Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award rs x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with total award money of ₹ 1,600. School B wants to spend ₹ 2,300 to award its 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 ₹ 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for the award.(CBSE 2014)

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) = 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+jM_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

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 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.