Question

Two schools P and Q want to award their selected students on the values of tolerance, kindness and leadership. School P 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 2,200. School Q wants to spend Rs 3,100 to award 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each value is Rs 1,200, using matrices, find the award money for each value.

Answer 1

The information given in the question can be written as:
3x + 2y + z = 2200                ... (1)
4x + y + 3z = 3100                ... (2)
x + y + z = 1200                    ... (3)
Here,

\[A = \begin{bmatrix}3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1\end{bmatrix}\] and \[B = \begin{bmatrix}2200 \\ 3100 \\ 1200\end{bmatrix}\] 

Now,

\[\Rightarrow \left| A \right| = 3 \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ 1 & 1\end{vmatrix} + 2 \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & 3 \\ 1 & 1\end{vmatrix} + \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 1 \\ 1 & 1\end{vmatrix}\]

\[\Rightarrow \left| A \right| = 3\begin{vmatrix}1 & 3 \\ 1 & 1\end{vmatrix} - 2\begin{vmatrix}4 & 3 \\ 1 & 1\end{vmatrix} + 1\begin{vmatrix}4 & 1 \\ 1 & 1\end{vmatrix}\]

\[\Rightarrow \left| A \right| = 3\left( 1 - 3 \right) - 2\left( 4 - 3 \right) + 1\left( 4 - 1 \right) = - 6 - 2 + 3 = - 5 \neq 0\]

So, A is invertible.
Let  \[C_{ij}\] be the cofactor of \[a_{ij}\] in \[A = \left[ a_{ij} \right]\] Then,

\[C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ 1 & 1\end{vmatrix} = - 2\]

\[C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & 3 \\ 1 & 1\end{vmatrix} = - 1\]

\[C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 1 \\ 1 & 1\end{vmatrix} = 3\]

\[C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = - 1\]

\[C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 1 \\ 1 & 1\end{vmatrix} = 2\]

\[C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = - 1\]

\[C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5\]

\[C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 1 \\ 4 & 3\end{vmatrix} = - 5\]

\[C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix} = - 5\]

∴ adj A = \[\begin{bmatrix}- 2 & - 1 & 3 \\ - 1 & 2 & - 1 \\ 5 & - 5 & - 5\end{bmatrix}^T = \begin{bmatrix}- 2 & - 1 & 5 \\ - 1 & 2 & - 5 \\ 3 & - 1 & - 5\end{bmatrix}\]

i.e. \[A^{- 1} = \frac{adj A}{\left| A \right|}\]

\[\Rightarrow A^{- 1} = - \frac{1}{5}\begin{bmatrix}- 2 & - 1 & 5 \\ - 1 & 2 & - 5 \\ 3 & - 1 & - 5\end{bmatrix}\]

Thus, the solution of the system of equations is given by

\[X = A^{- 1} B = - \frac{1}{5}\begin{bmatrix}- 2 & - 1 & 5 \\ - 1 & 2 & - 5 \\ 3 & - 1 & - 5\end{bmatrix}\begin{bmatrix}2200 \\ 3100 \\ 1200\end{bmatrix}\]

\[\Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 4400 - 3100 + 6000 \\ - 2200 + 6200 - 6000 \\ 6600 - 3200 - 6000\end{bmatrix}\]

\[\Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 1500 \\ - 2000 \\ - 2500\end{bmatrix}\]

\[\Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}300 \\ 400 \\ 500\end{bmatrix}\]

Hence, the money awarded for tolerance, kindness and leadership are Rs 300, Rs 400 and Rs 500, respectively.
Here, the determinant of the matrix A is non-zero. Therefore, x, y and z will have unique solutions: x = 300, y = 400 and z = 500.