# The system of equations: \begin{cases} x+2y-z=-3, \\ x+z = -7, \\ 2x-y-z = -8 \end {cases} has a...

## Question:

The system of equations:

$$\begin{cases} x+2y-z=-3, \\ x+z = -7, \\ 2x-y-z = -8 \end {cases}$$

has a unique solution. Find he solution using Gaussian elimination method or Gauss-Jordan elimination method

## Solution of System of Equations:

Given a system of equations, it is possible to obtain its solution by applying the Gauss - Jordan method. This method is based on operations on the rows and columns in order to obtain an extended identity matrix with the system solution.

We use operations on row and columns. The operation is shown in each step.

{eq}\left(\left.\begin{matrix} 1 & 2 & -1 \\ 1 & 0 & 1 \\ 2 & -1 & -1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -3 \\ -7 \\ -8 \end{matrix}\right) {/eq}

{eq}R_2\rightarrow R_2 - R_1 \\ R_3\rightarrow R_3 - 2 R_1 {/eq}

{eq}\left(\left.\begin{matrix} 1 & 2 & -1 \\ 0 & -2 & 2 \\ 0 & -5 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -3 \\ -4 \\ -2 \end{matrix}\right) {/eq}

We normalize the second row:

{eq}R_2\rightarrow \frac{R_2}{ -2 } {/eq}

{eq}\left(\left.\begin{matrix} 1 & 2 & -1 \\ 0 & 1 & -1 \\ 0 & -5 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -3 \\ 2 \\ -2 \end{matrix}\right) {/eq}

{eq}R_1\rightarrow R_1 - 2 R_2 \\ R_3\rightarrow R_3 + 5 R_2 {/eq}

{eq}\left(\left.\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & -4 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -7 \\ 2 \\ 8 \end{matrix}\right) {/eq}

We normalize the third row:

{eq}R_3\rightarrow \frac{R_3}{ -4 } {/eq}

{eq}\left(\left.\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -7 \\ 2 \\ -2 \end{matrix}\right) {/eq}

{eq}R_2\rightarrow R_1 - R_3 \\ R_2\rightarrow R_2 + R_3 {/eq}

{eq}\left(\left.\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -5 \\ 0 \\ -2 \end{matrix}\right) {/eq}

The solution of the system is:

{eq}x=-5\\ y=0\\ z=-2 {/eq}