Q. 234.5( 4 Votes )

A school wants to

Answer :

Let the award money given for Honesty = Rs x

The award money given for Regularity = Rs y


and the award money given for Hard work = Rs z


Since, total cash award is Rs 6,000.


x + y + z = 6,000 …(i)


Three times the award money for Hard work and Honesty amounts to Rs 11,000.


x + 3z = 11,000 …(ii)


Award money for Honesty and Hard work is double the one given for regularity.


x + z = 2y


x – 2y + z = 0 …(iii)


The above system of equations can be written in matrix form


AX = B



Here,



Now, we find the determinant |A|



|A| = (1)[0 – (-2)(3)] – (1)[1 – (-2)(1)] + (1)[3 – 0]


= 6 – [1 + 2] + 3


= 9 – 3


= 6 ≠ 0


Thus, A is a non – singular. Hence, it is invertible


We have to find A-1 and


Now, we have to find adj A and for that we have to find co-factors:













X = A-1B





Hence, x = 500, y = 2000, and z = 3500.


Thus, award money given for Honesty, Regularity and Hard work are Rs 500, Rs 2000 and Rs 3500 respectively.


School can include awards for obedience.


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