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Two schools P and Q want to award their selected students on the values of discipline, politeness and punctuality. The school P wants to award Rs x each, Rs y each and Rs z each for the three respective values to its 3, 2 and 1 students with a total award money of Rs 1,000. - Mathematics

Two schools P and Q want to award their selected students on the values of discipline, politeness and punctuality. The school P wants to award Rs x each, Rs y each and Rs z each for the three respective values to its 3, 2 and 1 students with a total award money of Rs 1,000. School Q wants to spend Rs 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize on each value is Rs 600, using matrices, find the award money for each value.
Apart from the above three values, suggest one more value for awards.

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Solution

 The information given in the question can be written as,

3x + 2y + z = 1000                ... (1)
4x + y + 3z = 1500                ... (2)
x + y + z = 600                      ... (3)

Where,

`A=[[3,2,1],[4,1,3],[1,1,1]] and B=[[1000],[1500],[600]]`

Now ,

`|A|=3(-1)^(1+1)|[1,3],[1,2]|+2(-1)^(1+2)|[4,3],[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 we have:

`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`

`adj A =[[-2,-1,3],[-1,2,-1],[5,-5,-5]]^T=[[-2,-1,5],[-1,2,-5],[3,-1,-5]]`

`i.e., A^(−1)=(adj A)/(∣A∣)`

`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]][[1000],[1500],[600]]`

`[[x],[y],[z]]=-1/5[[-2000-1500+3000],[-1000+3000-3000],[3000-1500-3000]]`

`[[x],[y],[z]]=-1/5[[-500],[-1000],[-1500]]`

`[[x],[y],[z]]=[[100],[200],[300]]`

Hence, the money awarded for discipline, politeness and punctuality are Rs 100, Rs 200 and Rs 300, respectively.

Here, the determinant of the matrix A is non-zero. Therefore, x, y and z will have unique solutions: x = 100, y = 200 and z = 300.

  Is there an error in this question or solution?
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