Q. 235.0( 1 Vote )

Two schools A and

Answer :

There are 3 values sincerity, truthfulness and helpfulness


Rs x for sincerity, Rs y for truthfulness and Rs z for helpfulness


The total amount for one prize on each value is 900rs


x + y + z = 900 …(i)


Now school A is giving prize in sincerity to 3 students, prize in truthfulness to 2 students and prize for helpfulness to 1 student and total award money is 1600rs


3x + 2y + z = 1600 …(ii)


School B is giving prize in sincerity to 4 students, prize in truthfulness to 1 student and prize for helpfulness to 3 student and total award money is 2300rs and prize for value remains the same Rs x, Rs y and Rs z respectively


4x + y + 3z = 2300 …(iii)


Representing equation (i), (ii) and (iii) in matrix



AX = B


Multiply by A-1 from left


A-1AX = A-1B


IX = A-1B


X = A-1B …(b)


Now solution for X will exist only if A-1 exists and A-1 will exist only if |A| ≠ 0


Let us verify if |A| ≠ 0



|A| = 5 – 5 – 5


|A| = -5 ≠ 0


Hence A-1 exist and hence solution for equation X = A-1B exist


Let us find A-1


Now A-1 is given by



Now adjoint(A) = CT where C is the cofactor matrix and CT is the transpose of the cofactor matrix


The cofactor is given by


Cij = (-1)i+jMij …(a)


Where M represents minor


Mij is the determinant of matrix leaving the ith row and jth column


The cofactor matrix C will be



Now transpose of C that is CT


For CT we will interchange the rows and columns



Using (a)



From matrix A the minors are


M11 = (3)(2) – 1 = 5


M12 = (3)(3) – 4 = 5


M13 = 3 – (4)(2) = -5


M21 = 3 – 1 = 2


M22 = 3 – 4 = -1


M23 = 1 – 4 = -3


M31 = 1 – 2 = -1


M32 = 1 – 3 = -2


M33 = 2 – 3 = -1


Hence CT will become



Hence


Using (iv)



Using (b)







Hence x = 200rs, y = 300rs and z = 400rs.


Hence for sincerity, the award money is 200rs, for truthfulness it is 300rs and for helpfulness is 400rs.


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