If `A=[(3,-4),(1,-1)]` prove that `A^k=[(1+2k,-4k),(k,1-2k)]` where `k` is any positive integer.

Explanation

We have.
`A^(2)=[(3,-4),(1,-1)][(3,-4),(1,-1)]=[(5,-8),(2,-3)]=[(1+2xx2,-4xx2),(2,1-2xx2)]`
and
`A^(3)=[(5,-8),(2,-3)][(3,-4),(1,-1)]=[(7,-12),(3,-5)]=[(1+2xx3,-4xx3),(3,1-2xx3)]`
Thus, it is true for indices 2 and 3. Now, assume
`A^(k)=[(1+2k,-4k),(k,1-2k)]` to be true.
Then,
`A^(k+1)=[(1+2k,-4k),(k,1-2k)][(3,-4),(1,-1)]`
`=[(3+2k,-4(k+1)),(k+1,-1-2k)]`
`=[(1+2(k+1),-4(k+1)),(k+1,1-2(k+1))]`
thus, if the law is true for `A^(k)`, it is also true for `A^(k+1)`. but it is true for `k=2, 3` etc. Hence, by induction, the required result follows.