Q. 355.0( 2 Votes )

Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping calm in tense situations, at the rate of ₹x, ₹y and ₹z per persons respectively. The first factory decided to honour respectively 2, 4 and 3 employees with total prize money of ₹29000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹30500. If the three prizes per person together cost ₹9500, then
i. represent the above situation by a matrix equation and form linear equations using matrix multiplication.

ii. Solve these equations using matrices.

iii. Which values are reflected in the questions? (CBSE 2013)

Answer :

Let the numbers are x, y,z be the prize amount per person for adaptability, carefulness and calmness respectively


As per the given data we get,


2x + 4y + 3z = 29000


5x + 2y + 3z = 30500


X + y + z = 9500


These three equations can be written as



A X = B


|A| = 2(2 – 3) – 4(5 – 3) + 3(5 – 2)


= 2(– 1) – 4(2) + 3(3)


= – 2 – 8 + 9


= – 1


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


C11 = (– 1)1 + 1 (2 – 3) = – 1


C12 = (– 1)1 + 2 (5 – 3) = – 2


C13 = (– 1)1 + 3 (5 – 2) = 3


C21 = (– 1)2 + 1 (4 – 3) = – 1


C22 = (– 1)2 + 2 (2 – 3) = – 1


C23 = (– 1)2 + 3 (2 – 4 ) = – 2


C31 = (– 1)3 + 1 (12 – 6) = 6


C32 = (– 1)3 + 2 (6 – 15) = – 9


C33 = (– 1)3 + 3 (4 – 20) = – 16


Adj A =


X = A – 1 B =


X =


X =


X =


=


Hence, x = 2500, y = 3000 and z = 4000

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