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Chapter 4 Class 12 Determinants

Serial order wise

Last updated at Jan. 23, 2020 by Teachoo

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 methodLet the cost of onion, wheat & rice per kg be x, y, z respectively. Given, Cost of 4 kg onion, 3 kg wheat & 2 kg rice is Rs 60. 4x + 3y + 2z = 60 Cost of 2 kg onion, 4 kg wheat & 6 kg rice is Rs 90. 2x + 4y + 6z = 90 Cost of 6 kg onion, 2 kg 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 Writing equation as AX = B [β 8(4&3&2@1&2&3@6&2&3)][β 8(π₯@π¦@π§)] = [β 8(60@45@70)] Hence A = [β 8(4&3&2@1&2&3@6&2&3)], X = [β 8(π₯@π¦@π§)] & B = [β 8(60@45@70)] Calculating |A| |π΄| = [β 8(4&3&2@1&2&3@6&2&3)] = 4 (6 β 6) β 1(9 β 4) + 6 (9 β 4) = 4 (0) β1 (5) + 6 (5) = β5 + 30 = 25 So, |A|β 0 β΄ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj A = [β 8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^β² = [β 8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [β 8(4&3&2@1&2&3@6&2&3)] Now, π΄11 = 2 Γ 3 β2 Γ 3 = 6 β 6 = 0 π΄12 = β[3 Γ 1β6 Γ 3] = β[3β18] = β(β15) = 15 π΄13 = 1 Γ 2 β 6 Γ 2 = 2 β 12 = β10 π΄21 = β[3 Γ 3β2 Γ 2 ] = β [9β4] = β5 π΄22 = 4 Γ 3 β 6 Γ 2 = 12 β 12 = 0 π΄23 = β [2 Γ 1β6 Γ 2] = β[2β12] = β (β10) = 10 π΄31 = 3 Γ 3 β 2 Γ 2 = 9 β 4 = 5 π΄32 = β[4 Γ 3β2 Γ 1] = [12β2]=β10 π΄33 = 4 Γ 2 β 3 Γ 1 = 8 β 3 = 5 Hence, adj (A) = [β 8(0&β5&5@15&0&β10@β10&10&5)] Now, Aβ1 = 1/|π΄| adj (A) Aβ1 = 1/25 [β 8(0&β5&5@15&0&β10@β10&10&5)] Aβ1 = 1/5 [β 8(0&β1&1@3&0&β2@β2&2&1)] Now, X = Aβ1 B [β 8(π₯@π¦@π§)] = 1/5 [ββ 8(0&β1&1@3&0&β2@β2&2&1)] [β 8(60@45@70)] [β 8(π₯@π¦@π§)] = 1/5 [β(β0β45+70@180+0β140@β120+90+70)] = 1/5 [β 8(25@40@40)] [β 8(π₯@π¦@π§)] = [β 8(5@8@8)] β΄ x = 5, y = 8, z = 8 Cost of onion, wheat & rice per kg is Rs 5, Rs 8 & Rs 8 Respectively.