Q. 234.0( 8 Votes )

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

Answer :

To prove: (aI + bA)^{n} = a^{n}I + na^{n-1}bA

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} = a^{n}I + na^{n-1}bA, n ϵ N

For n=1,

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

R.H.S: a^{1}I + 1a^{1-1}bA= aI + a^{0}bA= 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^{k}I + ka^{k-1}bA …… (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^{k}I + ka^{k-1}bA). (aI + bA) ……from (1)

= aI(a^{k}I + ka^{k-1}bA) + bA(a^{k}I + ka^{k-1}bA)

= aI(a^{k}I) + aI(ka^{k-1}bA) + bA(a^{k}I) + bA(ka^{k-1}bA)

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

= a^{k+1} I^{2} + ka^{1+k-1} bA + ba^{k}A + b^{2}ka^{k-1}A^{2} (IA = AI= A & I^{2} = I)

= a^{k+1} I + ka^{k} bA + ba^{k}A + b^{2}ka^{k-1}A^{2}

Calculating A^{2}

A^{2} = A.A

∴ A^{2} = O (O is the null matrix)

Putting value of A^{2} in L.H.S

L.H.S = a^{k+1} I + ka^{k} bA + ba^{k}A + b^{2}ka^{k-1}(O)

= a^{k+1} I + ka^{k} bA + ba^{k}A + 0

= a^{k+1} I + ka^{k} bA + ba^{k}A

= 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} = a^{n}I + na^{n-1}bA**

**, where I is the identity matrix of order 2 and n**

**ϵ**

**N.**

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