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If A^(n) = 0, then evaluate (i) I+A+A^(...

If `A^(n) = 0`, then evaluate
(i) `I+A+A^(2)+A^(3)+…+A^(n-1)`
(ii)`I-A + A^(2) - A^(3) +... + (-1) ^(n-1)` for odd 'n' where I is the identity matrix having the same
order of A.

Text Solution

Verified by Experts

(i) `A^(n) = 0 rArr A^(n) - I = -I`
`rArr A^(n) - I ^(n) = -I rArr I^(n) -A^(n) =I`
`rArr (I-A) (I + A+A^(2)+A^(3)+... + A^(n-1) ) = I`
`rArr (I+A+A^(2) +A^(3) +... + A^(n-1) )`
`= (I-A)^(-1) I = (I-A) ^(-1)`
(ii) `A^(n) = 0 rArr A^(n) + I = I`
`rArr A^(n) + I^(n) = I`
`rArr I^(n) + A^(n) = I`
`rArr (I+A) (I-A+A^(2) - A^(3) +...+ A^(n-1) )=I`
[`because` n is odd]
`rArr I-A+A^(2) - A^(3) +...+ A^(n-1) `
`=(I+A)^(-1) I = (I+A)^(-1)`
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