Evalute determinant of a 3x3 matrix

Chapter 4 Class 12 Determinants
Concept wise

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### Transcript

Ex 4.1, 4 If A = [■8(1&0&[email protected]&1&[email protected]&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&[email protected]&1&[email protected]&0&4)] = [■8(3 × 1&3 × 0&3 × [email protected] × 0&3 × 1&3 × [email protected] × 0&3 × 0&3 × 4)] = [■8(3&0&[email protected]&3&[email protected]&0&12)] And |3A| = |■8(3&0&[email protected]&3&[email protected]&0&12)| = 3 |■8(3&[email protected]&12)| – 0 |■8(0&[email protected]&12)| + 3 |■8(0&[email protected]&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&[email protected]&1&[email protected]&0&4)] = 1 |■8(1&[email protected]&4)| – 0 |■8(0&[email protected]&4)| + 1 |■8(0&[email protected]&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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.