Example 16 - 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