Let `A = [[0,1],[0,0]]` , show that `(aI + bA)^n = a^nI + na^(n-1) bA`, where I is the identity matrix of order 2 and`n in N`

Explanation

Let `p(n):(aI+ba)^(n)=a^(n)I+na^(n-1)ba`
step I for `n=1`
`LHS= (aI+ba)^(1) =aI+ba`
and RHS `= a^(1)I+1.a^(0) ba=aI+ba`
LHS=RHS
therefore, `p(1) is true.
step II Assume that `p(k) is true , then
`p(k): (aI+ba)^(k) I+ka^(k-1)ba`
step III for `n=k+1,` we have to prove that
`p(k+1):(aI+ba)^(k+1)k=a^(k+1) I+(k+I) a^(k)bA `
LHS `=(aI+bA)^(k+1) = (aI+bA)^(k) (aI+bA)`
`=a^(k+1) I^(2) + a^(k)b (IA) + ka^(k)b (AI)+k a^(k-1)b^(2) A^(2)`
`=a^(k+1) I+(k+1)a^(k)b A+0`
`[therefore AI=A,A^(2)=0and I^(2) = I]`
`=a^(k+1)I+(k+1)a^(k)bA=RHS`
therefore, `P(k+1)` is true.
Hence, by the principal of mathematical12 induction `p(n)` is true for all n `in` N.