Solution:
(a) Expressing above information in Equation format:
For A: 3x+5y-4z=6000
For B: 2x-3y+z=5000
For C: - x+4y+6z=13000
These Equation can be written as AX = B
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,
\[\left| A \right|=3 \left( - 18 - 4 \right) - 5\left( 12 + 1 \right) - 4\left(
8 - 3 \right)\]
\[ = - 66 - 65 - 20\]
\[ = - 151 \neq 0\]
\[\text{ So, }A^{- 1}\text{ exists .} \]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a
}}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 3 & 1 \\ 4 &
6\end{vmatrix} = - 22, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2
& 1 \\ - 1 & 6\end{vmatrix} = - 13, C_{13} = \left( - 1 \right)^{1 + 3}
\begin{vmatrix}2 & - 3 \\ - 1 & 4\end{vmatrix} = 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}5 & - 4 \\ 4 &
6\end{vmatrix} = - 46, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3
& - 4 \\ - 1 & 6\end{vmatrix} = 14, C_{23} = \left( - 1 \right)^{2 + 3}
\begin{vmatrix}3 & 5 \\ - 1 & 4\end{vmatrix} = - 17\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}5 & - 4 \\ - 3 &
1\end{vmatrix} = - 7, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3
& - 4 \\ 2 & 1\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3}
\begin{vmatrix}3 & 5 \\ 2 & - 3\end{vmatrix} = - 19\]
\[adj A = \begin{bmatrix}- 22 & - 13 & 5 \\ - 46 & 14 & - 17 \\
- 7 & - 11 & - 19\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5
& - 17 & - 19\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14
& - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ -
13 & 14 & - 11 \\ 5 & - 17 & -
19\end{bmatrix}\begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 453000 \\ - 151000 \\ - 302000\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3000 \\
1000 \\ 2000\end{bmatrix}\]
\[ \therefore x = 3000, y = 1000\text{ and }z = 2000\]
(b)Thus, the prices of the three commodities A, B and C are Rs
3000, Rs 1000 and Rs 2000 per unit, respectively .