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Misc 1 Let A = [■8(0&1@0&0)] , show that (aI + bA)n = anI + nan-1 bA , where I is the identity matrix of order 2 and n ∈ N We shall prove the result by using Mathematical induction Step 1: Given A = [■8(0&1@0&0)] Let I be the identity matrix of 2 × 2 i.e. I = [■8(1&0@0&1)] Let P(n) : (aI + bA)n = anI + nan-1 bA, n ∈ N Step 2: Prove for n = 1 For n = 1 L.H.S = (aI + bA)1 = aI + bA R.H.S = a1I + 1a1–1 bA = aI + a0 bA = aI + 1 . bA = aI + bA So, 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) : (aI + bA)k = akI + kak-1 bA , k ∈ N We will prove that P (k + 1) is true Putting k = k + 1. P (k + 1): (aI + bA)k+1 = ak+1 I + (k+1)ak + 1 – 1 bA. P (k + 1): (aI + bA)k+1 = ak+1 I + (k+1)ak bA Solving L.H.S (aI + bA)k+1 = (aI + bA)k . (aI + bA)1 = (aI + bA)k . (aI + bA) From (1) : (aI + bA)k = akI + kak − 1 bA = (akI + kak – 1 bA) (aI + bA) = (aI + bA) (ak I + kak – 1 bA) = aI (ak I + kak – 1 bA) + bA (ak I + kak – 1 bA) = aI (akI) + aI (kak – 1 bA) + bA (akI) + bA (kak – 1 bA) = (aka) (I × I) + kb (aak – 1) IA + bak (AI) + (bb) k ak – 1 (AA) = ak + 1 I2 + k ak – 1 + 1 b A + bak A + (b)2 kak – 1 (A2) = ak + 1 I + k ak b A + bak A + b2 k ak – 1 (A2) Finding value of A2 A2 = A . A = [■8(0&1@0&0)] [■8(0&1@0&0)] = [■8(0(0)+1(0)&0(1)+1(0)@0(0)+0(0)&0(1)+0(0))] = [■8(𝟎&𝟎@𝟎&𝟎)] ∴ A2 = O Putting A2 = O in L.H.S ak + 1 I + k ak b A + bak A + b2 k ak – 1 (A2) = ak + 1 I + k ak b A + bak A + b2 k ak – 1 (O) = ak + 1 I + k ak b A + bak A + O = ak + 1I + bAak (k + 1) = R.H.S Hence P(k + 1) is true Hence, P(k + 1) is true ∴ By the principle of mathematical induction P(n) is true for all n when n is natural number. Thus, (aI + bA)n = anI + nan-1 bA when I = [■8(1&0@0&1)], A = [■8(0&1@0&0)] For all n ∈ N

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.