Ex 4.1

Chapter 4 Class 12 Determinants
Serial order 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