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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