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.
Share with your friends
Call
Let the cost of 1 kg wheat and 1 kg rice be ₹x, ₹ y and ₹ z respectively
Given that , 4x+3y+2z=60
2x+4y+6z=90
6x+2y + 3z = 70
`rArr" "[{:(4,3,2),(2,4,6),(6,2,3):}][{:(x),(y),(z):}]=[{:(60),(90),(70):}]rArrAX=B……..(1)`
`"Here, A"=[{:(4,3,2),(2,4,6),(6,2,3):}]`
`rArr" "|A|=[{:(4,3,2),(2,4,6),(6,2,3):}]`
`= 4(12-12)-3(6-36)+2(4-24)`
`=0+90-4050ne0`
`therefore` A is invertible.
`"Now, "a_(11)=0, A_(12)=30, A_(13)=-20`
`A_(21)=-5, A_(22)=0, A_(23)=-20`
`A_(31)=10, A_(32)=-20, A_(33)=10`
`therefore" adj A"=[{:(0,30,-20),(-5,0,10),(10,-20,10):}]=[{:(0,-5,10),(30,0,-20),(-20,10,10):}]`
`"and A"^(-1)=1/|A|". adj A"=1/50[{:(0,-5,10),(30,0,-20),(-20,10,10):}]`
From equation (1),
`AX=B rArr X=A^(-1)B`
`[{:(x),(y),(z):}]=1/50[{:(0,-5,10),(30,0,-20),(-20,10,10):}][{:(60),(90),(70):}]`
`=1/50[{:(0-540+700),(1800+0-1400),(-1200+900+700):}]=[{:(5),(8),(8):}]`
`therefore" "x=5, y=8,z=8`
`therefore,` cost of 1 kg onion = Rs. 5
cost of 1 kg weat = Rs. 8
cost of 1 kg rice = Rs. 8
Talk Doctor Online in Bissoy App