# Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. School A wants to award Rs x each, Rs y each and Rs z each for the three respective values to 3, 2 and 1 students, respectively with a total award money of Rs 1,600. - Mathematics

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. School A wants to award Rs x each, Rs y each and Rs z each for the three respective values to 3, 2 and 1 students, respectively with a total award money of Rs 1,600. School B wants to spend Rs 2,300 to award 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is Rs 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for an award.

#### Solution

The information given in the question can be written as:

3x + 2y + z = 1600                ... (1)
4x + y + 3z = 2300                ... (2)
x + y + z = 900                      ... (3)

Here,

A=[[3,2,1],[4,1,3],[1,1,1]] and B=[[1600],[2300],[900]]

Now,

|A|=3(-1)^(1+1)|[1,3],[1,1]|+2(-1)^(1+2)|[4,3],[1,1]|+1(-1)^(1+3)|[4,1],[1,1]|

|A|=3|[1,3],[1,1]|-2|[4,3],[1,1]|+1|[4,1],[1,1]|

⇒|A|=3(1−3)−2(4−3)+1(4−1)=−6−2+3=−5≠0

So, A is invertible.
Let Cij be the cofactor of aij in A=[aij].
Then,

C_11=(−1)^(1+1)|[1,3],[1,1]|=-2

C_12=(−1)^(1+2)|[4,3],[1,1]|=-1

C_13=(−1)^(1+3)|[4,1],[1,1]|=3

C_21=(−1)^(2+1)|[2,1],[1,1]|=-1

C_22=(−1)^(2+2)|[3,1],[1,1]|=2

C_23=(−1)^(2+3)|[3,2],[1,1]|=-1

C_31=(−1)^(3+1)|[2,1],[1,3]|=5

C_32=(−1)^(3+2)|[3,1],[4,3]|=-5

C_33=(−1)^(3+3)|[3,2],[4,1]|=-5

cofactor of A=[[C_11,C_12,C_13],[C_21,C_22,C_23],[C_31,C_32,C_33]]

cofactor of A=[[-2,-1,3],[-1,2,-1],[5,-5,-5]]

therefore adjA=[[-2,-1,3],[-1,2,-1],[5,-5,-5]]^T=[[-2,-1,5],[-1,2,-5],[3,-1,-5]]

i.e A^(-1)=

A^(-1)=-1/5[[-2,-1,5],[-1,2,-5],[3,-1,-5]]

Thus, the solution of the system of equations is given by

X=A^(-1)B=-1/5[[-2,-1,5],[-1,2,-5],[3,-1,-5]]

=>[[x],[y],[z]]=-1/5[[-3200-2300+4500],[-1600+4600-4500],[4800-2300-4500]]

=>[[x],[y],[z]]=-1/5[[-1000],[-1500],[-2000]]

=>[[x],[y],[z]]=[[200],[300],[400]]

Hence, the money awarded for sincerity, truthfulness and helpfulness are Rs 200, Rs 300 and Rs 400, respectively.

Here, the determinant of the matrix A is non-zero. Therefore, x, y and z will have unique solutions: x = 200, y = 300 and z = 400.

Is there an error in this question or solution?
2013-2014 (March) All India Set 1

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