Example 16 - Chapter 4 Class 12 Determinants

Last updated at Dec. 16, 2024 by Teachoo

Example 16 - Solve system of equations 2x + 5y = 1 3x+2y=7 - Examples

part 2 - Example 16 - Examples - Serial order wise - Chapter 4 Class 12 Determinants

part 3 - Example 16 - Examples - Serial order wise - Chapter 4 Class 12 Determinants

part 4 - Example 16 - Examples - Serial order wise - Chapter 4 Class 12 Determinants

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Example 16 Solve the system of equations 2x + 5y = 1 3x + 2y = 7 Step 1 Write equation as AX = B [■8(2&5@3&2)] [■8(𝑥@𝑦)] = [■8(1@7)] A = [■8(2&5@3&2)] , X = [■8(𝑥@𝑦)] , B = [■8(1@7)] Step 2 Calculate |A|, A = [■8(2&5@3&2)] |A| = |■8(2&5@3&2)| = 2 × 2 – 5 × 3 = 4 – 15 = – 11 Since |𝑨| ≠ 0, System is consistent and the system of equations has a unique solution AX = B X = A-1 B Step 3 Calculating X = A-1 B Here, A-1 = 1/(|A|) adj (A) adj A = [■8(2&5@3&2)] = [■8(2&−5@−3&2)] Now, A-1 = 1/(|A|) adj A Putting values = 𝟏/(−𝟏𝟏) [■8(𝟐&−𝟓@−𝟑&𝟐)] & B = [■8(1@7)] Now X = A-1 B [■8(𝑥@𝑦)] = (−1)/11 [■8(2&−5@−3&2)] [■8(1@7)] [■8(𝑥@𝑦)] = (−1)/11 [■8(2(1)+(⤶7−5)7@−3(1)+2(7))] [■8(𝑥@𝑦)] = (−1)/11 [■8(2−35@−3+14)] = (−1)/11 [■8(−33@11)] = [■8(−33 × (−1)/11@11 × (−1)/11)] [■8(𝑥@𝑦)] = [■8(3@−1)] Hence x = 3 & y = – 1

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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