ย  Ex 4.5, 10 - Solve using matrix method 5x+2y=3 3x+2y=5 - Ex 4.5 - Ex 4.5

part 2 - Ex 4.5, 10 - Ex 4.5 - Serial order wise - Chapter 4 Class 12 Determinants
part 3 - Ex 4.5, 10 - Ex 4.5 - Serial order wise - Chapter 4 Class 12 Determinants part 4 - Ex 4.5, 10 - Ex 4.5 - Serial order wise - Chapter 4 Class 12 Determinants

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Ex 4.5, 10 Solve system of linear equations, using matrix method. 5x + 2y = 3 3x + 2y = 5 The system of equation is 5x + 2y = 3 3x + 2y = 5 Writing above equation as AX = B [โ– 8(5&2@3&2)] [โ– 8(๐‘ฅ@๐‘ฆ)] = [โ– 8(3@5)] Hence A = [โ– 8(5&2@3&2)], X = [โ– 8(๐‘ฅ@๐‘ฆ)] & B = [โ– 8(3@5)] Calculating |A| |A| = |โ– 8(5&2@3&2)| = 5(2) โ€“3(2) = 10 โ€“ 6 = 4 Since |A| โ‰  0 The System of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) A =[โ– 8(5&2@3&2)] adj A = [โ– 8(2&โˆ’2@โˆ’3&5)] Now, A-1 = 1/(|A|) adj A A-1 = 1/4 [โ– 8(2&โˆ’2@โˆ’3&5)] Thus, X = A-1 B [โ– 8(๐‘ฅ@๐‘ฆ)] = 1/4 [โ– 8(2&โˆ’2@โˆ’3&5)] [โ– 8(3@5)] = 1/4 [โ– 8(2(3)+(โคถ7โˆ’2)5@โˆ’3(3)+5 (5))] [โ– 8(๐‘ฅ@๐‘ฆ)] = 1/4 [โ– 8(6โˆ’10@โˆ’9+25)] = 1/4 [โ– 8(โˆ’4@16)] [โ– 8(๐‘ฅ@๐‘ฆ)] = [โ– 8(โˆ’1@4)] Hence, x = โ€“1 & y = 4

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