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Two Schools P And Q Want to Award Their Selected Students on the Values of Discipline, Politeness and Punctuality.

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Question

Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹x each, ₹y each and ₹z each the three respectively values to its 3, 2 and 1 students with a total award money of ₹1,000. School Q wants to spend ₹1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is ₹600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.

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Solution

Let the award money given for Discipline, Politeness and Punctuality be ₹x, ₹y and ₹z respectively.

Since, the total cash award is ₹600.
∴ x + y + z = 600                    ....(1)

Award money given by school P is ₹1,000.
∴ 3x + 2y + z = 1000              ....(2)

Award money given by school Q is ₹1,500.
∴ 4x + y + 3z = 1500              ....(3)

The above system of equations can be written in matrix form AX = B as

\[\begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}600 \\ 1000 \\ 1500\end{bmatrix}\]
\[\text{ Where, }A = \begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}600 \\ 1000 \\ 1500\end{bmatrix}\]
Now, 
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{vmatrix}\]
\[ = 1\left( 6 - 1 \right) - 1\left( 9 - 4 \right) + 1(3 - 8)\]
\[ = 5 - 5 - 5\]
\[ = - 5\]
\[\text{ Let }C_{ij}\text{ be the cofactors of elements }a_{ij}\text{ in }A = \left[ a_{ij} \right] .\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 1 \\ 4 & 3\end{vmatrix} = - 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix} = - 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 4 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 4 & 1\end{vmatrix} = 3\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = - 1\]
\[adj A = \begin{bmatrix}5 & - 5 & - 5 \\ - 2 & - 1 & 3 \\ - 1 & 2 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\begin{bmatrix}600 \\ 1000 \\ 1500\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}3000 - 2000 - 1500 \\ - 3000 - 1000 + 3000 \\ - 3000 + 3000 - 1500\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 500 \\ - 1000 \\ - 1500\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 500}{- 5}, y = \frac{- 1000}{- 5}\text{ and }z = \frac{- 1500}{- 5}\]
\[ \therefore x = 100, y = 200\text{ and }z = 300 .\]

Hence, the award money for each value of Discipline, Politeness and Punctuality is ₹100, ₹200 and ₹300.
​One more value which should be considered for award is Honesty.

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Chapter 7: Solution of Simultaneous Linear Equations - Exercise 8.1 [Page 17]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 7 Solution of Simultaneous Linear Equations
Exercise 8.1 | Q 18 | Page 17
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