Question

Use the augmented matrix method to solve the following system of equations. Your answers may be given as decimals or fractions.

Answer 1

Answer:

$x\ =\ \dfrac{21}{5}$

$y\ =\ \dfrac{3}{5}$

z = 2

Step-by-step explanation:

Given equations are

x - 2y - z = 2

x + 3y - 2z = 4

-x + 2y + 3z = 2

from the given equations the augmented matrix can be written as

$\left[\begin{array}{ccc}1&-2&-1:2\\1&3&-2:4\\-1&2&3:2\end{array}\right]$

$R_2=>R_2-R_1\ and\ R_3=>R_3+R_1$

$=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&5&-1:2\\0&0&2:4\end{array}\right]$

$R_2=>\dfrac{R_2}{5}$

$=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\0&0&2:4\end{array}\right]$

$R_1=>R_1+2.R_2$

$=\ \left[\begin{array}{ccc}1&0&-1-\dfrac{2}{5}:2+\dfrac{4}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]$

$=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]$

$R_3=>\dfrac{R_3}{2}$

$=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&1:2\end{array}\right]$

$R_1=>R_1+\dfrac{7}{5}R_3\ and\ R_2+\dfrac{1}{5}R_3$

$=\ \left[\begin{array}{ccc}1&0&0:\dfrac{14}{5}+\dfrac{7}{5}\\\\0&1&0:\dfrac{2}{5}+\dfrac{1}{5}\\\\0&0&1:2\end{array}\right]$

So, from the above augmented matrix, we can write

$x\ =\ \dfrac{21}{5}$

$y\ =\ \dfrac{3}{5}$

z = 2