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

To prove: (aI + bA)ⁿ = aⁿI + na^n-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 + na^n-1bA, n ϵ N

For n=1,

L.H.S: (aI + bA)¹ = aI + bA

R.H.S: a¹I + 1a^1-1bA= 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 = a^kI + ka^k-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

= (a^kI + ka^k-1bA). (aI + bA) ……from (1)

= aI(a^kI + ka^k-1bA) + bA(a^kI + ka^k-1bA)

= aI(a^kI) + aI(ka^k-1bA) + bA(a^kI) + bA(ka^k-1bA)

= (a.a^k) (I×I) + kb(a.a^k-1)(IA) + (ba^k)(AI) + (bb) ka^k-1(AA)

= a^k+1 I² + ka^1+k-1 bA + ba^kA + b²ka^k-1A² (IA = AI= A & I² = I)

= a^k+1 I + ka^k bA + ba^kA + b²ka^k-1A²

Calculating A²

A² = A.A

∴ A² = O (O is the null matrix)

Putting value of A² in L.H.S

L.H.S = a^k+1 I + ka^k bA + ba^kA + b²ka^k-1(O)

= a^k+1 I + ka^k bA + ba^kA + 0

= a^k+1 I + ka^k bA + ba^kA

= a^k+1 I + (k+1)a^k bA

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

R.H.S = a^k+1 I + (k+1)a^k 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 + na^n-1bA, where I is the identity matrix of order 2 and n ϵ N.