Q. 325.0( 1 Vote )

# The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping 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. A part from these values, namely, honesty cooperation and supervision, suggest one more value which the management must include for awards. (CBSE 2013)

Answer :

Let the numbers are x, y, z

3x + 5y – 4 z = 6000 …… (i)

Also,

2x – 3y + z = 5000 ……(ii)

Again,

– x + 4y + 6z = 13000 …… (iii)

A X = B

|A| = 1(3 + 6) – 1(2 – 3) + 1(– 4 – 3)

= 1(9) – 1(– 1) – 7

= 9 + 1 – 7

= 3

Hence, the unique solution given by x = A – 1B

C11 = (– 1)1 + 1 (3 + 6) = 9

C12 = (– 1)1 + 2 (2 – 3) = 1

C13 = (– 1)1 + 3 (– 4 – 3) = – 7

C21 = (– 1)2 + 1 (1 + 2) = – 3

C22 = (– 1)2 + 2 (1 – 1) = 0

C23 = (– 1)2 + 3 (– 2 – 1 ) = 3

C31 = (– 1)3 + 1 (3 – 3) = 0

C32 = (– 1)3 + 2 (3 – 2) = – 1

C33 = (– 1)3 + 3 (3 – 2) = 1

Adj A =

X = A – 1 B =

X =

X =

X =

=

Hence, x = 3, y = 4 and z = 5

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