Ex 4.6, 7 - Chapter 4 Class 12 Determinants (Term 1)

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Ex 4.6, 7 Solve system of linear equations, using matrix method. 5x+ 2y = 4 7x + 3y = 5 The system of equations is 5x + 2y = 4 7x + 3y = 5 Writing equation as AX = B [■8(5&[email protected]&3)] [■8(𝑥@𝑦)] = [■8([email protected])] Hence A = [■8(5&[email protected]&3)], X = [■8(𝑥@𝑦)] & B = [■8([email protected])] Calculating |A| |A| = |■8(5&[email protected]&3)| = 5(3) – 7(2) = 15 – 14 = 1 ≠ 0 Since |A|≠ 0, System of equations is consistent & has a unique solution Now, AX = B Solution is X = A-1 B Calculating A−1 A-1 = 1/(|A|) adj (A) A = [■8(5&[email protected]&3)] adj A = [■8(5&[email protected]&3)] = [■8(3&−[email protected]−7&5)] Now, A-1 = 1/(|A|) adj A Putting values = 1/1 [■8(3&−[email protected]−7&5)] = [■8(3&−[email protected]−7&5)] Now X = A-1 B [■8(𝑥@𝑦)] = [■8(3&−[email protected]−7&5)] [■8([email protected])] [■8(𝑥@𝑦)] = [■8(3(4)+(⤶7−2)[email protected]−7(4)+5(5))] [■8(𝑥@𝑦)] = [■8([email protected]−3)] Hence, x = 2 & y = –3