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If A=[[1,1,1],[1,1,1],[1,1,1]], prove th...

If `A=[[1,1,1],[1,1,1],[1,1,1]]`, prove that `A^(n)=[[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)]]`,n `in` N.

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Given `A=[[1, 1, 1],[ 1, 1, 1],[ 1, 1, 1]]`
To prove that `A^n=[[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)]], n in N.`

let n=1

`A^1=[[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)]] =>[[3^0,3^0,3^0],[3^0,3^0,3^0],[3^0,3^0,3^0]]`

`=>[[1, 1, 1],[ 1, 1, 1],[ 1, 1, 1]]=A`
The result is true for n=1
...
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