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^{-1}B

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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