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
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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