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

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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