





Ex 4.6
Ex 4.6, 2 Deleted for CBSE Board 2022 Exams
Ex 4.6, 3 Important Deleted for CBSE Board 2022 Exams
Ex 4.6, 4 Deleted for CBSE Board 2022 Exams
Ex 4.6, 5 Important Deleted for CBSE Board 2022 Exams
Ex 4.6, 6 Deleted for CBSE Board 2022 Exams
Ex 4.6, 7
Ex 4.6, 8
Ex 4.6, 9
Ex 4.6, 10 Important
Ex 4.6, 11
Ex 4.6, 12 You are here
Ex 4.6, 13 Important
Ex 4.6, 14 Important
Ex 4.6, 15 Important
Ex 4.6, 16 Important
Ex 4.6, 12 Solve system of linear equations, using matrix method. x y + z = 4 2x + y 3z = 0 x + y + z = 2 The system of equations is x y + z = 4 2x + y 3z = 0 x + y + z = 2 Step 1 Write equation as AX = B 1 1 1 2 1 3 1 1 1 = 4 0 2 Hence A = 1 1 1 2 1 3 1 1 1 , X = & B = 4 0 2 Step 2 Calculate |A| |A| = 1 1 1 2 1 3 1 1 1 = 1 1 3 1 1 ( 1) 2 3 1 1 + 1 2 1 1 1 = ( 1 + 3) + 1 ( 2 + 3) + 1 (2 1) = 1 (4) + 1 (5) + 1 (1) = 4 + 5 + 1 = 10 Since |A| 0 The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculate 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 = 1 1 1 2 1 3 1 1 1 M11 = 1 3 1 1 = 1 + 3 = 4 M12 = 2 3 1 1 = 2 + 3 = 5 M13 = 2 1 1 1 = 2 1= 1 M21 = 1 1 1 1 = 1 1 = 2 M22 = 1 1 1 1 = 1 1 = 0 M23 = 1 1 1 1 = 1 + 1 = 2 M31 = 1 1 1 3 = 3 1 = 2 M32 = 1 1 2 3 = 3 4 = 5 M33 = 1 1 2 1 = 3 + 2 = 3 A11 = ( 1) 1+1 M11= ( 1)2 . 4= 4 A12 = ( 1) 1+2 M12 = ( 1) 3 . 5 = 5 A13 = ( 1) 1+3 M13= ( 1) 4 . (1) = 1 A21 = 1 2+1 M21= ( 1) 3 . (-2) = 2 A22 = ( 1) 2+2 M22 = ( 1)4 . 0 = 0 A23 = ( 1) 2+3 . M23 = ( 1) 5 . ( 2) = 2 A31 = ( 1) 3+1 . M31 = ( 1) 4 . (2) = 2 A32 = ( 1) 3+2 . M32 = ( 1) 5 . ( 5) = 5 A33 = ( 1) 3+3 . M33 = ( 1)6 . 3 = 3 Thus , adj A = 4 2 2 5 0 5 1 2 3 & |A| = 10 So, A-1 = 1 |A| adj A A-1 = 1 10 4 2 2 5 0 5 1 2 3 & B = 4 0 2 Now, solving X = A-1 B = 1 10 4 2 2 5 0 5 1 1 3 4 0 2 = 1 10 4 4 +2 0 +2(2) 0 4 +0 0 +5(2) 2 4 +1 0 +3(2) = 1 10 16+0+4 20+0+10 4+0+6 = 1 10 20 10 10 = 2 1 1 Hence, x = 2 , y = 1, & z = 1