Question

Two schools A and B want to award their selected students on the values of their sincerity, truthfulness and helpfulness. The 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.

Answer 1

Hint: Let A be a invertible square matrix (invertible if its determinant value is a non-zero number), then its inverse matrix \[{{A}^{-1}}\] will be equal to the adjoint matrix of A
(which is transpose of cofactor matrix) divided by its determinant value, i.e., \[{{A}^{-1}}=\dfrac{1}{|A|}(adj(A))\]. Let X be a variable matrix, then for AX=B where B is also a matrix, we can find the value of variables of matrix X by the equation \[\Rightarrow X={{A}^{-1}}B\].

Complete step-by-step solution:
Let the numbers x, y and z be the prize amount per person for sincerity, truthfulness and helpfulness respectively.
As per the given question, we can write
3x + 2y + z = 1600
4x + y + 3z = 2300
x + y + z = 900
From these three equations, we can write in matrix form
\[\left[ \begin{matrix}
   3 & 2 & 1 \\
   4 & 1 & 3 \\
   1 & 1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   1600 \\
   2300 \\
   900 \\
\end{matrix} \right]\]
i.e., AX=B
hence, we get the solution from \[\Rightarrow X={{A}^{-1}}B\].
Now let’s find the determinant of A,
|A|=3(1 – 3) – 2(4 – 3) + 1(4 – 1)
= - 6 – 2 + 3 = -5
\[\Rightarrow |A|=-5\]
Now, we find the cofactors:
\[{{C}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix}
   1 & 3 \\
   1 & 1 \\
\end{matrix} \right|=1-3=-2\]
\[{{C}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix}
   4 & 3 \\
   1 & 1 \\
\end{matrix} \right|=-(4-3)=-1\]
\[{{C}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix}
   4 & 1 \\
   1 & 1 \\
\end{matrix} \right|=(4-1)=3\]
\[{{C}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix}
   2 & 1 \\
   1 & 1 \\
\end{matrix} \right|=-(2-1)=-1\]
\[{{C}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix}
   3 & 1 \\
   1 & 1 \\
\end{matrix} \right|=(3-1)=2\]
\[{{C}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right|=-(3-2)=-1\]
\[{{C}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix}
   2 & 1 \\
   1 & 3 \\
\end{matrix} \right|=(6-1)=5\]
\[{{C}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix}
   3 & 1 \\
   4 & 3 \\
\end{matrix} \right|=-(9-4)=-5\]
\[{{C}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix}
   3 & 2 \\
   4 & 1 \\
\end{matrix} \right|=3-8=-5\]
Adj(A) = \[{{\left[ \begin{matrix}
   -2 & -1 & 3 \\
   -1 & 2 & -1 \\
   5 & -5 & -5 \\
\end{matrix} \right]}^{T}}\]
=\[\left[ \begin{matrix}
   -2 & -1 & 5 \\
   -1 & 2 & -5 \\
   3 & -1 & -5 \\
\end{matrix} \right]\]
Then we write,
\[X={{A}^{-1}}B=\dfrac{1}{|A|}(adj(A))B\]
\[\begin{align}
  & X=-\dfrac{1}{5}\left[ \begin{matrix}
   -2 & -1 & 5 \\
   -1 & 2 & -5 \\
   3 & -1 & -5 \\
\end{matrix} \right]\left[ \begin{matrix}
   1600 \\
   2300 \\
   900 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   640+460-900 \\
   -320-920+900 \\
   -960+460+900 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   200 \\
   300 \\
   400 \\
\end{matrix} \right] \\
\end{align}\]
Hence, x=200, y=300 and z=400 are the values of the amount awarded.

Note: Here the word value means excellence in extracurricular activities should be another value considered for an award. While solving such problems, we generally make mistakes in considering \[{{(-1)}^{i+j}}\] in calculation of cofactors. We must make sure that we take the transpose of the cofactor matrix.