 # A company produces three products every day.

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.

Solution:

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

According to the question,

$x+y+z=45 \quad \ldots(1)$

$-x+z=8$                ....(2)

$x+z=2 y$      (Since the production of first and third product is twice the production of second product)

$x-2 y+z=0$           ....(3)

The given system of equation can be written in matrix form as follows:

$\left[\begin{array}{ccc}1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}45 \\ 8 \\ 0\end{array}\right]$

$A X=B$

$A=\left[\begin{array}{ccc}1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1\end{array}\right] X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] B=\left[\begin{array}{c}45 \\ 8 \\ 0\end{array}\right]$

Now,

$|A|=1(-0+2)-1(-1-1)+1(2-0)$

$=2+2+2$

$=6$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}0 & 1 \\ -2 & 1\end{array}\right|=2, C_{12}=(-1)^{1+2}\left|\begin{array}{cc}-1 & 1 \\ 1 & 1\end{array}\right|=2, C_{13}=(-1)^{1+3}\left|\begin{array}{cc}-1 & 0 \\ 1 & -2\end{array}\right|=2$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right|=-3, C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|=0, C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & 1 \\ 1 & -2\end{array}\right|=3$

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right|=1, C_{32}=(-1)^{3+2}\left|\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right|=-2, C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right|=1$

$\operatorname{adj} A=\left[\begin{array}{ccc}2 & 2 & 2 \\ -3 & 0 & 3 \\ 1 & -2 & 1\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{6}\left[\begin{array}{ccc}2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow X=\frac{1}{6}\left[\begin{array}{ccc}2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1\end{array}\right]\left[\begin{array}{c}45 \\ 8 \\ 0\end{array}\right]$

$\Rightarrow X=\frac{1}{6}\left[\begin{array}{c}90-24+0 \\ 90+0+0 \\ 90+24+0\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}66 \\ 90 \\ 114\end{array}\right]$

$\therefore x=11, y=15$ and $z=19$

Thus, the production level of first, second and third product is 11,15 and 19 , respectively.