If `A=[1 1 1 1 1 1 1 1 1]` , then 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)]` for every positive integer `ndot`

Answer

N/a

Explanation

`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)):}]`
for ` n=1`
` a=[{:(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):}]`
which is true ,
`therefore A^(n)` is true for n=1,
Let `A^(n)` be true for n=K
`therefore A^(k)=[{:(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1)),3^(k-1)),(3^(k-1),3^(k-1) ,3^(k-1)):}].. .(1)`
for n-K+1,
`A^(k+1) =A^k) .A`
`=[{:( 3^(k-1).3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)):}][{:(1,1,1),(1,1,1),(1,1,1):}]`
`=[{:(3^(k-1)+3^(k-1)+3^(k-1),3^(k-1)+3^(k-1)+3^(k-1),3^(k-1)+3^(k-1)+3^(k-1)),(3^(k-1)+3^(k-1)+3^(k-1),3^(k-1)+3^(k-1)+3^(k-1),3^(k-1)+3^(k-1)+3^(k-1)),(3^(k-1)+3^(k-1)+3^(k-1),3^(k-1)+3^(k-1)+3^(k-1),3^(k-1)+3^(k-1)+3^(k-1)):}]`
`=[{:(3.3^(k-1),3.3^(k-1) ,3.3^(k-1)),(3.3^(k-1),3.3^(k-1),3.3^(k-1)),(3.3^(k-1),3.3^(k-1),3.3^(k-1)):}]=[{:(3^(k),3^(k),3^(k)),(3^(k),3^(k),3^(k)),(3^(k),3^(k),3^(k)):}]`
`therefore A^(n) also true for all the number n,
hence proved .