Let `A=[(1,2,2),(2,1,2),(2,2,1)]` . Then `A^2-4A-5I_3=O` b. `A^(-1)=1/5(A-4I_3)` c. `A^3` is not invertible d. `A^2` is invertible

Options

A. `A^(2)-4A-5I_(3)=O`
B. `A^(-1)=1/5 (A-4I_(3))`
C. `A^(3)` is not invertible
D. `A^(2)` is invertible

Explanation

`A^(2)=[(1,2,2),(2,1,2),(2,2,1)][(1,2,2),(2,1,2),(2,2,1)]=[(9,8,8),(8,9,8),(8,8,9)]`
We have,
`A^(2)-4A-5I_(3)`
`=[(9,8,8),(8,9,8),(8,8,9)]-4 [(1,2,2),(2,1,2),(2,2,1)]-5 [(1,0,0),(0,1,0),(0,0,1)]=O`
or `5I_(3)=A^(2)-4A=A(A-4I_(3))`
or `I_(3)=A. 1/5 (A-4I_(3))` or `A^(-1) =1/5 (A-4I_(3))`
Note that `|A|=5`. Since `|A^(3)|=|A|^(3)=5^(3) ne 0, A^(3)` is inveritible. Similarly, `A^(2)` is invertrible.