Ex 4.6, 13 - Chapter 4 Class 12 Determinants

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Ex 4.6, 13 Solve system of linear equations, using matrix method. 2x + 3y + 3z = 5 x 2y + z = 4 3x y 2z = 3 The system of equations is 2x + 3y + 3z = 5 x 2y + z = 4 3x y 2z = 3 Step 1 Write equation as AX = B 2 3 3 1 2 1 3 1 2 = 5 4 3 Hence A = 2 3 3 1 2 1 3 1 2 = & B = 5 4 3 Step 2 Calculate |A| |A| = 2 3 3 1 2 1 3 1 2 = 2 2 1 1 2 3 1 1 3 2 + 2 1 2 3 1 = 2 ( 4 + 1) 3 ( 2 3) + 3 ( 2 + 6) = 2 (5) 3 ( 5) + 3 (5) = 10 + 15 + 5 = 40 Since |A| 0 The system of equations is consistent & has a unique solution AX = B X = A-1 B Step 3 Calculating X = A-1 B Calculating A-1 Now, A-1 = 1 |A| adj (A) adj A = A11 A12 A13 A21 A22 A23 A31 A32 A33 = A11 A21 A31 A12 A22 A32 A13 A23 A33 A = 2 3 3 1 2 1 3 1 2 M11 = 2 1 1 2 = 4 + 1 = 5 M12 = 1 1 3 2 = 2 3 = 5 M13 = 1 2 3 1 = 1 + 6 = 5 M21 = 3 3 1 2 = 6 + 3 = 3 M22 = 2 3 3 2 = 1 9 = 13 M23 = 2 3 3 1 = 2 9 = 11 M31 = 3 3 2 1 = 3 + 6 = 9 M32 = 2 3 1 1 = 2 3 = 1 M33 = 2 3 1 2 = 4 + 3 = 7 A11 = ( 1) 1+1 M11 = ( 1)2 . (5) = 5 A12 = ( 1) 1+2 M12 = ( 1) 3 . ( 5) = 5 A13 = ( 1) 1+3 M13 = ( 1) 4 . (5) = 5 A21 = ( 1) 2+1 M21 = ( 1) 3 . ( 3) = 3 A22 = ( 1) 2+2 M22 = ( 1)4 . ( 13) = 13 A23 = ( 1) 2+3 . M23 = ( 1) 5 . ( 11) = 11 A31 = ( 1) 3+1 . M31 = ( 1) 4 . (9) = 9 A32 = ( 1) 3+2 . M32 = ( 1) 5 . ( 1) = 1 A33 = ( 1) 3+3 . M33 = ( 1)6 . ( 7) = 7 Thus, adj A = 5 3 9 5 13 1 5 11 7 & |A| = 40 So, A-1 = 1 |A| adj A A-1 = 1 40 5 3 9 5 13 1 5 11 7 & B = 5 4 3 Now, solving X = A-1 B = 1 40 5 3 9 5 13 1 5 11 7 5 4 3 = 1 40 5 5 +3 4 +9(3) 5 5 +( 13) 4 +1(3) 5 5 +11 4 +( 7)(3) = 1 40 25 12+27 25+52+3 25 44 21 = 1 40 25 12+27 25+52+3 25 44 21 = 1 40 40 80 40 = 1 2 1 Hence x = 1 , y = 2, & z = 1 But X = So, = 1 2 1 Hence x = 1 , y = 2, & z = 1