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Let A =[[1,2,2],[2,1,2],[2,2,1]], then...

Let `A =[[1,2,2],[2,1,2],[2,2,1]]`, then

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

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
`because 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 - 5 I_(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]]`
`=[[0,0,0],[0,o,0],[0,0,0]]= 0`
`rArr 5I_(3) =A^(2) - 4 A = A(A-4I_(3))`
` rArr I_(3) = 1/5 A(A-4I_(3))`
`therefore A^(-1) = 1/5 A (A-4I_(3))`
Since, `abs(A) = 5`
`therefore abs(A^(3)) = abs(A)^(3) = 125 ne 0 `
` rArr A^(3)` is invertible
Similarly, `A^(2)` is invertible.
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