A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 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.
Asked by Topperlearning User | 3rd May, 2016, 05:12: PM
Let the award money given for Honesty, Regularity and Hard work be Rs x, Rs y and Rs z respectively.
Since total cash award is Rs 6,000.
x + y + z = 6,000 ...(1)
Three times the award money for Hard work and Honesty amounts to Rs 11,000.
x + 3z = 11,000
x + 0.y + 3 z = 11,000...(2)
Award money for Honesty and Hard work is double the one given for regularity.
x + z = 2y
x - 2y + z = 0...(3)
The above system of equations can be written in matrix form AX = B as:
Thus, A is non-singular. Hence, it is invertible.
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.
Answered by | 3rd May, 2016, 07:12: PM
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