Name: Solution of Two Simultaneous Linear Equations
Uploaded: 2019-07-19
Duration: 9 min 55 s
Description: Two linear equations in 2 variables taken together are called simultaneous linear equations. The solution to these equations is the ordered pair which satisf...

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.

Accepted Answer

Given that 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.
According to the question, $x + y + z = 45 . . . (1)$ , $- x + z = 8 . . . (2)$ , $x - 2 y + z = 0 . . . (3)$

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

$[\begin{matrix} 1 & 1 & 1 \\ - 1 & 0 & 1 \\ 1 & - 2 & 1 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 45 \\ 8 \\ 0 \end{matrix}]$

$A X = B$

$A = [\begin{matrix} 1 & 1 & 1 \\ - 1 & 0 & 1 \\ 1 & - 2 & 1 \end{matrix}] X = [\begin{matrix} x \\ y \\ z \end{matrix}] B = [\begin{matrix} 45 \\ 8 \\ 0 \end{matrix}]$

Now,

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

$= 2 + 2 + 2$

$= 6$

Let $C_{i j}$ be the cofactors of the elements $a_{i j}$ in $A (a_{i j})$ . Then,

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

$C_{21} = {(- 1)}^{2 + 1} |\begin{matrix} 1 & 1 \\ - 2 & 1 \end{matrix}| = - 3, C_{22} = {(- 1)}^{2 + 2} |\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}| = 0, C_{23} = {(- 1)}^{2 + 3} |\begin{matrix} 1 & 1 \\ 1 & - 2 \end{matrix}| = 3$

$C_{31} = {(- 1)}^{3 + 1} |\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}| = 1, C_{32} = {(- 1)}^{3 + 2} |\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix}| = - 2, C_{33} = {(- 1)}^{3 + 3} |\begin{matrix} 1 & 1 \\ - 1 & 0 \end{matrix}| = 1$

$a d j A = {[\begin{matrix} 2 & 2 & 2 \\ - 3 & 0 & 3 \\ 1 & - 2 & 1 \end{matrix}]}^{T}$

$= [\begin{matrix} 2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1 \end{matrix}]$

$\Rightarrow A^{- 1} = \frac{1}{|A|} a d j A$

$= \frac{1}{6} [\begin{matrix} 2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1 \end{matrix}]$

$X = A^{- 1} B$

$\Rightarrow X = \frac{1}{6} [\begin{matrix} 2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1 \end{matrix}] [\begin{matrix} 45 \\ 8 \\ 0 \end{matrix}]$

$\Rightarrow X = \frac{1}{6} [\begin{matrix} 90 - 24 + 0 \\ 90 + 0 + 0 \\ 90 + 24 + 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{6} [\begin{matrix} 66 \\ 90 \\ 114 \end{matrix}]$

$\Rightarrow [\begin{matrix} 11 \\ 15 \\ 19 \end{matrix}]$

$∴ x = 11, y = 15$ and $z = 19$

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

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