Check sibling questions

Ex 4.1, 4 - Show that  |3A|  = 27|A|, if A =  [1 0 1 - Ex 4.1

Ex 4.1, 4 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.1, 4 - Chapter 4 Class 12 Determinants - Part 3

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Transcript

Ex 4.1, 4 If A = [■8(1&0&1@0&1&2@0&0&4)] , then show that |3A| = 27 |A| We have to prove |3A| = 27 |A| Taking L.H.S |3A| First Calculating 3A 3A = 3 [■8(1&0&1@0&1&2@0&0&4)] = [■8(3 × 1&3 × 0&3 × 1@3 × 0&3 × 1&3 × 2@3 × 0&3 × 0&3 × 4)] = [■8(3&0&3@0&3&6@0&0&12)] And |3A| = |■8(3&0&3@0&3&6@0&0&12)| = 3 |■8(3&6@0&12)| – 0 |■8(0&6@0&12)| + 3 |■8(0&3@0&0)| = 3(3(12)– 0(6)) – 0 (0(12) – 0(6)) +3 (0(0) – 0(3) = 3(36 – 0) – 0(0) + 3(0) = 3(36) + 0 + 0 = 108 Taking R.H.S 27|A| |A| = [■8(1&0&1@0&1&2@0&0&4)] = 1 |■8(1&2@0&4)| – 0 |■8(0&2@0&4)| + 1 |■8(0&1@0&0)| = 1(1(4) – 0(2)) – 0 (0(4) – 0(2)) + 1(0 – 0(1)) = 1 (4 – 0) – 0 (0) + 1(0) = 4 Now, 27|A| = 27 (4) = 108 = |3A| Hence L.H.S = R.H.S Hence proved

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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.