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

Let Matrix D represents the number of students receiving prizes for the three categories:
X = ⎡⎢⎣xyz⎤⎥⎦ where x, y and z are rupees mentioned as it is in the question, for sincerity, truthfulness and helpfulness respectively.
E = ⎡⎢⎣16002300900⎤⎥⎦ is a matrix representing the total award money for school A, B and for one prize for each value.
We can represent the given question in matrix multiplication as:
DX = E ⇒ ∣∣∣∣321413111∣∣∣∣∣∣∣∣xyz∣∣∣∣ = ∣∣∣∣16002300900∣∣∣∣
Solution of the matrix equation exists if |D| ≠ 0 i.e., ∣∣∣∣321413111∣∣∣∣ = 3(1-3) - 2(4-3) + 1(4-1) = -6 -2 + 3 = -5
Therefore, the solution of the matrix equation is X = D⁻¹E
To find D⁻¹:
D⁻¹ = (1/|D|) adj(D)
Cofactor matrix of D = ∣∣∣∣-21-1-5-2∣∣∣∣
Transpose of cofactor matrix = Adj D ⇒ ∣∣∣∣-2-12-5-2∣∣∣∣
Therefore, D⁻¹ = (1/-5) ⎡⎢⎣-2-12-5-2⎤⎥⎦
Now, X = D⁻¹E = (1/-5) ⎡⎢⎣-2-12-5-2⎤⎥⎦ ∣∣∣∣16002300900∣∣∣∣ = ∣∣∣∣200300400∣∣∣∣
Therefore, x = 200, y = 300, z = 400.