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^{-1}AX = A^{-1}B

⇒ IX = A^{-1}B

⇒ X = A^{-1}B …(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^{-1}B exist

Let us find A^{-1}

Now A^{-1} is given by

Now adjoint(A) = C^{T} where C is the cofactor matrix and C^{T} is the transpose of the cofactor matrix

The cofactor is given by

⇒ C_{ij} = (-1)^{i+j}M_{ij} …(a)

Where M represents minor

M_{ij} is the determinant of matrix leaving the i^{th} row and j^{th} column

The cofactor matrix C will be

Now transpose of C that is C^{T}

For C^{T} we will interchange the rows and columns

Using (a)

From matrix A the minors are

M_{11} = (3)(2) – 1 = 5

M_{12} = (3)(3) – 4 = 5

M_{13} = 3 – (4)(2) = -5

M_{21} = 3 – 1 = 2

M_{22} = 3 – 4 = -1

M_{23} = 1 – 4 = -3

M_{31} = 1 – 2 = -1

M_{32} = 1 – 3 = -2

M_{33} = 2 – 3 = -1

Hence C^{T} 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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