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Question
The prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. Cpurchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.
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Solution
The prices of three commodities P, Q and R are Rs x, Rs y and Rs z per unit, respectively .
According to the question,
\[3x + 5y - 4z = 6000\]
\[2x - 3y + z = 5000\]
\[ - x + 4y + 6z = 13000\]
The given system of equations can be written in matrix form as follows:
\[ \begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[AX = B\]
Here,
\[A = \begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix} X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} B = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
Now,
\[\left| A \right|=3 \left( - 18 - 4 \right) - 5\left( 12 + 1 \right) - 4\left( 8 - 3 \right)\]
\[ = - 66 - 65 - 20\]
\[ = - 151 \neq 0\]
\[\text{ So, }A^{- 1}\text{ exists .} \]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 3 & 1 \\ 4 & 6\end{vmatrix} = - 22, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ - 1 & 6\end{vmatrix} = - 13, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 3 \\ - 1 & 4\end{vmatrix} = 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}5 & - 4 \\ 4 & 6\end{vmatrix} = - 46, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 4 \\ - 1 & 6\end{vmatrix} = 14, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 5 \\ - 1 & 4\end{vmatrix} = - 17\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}5 & - 4 \\ - 3 & 1\end{vmatrix} = - 7, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 4 \\ 2 & 1\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 5 \\ 2 & - 3\end{vmatrix} = - 19\]
\[adj A = \begin{bmatrix}- 22 & - 13 & 5 \\ - 46 & 14 & - 17 \\ - 7 & - 11 & - 19\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 453000 \\ - 151000 \\ - 302000\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3000 \\ 1000 \\ 2000\end{bmatrix}\]
\[ \therefore x = 3000, y = 1000\text{ and }z = 2000\]
According to the question,
\[3x + 5y - 4z = 6000\]
\[2x - 3y + z = 5000\]
\[ - x + 4y + 6z = 13000\]
The given system of equations can be written in matrix form as follows:
\[ \begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[AX = B\]
Here,
\[A = \begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix} X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} B = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
Now,
\[\left| A \right|=3 \left( - 18 - 4 \right) - 5\left( 12 + 1 \right) - 4\left( 8 - 3 \right)\]
\[ = - 66 - 65 - 20\]
\[ = - 151 \neq 0\]
\[\text{ So, }A^{- 1}\text{ exists .} \]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 3 & 1 \\ 4 & 6\end{vmatrix} = - 22, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ - 1 & 6\end{vmatrix} = - 13, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 3 \\ - 1 & 4\end{vmatrix} = 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}5 & - 4 \\ 4 & 6\end{vmatrix} = - 46, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 4 \\ - 1 & 6\end{vmatrix} = 14, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 5 \\ - 1 & 4\end{vmatrix} = - 17\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}5 & - 4 \\ - 3 & 1\end{vmatrix} = - 7, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 4 \\ 2 & 1\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 5 \\ 2 & - 3\end{vmatrix} = - 19\]
\[adj A = \begin{bmatrix}- 22 & - 13 & 5 \\ - 46 & 14 & - 17 \\ - 7 & - 11 & - 19\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 453000 \\ - 151000 \\ - 302000\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3000 \\ 1000 \\ 2000\end{bmatrix}\]
\[ \therefore x = 3000, y = 1000\text{ and }z = 2000\]
Thus, the prices of the three commodities P, Q and R are Rs 3000, Rs 1000 and Rs 2000 per unit, respectively .
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