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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PREVIOUSFind A – 1, if . Hence, solve the following system of linear equations:
x + 2y + 5z = 10, x – y – z = – 2,
2x + 3y – z = – 11 (CBSE 2012)NEXTA school wants to award its student for the values of Honesty, Regularity and Hard work with a total cash award of ₹6000. Three times the award money for Hard work added to that given for honesty amounts to ₹11000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award for each value, using the matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, and suggest one more value which the school must include for awards.(CBSE 2013)
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