Q29 of 45 Page 1

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

As x members are awarded for honesty, y for helping others and z for supervising the workers.


According to question:


x + y + z = 12 …(1)


3(y + z) + 2x = 33 or 2x + 3y + 3z = 33 …(2)


And x + z = 2y or x – 2y + z = 0 …(3)


The above equations can be represented in matrix form as given below –



⇒ …(4)


Let,


As A-1 =


∴ |A| =


Expanding about first row-


|A| = 1(3+6)-1(2-3)+1(-4-3) = 13-7 = 3


As |A| = 3 ≠ 0 , so solution is possible and unique.


Adj(A) can be determined by finding the co-factor matrix of A and taking its transpose.


Adj(A) =


∴ A-1 =


From equation 4 we have -



By matrix multiplication we get –



⇒ ⇒


∴ 3 members are awarded for honesty, 4 for helping others and 5 for supervising the workers.


Apart from these, Society can include cleanliness for awarding the members.


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