91影视

The demand vector for the industry $I_i$is represented by the vector$\stackrel{鈫�}{{蠀}_i}$.

The vector$\stackrel{鈫�}{{蠀}_2}$ represents the manufacturing industry and the vector $\stackrel{鈫�}{{蠀}_3}$represents the energy industry.

As manufacturing and the energy industries does not make any kind of demand on agriculture industry, thus, the vectors of manufacturing and the energy industries equal zero.

The vector values of the agriculture, manufacturing and the energy industries are,

$\begin{array}{l}\stackrel{鈫�}{b}=\left[\begin{array}{c}13.2\\ 17.6\\ 1.8\end{array}\right],\stackrel{鈫�}{{蠀}_1}=\left[\begin{array}{c}0.293\\ 0.014\\ 0.044\end{array}\right],\\ \stackrel{鈫�}{{蠀}_2}=\left[\begin{array}{c}0\\ 0.207\\ 0.01\end{array}\right],\stackrel{鈫�}{{蠀}_3}=\left[\begin{array}{c}0\\ 0.017\\ 0.216\end{array}\right]\end{array}$

The demand function is written as,$\stackrel{鈫�}{b}=x_1\stackrel{鈫�}{{蠀}_1}+x_2\stackrel{鈫�}{{蠀}_2}+x_3\stackrel{鈫�}{{蠀}_3}$$\stackrel{鈫�}{b}=x_1\stackrel{鈫�}{{蠀}_1}+x_2\stackrel{鈫�}{{蠀}_2}+x_3\stackrel{鈫�}{{蠀}_3}$.

role="math" localid="1659671682712" $\begin{array}{c}\left[\begin{array}{c}13.2\\ 17.6\\ 1.8\end{array}\right]=x_1\left[\begin{array}{c}0.293\\ 0.014\\ 0.044\end{array}\right]+x_2\left[\begin{array}{c}0\\ 0.207\\ 0.01\end{array}\right]+x_3\left[\begin{array}{c}0\\ 0.017\\ 0.216\end{array}\right]\\ \left[\begin{array}{c}13.2\\ 17.6\\ 1.8\end{array}\right]=\left[\begin{array}{ccc}0.293& 0& 0\\ 0.014& 0.207& 0.017\\ 0.044& 0.01& 0.216\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]\end{array}$

Therefore, the matrix form is,

$\left[\begin{array}{cccc}0.293& 0& 0& 13.2\\ 0.014& 0.207& 0.017& 17.6\\ 0.044& 0.01& 0.216& 1.8\end{array}\right]$

Consider the matrix.

$\left[\begin{array}{cccc}0.293& 0& 0& 13.2\\ 0.014& 0.207& 0.017& 17.6\\ 0.044& 0.01& 0.216& 1.8\end{array}\right]$

Using row Echelon form to reduce the matrix.

$\left[\begin{array}{cccc}1& 0& 0& 45.05\\ 0& 1& 0& 81.95\\ 0& 0& 1& 鈭�4.63\end{array}\right]$

The values are: $x_1鈮�45.05,x_2鈮�81.95,x_3鈮�|鈭�4.63|$

As the output cannot be negative, thus, the values will be,

$x_1鈮�45.05,x_2鈮�81.95,x_3鈮�4.63$