Let `A=[0 1 0 0]`show that `(a I+b A)^n=a^n I+n a^(n-1)b A`, where I is the identitymatrix of order 2 and `n in N`.

Answer

N/a

Explanation

Here `A=[{:(0,1),(1,0):}]`
`Let P(n) : (al+bA)^(n) -a^(n) I +na^(n-1) bA`
For` n=1`
`(p) =(al+bA)^(n) =a^(n)I+na^(n-1),bA=aI +bA `
which is true
` therefore p(n) ` is true for n=1.
Let p(n) be true for n=k.
`therefore p(k) :(aI+bA)^k) =a^(k)IKa^(k-1) bA .. .(1)`
for `n=k+1`
`p(k+1) : (aI+bA )^(k+1)=(aI+bA)^(k) .(aI+bA)`
`=(a^(k)I+ka^(k-1) bA),(aI+bA)` from equatiom (1)
`=a^(k+1)I+a^(k) IbA +Ka^(k)bAI +ka^(k-1)b^(2) A^(2)`
`=a^(k+1)I+a^(k)bA+Ka^(k)bA+0`
`{:' A^(2)=[{:(0,1),(0,0):}][{:(0,1),(0,0):}]=[{:(0,0),(0,0):}]=o}`
`=a^(K+1) I+(K+1)a^(k)bA`
`implies p(n) ` is also true for `n=K+1.`
therefore ,p(n) is also true for all natural numbers n, by the principle of mathematical induction.