A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of ₹ 6,000. Three times the award money for Hard work added to that given for honesty amounts to ₹ 11,000. 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 money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Question

Philoid · Accepted 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.