Question

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

x - 2y - z = 2

x + 3y - 2z = 4

-x + 2y + 3z = 2
x =
y =
z =

Answer 1

`x\ =\ (21)/(5)`

`y\ =\ (3)/(5)`

z = 2

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=>(R_2)/(5)`

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

`R_1=>R_1+2.R_2`

`=\ \left[\begin{array}{ccc}1&0&-1-(2)/(5):2+(4)/(5)\\\\0&1&(-1)/(5):(2)/(5)\\\\0&0&2:4\end{array}\right]`

`=\ \left[\begin{array}{ccc}1&0&(-7)/(5):(14)/(5)\\\\0&1&(-1)/(5):(2)/(5)\\\\0&0&2:4\end{array}\right]`

`R_3=>(R_3)/(2)`

`=\ \left[\begin{array}{ccc}1&0&(-7)/(5):(14)/(5)\\\\0&1&(-1)/(5):(2)/(5)\\\\0&0&1:2\end{array}\right]`

`R_1=>R_1+(7)/(5)R_3\ and\ R_2+(1)/(5)R_3`

`=\ \left[\begin{array}{ccc}1&0&0:(14)/(5)+(7)/(5)\\\\0&1&0:(2)/(5)+(1)/(5)\\\\0&0&1:2\end{array}\right]`

So, from the above augmented matrix, we can write

`x\ =\ (21)/(5)`

`y\ =\ (3)/(5)`

z = 2