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प्रश्न
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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उत्तर
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,
\[\left| A \right|=3 \left( - 18 - 4 \right) - 5\left( 12 + 1 \right) - 4\left( 8 - 3 \right)\]
\[ = - 66 - 65 - 20\]
\[ = - 151 \neq 0\]
\[\text{ So, }A^{- 1}\text{ exists .} \]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 3 & 1 \\ 4 & 6\end{vmatrix} = - 22, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ - 1 & 6\end{vmatrix} = - 13, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 3 \\ - 1 & 4\end{vmatrix} = 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}5 & - 4 \\ 4 & 6\end{vmatrix} = - 46, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 4 \\ - 1 & 6\end{vmatrix} = 14, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 5 \\ - 1 & 4\end{vmatrix} = - 17\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}5 & - 4 \\ - 3 & 1\end{vmatrix} = - 7, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 4 \\ 2 & 1\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 5 \\ 2 & - 3\end{vmatrix} = - 19\]
\[adj A = \begin{bmatrix}- 22 & - 13 & 5 \\ - 46 & 14 & - 17 \\ - 7 & - 11 & - 19\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 453000 \\ - 151000 \\ - 302000\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3000 \\ 1000 \\ 2000\end{bmatrix}\]
\[ \therefore x = 3000, y = 1000\text{ and }z = 2000\]
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