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.

#### Solution

Let the award money given for Honesty, Regularity and Hard work be *x*, *y* and *z* respectively.

Since total cash award is Rs 6,000.

∴ *x *+ *y* + *z* = Rs 6,000 ...(1)

Three times the award money for Hard work and Honesty is Rs 11,000.

∴ *x *+ 3 *z* = Rs 11,000

⇒ *x *+ 0.*y* + 3 *z* = Rs 11,000 ...(2)

Award money for Honesty and Hard work is double the one given for regularity.

∴ *x *+ *z* = 2*y*

⇒ *x *− 2*y* + *z* = 0 ...(3)

The above system can be written in matrix form as,

`[[1,1,1],[1,0,3],[1,-2,1]][[x],[y],[z]]=[[6000],[11000],[0]]`

Or *AX* = *B*, where

`A=[[1,1,1],[1,0,3],[1,-2,1]], X=[[x],[y],[z]] and B=[[6000],[11000],[0]]`

`|A|=6!=0`

Thus, *A* is non-singular. Hence, it is invertible.

`Adj A=[[6,-3,3],[2,0,-2],[-2,3,1]]`

`A^(-1)=1/|A|(adj A)=1/6[[6,-3,3],[2,0,-2],[-2,3,1]]`

`X=A^(-1)B=1/6[[6,-3,3],[2,0,-2],[-2,3,1]][[6000],[11000],[0]]`

`[[x],[y],[z]]=[[500],[2000],[3500]]`

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 sincerity for awards.