   1. Chapter 3 Class 12 Matrices
2. Concept wise
3. Proof using mathematical induction

Transcript

Misc. 3 If A = [ 8(3& 4@1& 1)] , then prove An = [ 8(1+2n& 4n@n&1 2n)] where n is any positive integer We shall prove the result by using mathematical induction. Step 1: P(n): If A= [ 8(3& 4@1& 1)] , then An = [ 8(1+2n& 4n@n&1 2n)] , n N Step 2: Prove for n = 1 For n = 1 L.H.S = A1 = A = [ 8(3& 4@1& 1)] R.H.S = [ 8(1+2(1)& 4(1)@1&1 2(1))]=[ 8(1+2& 4@1&1 2)]" = " [ 8(3& 4@1& 1)] L.H.S = R.H.S P(n) is true for n = 1 Step 3: Assume P(k) to be true and then prove P(k+1) is true Assume that P (k) is true P(k) : If A= [ 8(3& 4@1& 1)] , then Ak = [ 8(1+2k& 4k@k&1 2k)] We will have to prove that P( k +1) is true P(k+1) : If A= [ 8(3& 4@1& 1)] , then Ak+1 = [ 8(1+2(k+1)& 4(k+1)@(k+1)&1 2(k+1))] Taking L.H.S Ak+1 = Ak . A = [ 8(1+2k& 4k@k&1 2k)] [ 8(3& 4@1& 1)] = [ 8((1+2k)3 4k(1)&(1+2k)( 4) 4k( 1)@k(3)+(1 2k)1&k( 4)+(1 2k)( 1))] = [ 8(3+6k 4k& 4 8k+4k@3k+1 2k& 4k 1+2k)] = [ 8(3+2k& 4 4k@1+k& 1 2k)] = [ 8(1+2(k+1)& 4(k+1)@1+k&1 2(k+1))] = R.H.S Thus P (k+1) is true By the principal of mathematical induction , P(n) is true for n N Hence, if A= [ 8(3& 4@1& 1)] , then An = [ 8(1+2n& 4n@n&1 2n)] , n N

Proof using mathematical induction 