lambda }&2&{ - 2}\\ 2&{5 - \lambda }&{ - 4}\\ 3&7&{ - 5 - \lambda } \end{array}} \right] = 0\)

(1 - λ) [(5 - λ) (-5 - λ) + 28] -2 [2 (-5 - λ) + 12] – 2 [14 - 15 + 3λ] = 0

(1 – λ) (λ² + 3) – 2 (2 – 2λ) – 2 (-1 + 3λ) = 0

 λ² + 3 – λ³ – 3λ – 4 + 4 λ + 2 - 6 λ = 0

- λ³ + λ² - 5λ + 1 = 0

By cayley hamilton theorem,

-A³ + A² – 5A + I = 0

A³ - A² + 5A - I = 0

On dividing it by A

∴ A² – A + 5I = A^-1