If A={:[(p,0),(0,p)]:}" then "A^(n+1) is...

If `A={:[(p,0),(0,p)]:}" then "A^(n+1)` is _________ .

`{:[(p,0),(0,p)]:}`

`{:[(p^(n+1),0),(0,p^(n+1))]:}`

`{:[(np,0),(0,np)]:}`

None of these

Text Solution

Verified by Experts

The correct Answer is:

Given `A={:[(p,0),(0,P)]:}`
`A=P={:[(1,0),(0,1)]:}`
`A=PI`
`A^(n+1)=P^(n+1)xxI^(n+1)=P^(n+1)I(becauseI^(n)=I)`
`=P^(n+1){:[(1,0),(0,1)]:}={:[(p^(n+),0),(0,P^(n+1))]:}`.

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