Question

Let , show that , where I is the identity matrix of order 2 and n ∈ N

Answer 1

Use the mathematical induction to show the given expression,

Step 1:

Consider I is the identity matrix of the order 2×2 ,

I=[ 1 0 0 1 ]

Let,

P( n ): ( aI + bA ) n = a n I + na n−1 bA, n∈N

Step 2:

For n=1 , the value of the left hand side (L.H.S.) is,

LHS= ( aI + bA ) 1 =aI + bA

And the value of the right hand side (R.H.S.) is,

RHS= a 1 I + 1a 1−1 bA = aI + a 0 bA =aI+bA

Here,

LHS=RHS

Hence, P( n ) is true for n=1 .

Step 3:

Consider the value of P( k ) is true so that,

P( k ): ( aI+bA ) k = a k I + ka k−1 bA, k∈N (1)

Step 4:

Proving that P(k + 1) is true.

Replace k with ( k+1 ) in the equation (1),

( aI + bA ) k+1 = a k+1 I+( k+1 ) a ( k+1 )−1 bA, k+1∈N ( aI + bA ) k+1 ︸ LHS = a k+1 I+( k+1 ) a k bA ︸ RHS

Consider left hand side of the above expression,

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

From equation (1),

( aI + bA ) k+1 =( a k I + ka k−1 bA )( aI + bA ) =( aI + bA )( a k I + ka k−1 bA ) =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 )

Further, simplify the above expression

( aI + bA ) k+1 = a k+1 I+ ka k−1+1 bA+ ba k A+( b 2 ) ka k−1 ( A 2 )      ( IA=A ) = a k+1 I+ ka k bA+ ba k A+ b 2 a k-1 ( A 2 ) (2)

Here, the value of the matrix A is given as,

A 2 =[ 0 1 0 0 ][ 0 1 0 0 ] =[ 0 0 0 0 ] =O

Substitute the value of A 2 in the equation (2),

a k+1 I+ ka k bA+ ba k A+ b 2 a k-1 ( O )= a k+1 I+ ka k bA+ ba k A = a k+1 I+ ba k A( k+1 ) =RHS

Hence, P( k+1 ) is true.

Therefore, by the mathematical induction this will prove that P( n ) is true for all n∈N .