Question

A mixture is to be made of three foods A, B, C. The three foods A, B, C contain nutrients P, Q, R as shown below :

Food	Grams per kg of nutrient
	P	Q	R
A	1	2	5
B	3	1	1
C	4	2	1

How to form a mixture which will have 8 grams of P, 5 grams of Q and 7 grams of R?

Answer 1

Let food needed be x kg of A, y kg of B and z kg of C. Therefore x kg of A contains 1 gram of nutrient P. So, x kg of A will contain x grams of nutrient P. Similarly, the amount of nutrient P in y kg of food B and z kg of food C are 3y and 4z grams respectively. So, total quantity of nutrient P in x kg of food A, y kg of food B and z kg of food C is x + 3y + 4z grams.

x + 3y + 4z = 8

Similarly,

 2x + y + 2z = 5 [For Q]

and 5x + y + z = 7 [For R]

The above system of simultaneous linear equations can be written in matrix form as AX = B.

So, A^–1 exists and system have unique solution. 

Let C_ij be the cofactor of a_ij in A = [a_ij]. Then,

C₁₁ = –1; C₁₂ = 8; C₁₃ = –3 C₂₁ = 1 ; C₂₂ = –19; C₂₃ = 14 C₃₁ = 2 ; C₃₂ = 6; C₃₃ = –5

Putting value of X, A^-1 and B in X = A^-1B, we get

Thus, the mixture is formed by mixing 1 kg of each of the food A, B and C.