Misc. 16 - Chapter 4 Class 12 Determinants

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Misc. 16 Solve the system of the following equations 2﷮x﷯ + 3﷮y﷯ + 10﷮z﷯ = 4 4﷮x﷯ + 6﷮y﷯ + 5﷮z﷯ = 1 6﷮x﷯ + 9﷮y﷯ + 20﷮z﷯ = 2 The system of equations are 2﷮x﷯ + 3﷮y﷯ + 10﷮z﷯ = 4 4﷮x﷯ + 6﷮y﷯ + 5﷮z﷯ = 1 6﷮x﷯ + 9﷮y﷯ + 20﷮z﷯ = 2 Now let 𝟏﷮𝒙﷯ = u , 𝟏﷮𝒚﷯ = v , & 𝟏﷮𝒛﷯ = w The system of equations become 2u + 3v + 10w = 4 4u – 6v + 5w = 1 6u + 9v – 20w = 2 Step 1 Write equation as AX = B 2﷮3﷮10﷮4﷮−6﷮5﷮6﷮9﷮−20﷯﷯ 𝑢﷮𝑣﷮𝑤﷯﷯ = 4﷮1﷮2﷯﷯ Hence A = 2﷮3﷮10﷮4﷮−6﷮5﷮6﷮9﷮−20﷯﷯ , X = 𝑢﷮𝑣﷮𝑤﷯﷯ & B = 4﷮1﷮2﷯﷯ Step 2 Calculate |A| |A| = 2﷮3﷮10﷮4﷮−6﷮5﷮6﷮9﷮−20﷯﷯ = 2 −6﷮5﷮9﷮−20﷯﷯ – 3 4﷮5﷮6﷮−20﷯﷯ + 10 4﷮−6﷮6﷮9﷯﷯ = 2 ( 120 – 45) – 3 ( – 80 – 30) + 10 ( 36 + 36) = 2 (75) – 3 (– 110) + 10 (72) = 150 + 330 + 720 = 1200 ∴ |A|≠ 0 So, the system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculating X = A-1 B 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﷮10﷮4﷮−6﷮5﷮6﷮9﷮−20﷯﷯ M11 = −6﷮5﷮9﷮−20﷯﷯ = 120 – 45 = 75 M12 = 4﷮5﷮6﷮−20﷯﷯ = ( – 80 – 30) = – 110 M13 = 4﷮−6﷮6﷮9﷯﷯ = 36 – 36 = 72 M21 = 3﷮10﷮9﷮−20﷯﷯ = − 60 – 90 = – 150 M22 = 2﷮10﷮6﷮−20﷯﷯ = – 40 – 60 = – 100 M23 = 2﷮3﷮6﷮9﷯﷯ = 18 – 18 = 0 M31 = 3﷮10﷮−6﷮5﷯﷯ = 15 + 60 = 75 M32 = 2﷮10﷮4﷮5﷯﷯ = 10 – 40 = – 30 M33 = 2﷮3﷮4﷮−6﷯﷯ = – 12 – 12 = – 24 A11 = ( –1)﷮1+1﷯ M11 = ( –1)2 . 75 = 75 A12 = ( –1)﷮1+2﷯ M12 = ( –1)﷮3﷯ . ( – 110) = 110 A13 = ( −1)﷮1+3﷯ M13 = ( −1)﷮4﷯ . (72) = 72 A21 = ( −1)﷮2+1﷯ M21 = ( −1)﷮3﷯ . ( – 150) = 150 A22 = ( −1)﷮2+2﷯ M22 = ( –1)4 . ( – 100) = – 100 A23 = (−1)﷮2+3﷯. M23 = (−1)﷮5﷯. 0 = 0 A31 = (−1)﷮3+1﷯. M31 = (−1)﷮4﷯ . 75 = 75 A32 = (−1)﷮3+2﷯ . M32 = (−1)﷮5﷯. ( – 30) = 30 A33 = (−1)﷮3+3﷯ . M33 = ( –1)6 . – 24 = – 24 Thus adj A = 75﷮150﷮75﷮110﷮−110﷮30﷮72﷮0﷮−24﷯﷯ Now, A-1 = 1﷮|A|﷯ adj A A-1 = 1﷮1200﷯ 75﷮150﷮75﷮110﷮−110﷮30﷮72﷮0﷮−24﷯﷯ X = A-1 B Putting Values 𝑢﷮𝑣﷮𝑤﷯﷯= 1﷮1200﷯ 75﷮150﷮75﷮110﷮−110﷮30﷮72﷮0﷮−24﷯﷯ 4﷮1﷮2﷯﷯ 𝑢﷮𝑣﷮𝑤﷯﷯= 1﷮1200﷯ 75(4)﷮+150(1)﷮+75(4)﷮110(4)﷮+(−110)(1)﷮+30(1)﷮72(4)﷮+0(1)﷮+ −24﷯2﷯﷯ 𝑢﷮𝑣﷮𝑤﷯﷯ = 1﷮1200﷯ 300+150+150﷮440−100+60﷮288+0−48﷯﷯ = 1﷮1200﷯ 600﷮400﷮140﷯﷯ ∴ 𝑢﷮𝑣﷮𝑤﷯﷯ = 1﷮2﷯﷮ 1﷮3﷯﷮ 1﷮5﷯﷯﷯ Hence u = 1﷮2﷯ , v = 1﷮3﷯ , & w = 1﷮5﷯ Hence x = 2, y = 3 & z = 5