# Ex 4.6, 16

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

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 432123623 𝑥𝑦𝑧 = 604570 Hence A = 432123623, X = 𝑥𝑦𝑧 & B = 604570 Step 2 Calculate |A| 𝐴 = 432123623 = 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 = A11A12A13A21A22A23A31A32A33′ = A11A21A31A12A22A32A13A23A33 A = 1−1234−52−13 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) = 0−55150−10−10105 Now, A−1 = 1 𝐴 adj (A) A−1 = 125 0−55150−10−10105 = 15 0−1130−2−221 Now, X = A−1 B 𝑥𝑦𝑧 = 15 − 0−1130−2−221 604570 𝑥𝑦𝑧 = 15 −0−45+70180+0−140−120+90+70 = 15 254040 𝑥𝑦𝑧 = 588 ∴ x = 5, y = 8, z = 8 Cost of onion, wheat & rice per kg is Rs 5, Rs 8 & Rs 8 Respectively.

Chapter 4 Class 12 Determinants

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .