Question

A company produces three products every day. Their production on a certain day is 45 tons. It is found that the production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product. Determine the production level of each product using matrix method.

Answer 1

Let x, y and z be the production level of the first, second and third product, respectively .

According to the question,
\[x + y + z = 45 . . . (1)\]
\[ - x + z = 8 . . . (2)\]
\[x + z = 2y \left( \text{ Since the production of first and third product is twice the production of second product }\right)\]
\[x - 2y + z = 0 . . . (3)\]
The given system of equation can be written in matrix form as follows:
\[ \begin{bmatrix}1 & 1 & 1 \\ - 1 & 0 & 1 \\ 1 & - 2 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}45 \\ 8 \\ 0\end{bmatrix}\]
\[AX = B\]
\[A = \begin{bmatrix}1 & 1 & 1 \\ - 1 & 0 & 1 \\ 1 & - 2 & 1\end{bmatrix} X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} B = \begin{bmatrix}45 \\ 8 \\ 0\end{bmatrix}\]
Now,
\[\left| A \right|=1 \left( - 0 + 2 \right) - 1\left( - 1 - 1 \right) + 1\left( 2 - 0 \right)\]
\[ = 2 + 2 + 2\]
\[ = 6\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & 1 \\ - 2 & 1\end{vmatrix} = 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}- 1 & 1 \\ 1 & 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}- 1 & 0 \\ 1 & - 2\end{vmatrix} = 2\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ - 2 & 1\end{vmatrix} = - 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = 3\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 0 & 1\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ - 1 & 1\end{vmatrix} = - 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ - 1 & 0\end{vmatrix} = 1\]
\[adj A = \begin{bmatrix}2 & 2 & 2 \\ - 3 & 0 & 3 \\ 1 & - 2 & 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{6}\begin{bmatrix}2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{6}\begin{bmatrix}2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1\end{bmatrix}\begin{bmatrix}45 \\ 8 \\ 0\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{6}\begin{bmatrix}90 - 24 + 0 \\ 90 + 0 + 0 \\ 90 + 24 + 0\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}66 \\ 90 \\ 114\end{bmatrix}\]
\[ \]
\[ \therefore x = 11, y = 15\text{ and }z = 19\]
Thus, the production level of first, second and third product is 11, 15 and 19, respectively .