# A school wants to award its student for the values of Honesty, Regularity and Hard work with a total cash award of ₹6000. Three times the award money for Hard work added to that given for honesty amounts to ₹11000. 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 for each value, using the matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, and suggest one more value which the school must include for awards.(CBSE 2013)

Let the numbers are x, y, z be the cash awards for Honesty, Regularity and Hard Work

x + y + z = 6000 …… (i)

Also,

x + 3z = 11000 …… (ii)

Again,

x – 2y + z = 0 …… (iii)

A X = B

|A| = 1(0 + 6) – 1(1 – 3) + 1(– 2 – 0)

= 1(6) – 1(– 2) – 2

= 6 + 2 – 2

= 6

Hence, the unique solution given by x = A – 1B

C11 = (– 1)1 + 1 (0 + 6) = 6

C12 = (– 1)1 + 2 (1 – 3) = 2

C13 = (– 1)1 + 3 (– 2 – 0) = – 2

C21 = (– 1)2 + 1 (1 + 2) = – 3

C22 = (– 1)2 + 2 (1 – 1) = 0

C23 = (– 1)2 + 3 (– 2 – 1 ) = 3

C31 = (– 1)3 + 1 (3 – 0) = 3

C32 = (– 1)3 + 2 (3 – 1) = – 2

C33 = (– 1)3 + 3 (0 – 1) = – 1

X = A – 1 B =

X =

X =

=

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

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