Ex 4.6, 16

Transcript

Ex 4.6, 16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method Let the cost of onion, wheat & rice per kg be x, y, z respectively. Given, Cost of 4kg onion, 3 kg wheat & 2 kg rice is Rs 60. 4x + 3y + 2z = 60 Cost of 2kg onion, 4kg wheat & 6kg rice is Rs 90. 2x + 4y + 6z = 90 Cost of 6kg onion, 2kg wheat & 3 kg Rice is Rs 70 6x + 2y + 3z = 70 So the system of equation is 4x + 3y + 2z = 60 x + 2y + 3z = 45 6x + 2y + 3z = 70 Step 1 Write equation as AX = B 4﷮3﷮2﷮1﷮2﷮3﷮6﷮2﷮3﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 60﷮45﷮70﷯﷯ Hence A = 4﷮3﷮2﷮1﷮2﷮3﷮6﷮2﷮3﷯﷯, X = 𝑥﷮𝑦﷮𝑧﷯﷯ & B = 60﷮45﷮70﷯﷯ Step 2 Calculate |A| 𝐴﷯ = 4﷮3﷮2﷮1﷮2﷮3﷮6﷮2﷮3﷯﷯ = 4 (6 − 6) − 1(9 − 4) + 6 (9 − 4) = 4 (0) − 1 (5) + 6 (5) = − 5 + 30 = 25 So, |A|≠ 0 ∴ The system of equation is consistent & has a unique solutions Now, 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 = 1﷮−1﷮2﷮3﷮4﷮−5﷮2﷮−1﷮3﷯﷯ 𝐴11 = 2 × 3 − 2 × 3 = 6 − 6 = 0 𝐴12 = − 3×1−6×3﷯ = − 3−18﷯ = − (−15) = 15 𝐴22 = 4 × 3 − 6 × 2 = 12 − 12 = 0 𝐴23 = − 2×1−6×2﷯ = − 2−12﷯ = − (−10) = 10 𝐴31 = 3 × 3 − 2 × 2 = 9 − 4 = 5 𝐴32 = − 4×3−2×1﷯ = 12−2﷯=−10 𝐴33 = 4 × 2 − 3 × 1 = 8 − 3 = 5 Hence, adj (A) = 0﷮−5﷮5﷮15﷮0﷮−10﷮−10﷮10﷮5﷯﷯ Now, A−1 = 1﷮ 𝐴﷯﷯ adj (A) A−1 = 1﷮25﷯ 0﷮−5﷮5﷮15﷮0﷮−10﷮−10﷮10﷮5﷯﷯ = 1﷮5﷯ 0﷮−1﷮1﷮3﷮0﷮−2﷮−2﷮2﷮1﷯﷯ Now, X = A−1 B 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮5﷯ − 0﷮−1﷮1﷮3﷮0﷮−2﷮−2﷮2﷮1﷯﷯ 60﷮45﷮70﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮5﷯ −0−45+70﷮180+0−140﷮−120+90+70﷯﷯ = 1﷮5﷯ 25﷮40﷮40﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 5﷮8﷮8﷯﷯ ∴ x = 5, y = 8, z = 8 Cost of onion, wheat & rice per kg is Rs 5, Rs 8 & Rs 8 Respectively.