Answer :

To prove: (aI + bA)n = anI + nan-1bA

Proof: Given A


We will be proving the above equation using mathematical induction.


Steps involved in mathematical induction are-


1. Prove the equation for n=1


2. Assume the equation to be true for n=k, where k ϵ N


3. Finally prove the equation for n=k+1


I is the identity matrix of order 2,


i.e. I


Let P(n): (aI + bA)n = anI + nan-1bA, n ϵ N


For n=1,


L.H.S: (aI + bA)1 = aI + bA


R.H.S: a1I + 1a1-1bA= aI + a0bA= aI + bA


So, L.H.S = R.H.S


P(n) is true for n=1


Now assuming P(n) to be true for n=k, where k ϵ N


P(k) : (aI + bA)k = akI + kak-1bA …… (1)


Now proving for n=k+1, i.e. P(k+1) is also true


L.H.S = (aI + bA)k+1


= (aI + bA)k . (aI + bA)1


= (akI + kak-1bA). (aI + bA) ……from (1)


= aI(akI + kak-1bA) + bA(akI + kak-1bA)


= aI(akI) + aI(kak-1bA) + bA(akI) + bA(kak-1bA)


= (a.ak) (I×I) + kb(a.ak-1)(IA) + (bak)(AI) + (bb) kak-1(AA)


= ak+1 I2 + ka1+k-1 bA + bakA + b2kak-1A2 (IA = AI= A & I2 = I)


= ak+1 I + kak bA + bakA + b2kak-1A2


Calculating A2


A2 = A.A



A2 = O (O is the null matrix)


Putting value of A2 in L.H.S


L.H.S = ak+1 I + kak bA + bakA + b2kak-1(O)


= ak+1 I + kak bA + bakA + 0


= ak+1 I + kak bA + bakA


= ak+1 I + (k+1)ak bA


Putting n=k+1 in R.H.S


R.H.S = ak+1 I + (k+1)ak bA


L.H.S = R.H.S


All conditions are proved. Hence P(k+1) is true.


By mathematical induction we have proved that P(n) is true for all n ϵ N.


Thus, (aI + bA)n = anI + nan-1bA, where I is the identity matrix of order 2 and n ϵ N.


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