Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P want to award ₹x each, ₹y each and ₹z each for the three respective values to 3,2 and 1 students respectively with total award money of ₹2,200. School Q wants to spend ₹3,100 to award its 4,1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for the award
x, y and z be the prize amount per student for Discipline, Politeness and Punctuality respectively.
3x + 2y + z = 2200
4x + y + 3z = 3100
x + y + z = 1200
These three equations can be written as
A X = B
|A| = 3(1 – 3) – 2(4 – 3) + 1(4 – 1)
= 3(– 2) – 2(1) + 1(3)
= – 6 – 2 + 3
= – 5
Hence, the unique solution given by x = A – 1B
C11 = (– 1)1 + 1 (1 – 3) = – 2
C12 = (– 1)1 + 2 (4 – 3) = – 1
C13 = (– 1)1 + 3 (4 – 1) = 3
C21 = (– 1)2 + 1 (2 – 1) = – 1
C22 = (– 1)2 + 2 (3 – 1) = 2
C23 = (– 1)2 + 3 (3 – 2 ) = – 1
C31 = (– 1)3 + 1 (6 – 1) = 5
C32 = (– 1)3 + 2 (9 – 4) = – 5
C33 = (– 1)3 + 3 (3 – 8) = – 5
Adj A = 
X = A – 1 B = 
X = 
X = 
X = 
= 
Hence, x = 300, y = 400 and z = 500
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