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Q : 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.
 

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So, let us assume the cost of onion, wheat, and rice be x, y and z respectively.

Then we have the equations for the given situation :

4x+3y+2z = 60

2x+4y+6z = 90

6x+2y+3y = 70

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix A.

We have;

A = \begin{bmatrix} 4 &3 &2 \\ 2& 4 &6 \\ 6 & 2 & 3 \end{bmatrix}, X= \begin{bmatrix} x\\y \\ z \end{bmatrix}  and\ B = \begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}.

|A| = 4(12-12) -3(6-36)+2(4-24) = 0 +90-40 = 50 \neq 0

Now, we will find the cofactors of A;

A_{11} = (-1)^{1+1}(12-12) = 0                 A_{12} = (-1)^{1+2}(6-36) = 30

A_{13} = (-1)^{1+3}(4-24) = -20                 A_{21} = (-1)^{2+1}(9-4) = -5

A_{22} = (-1)^{2+2}(12-12) = 0                A_{23} = (-1)^{2+3}(8-18) = 10

A_{31} = (-1)^{3+1}(18-8) = 10               A_{32} = (-1)^{3+2}(24-4) = -20

A_{33} = (-1)^{3+3}(16-6) = 10

Now we have adjA;

adjA = \begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}

A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}s

So, the solutions can be found by X = A^{-1}B = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}

\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-450+700\\1800+0-1400 \\ -1200+900+700 \end{bmatrix} =\frac{1}{50} \begin{bmatrix} 250\\400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5\\8 \\ 8 \end{bmatrix}

Hence the solutions of the given system of equations;

x =5,\ y =8,\ and\ \ z=8.

Therefore, we have the cost of onions is Rs. 5 per Kg, the cost of wheat is Rs. 8 per Kg, and the cost of rice is Rs. 8 per kg.

 

 

Posted by

Divya Prakash Singh

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