Row Operations to solve system using Gaussian elimination

[Courtney.Hampton9]

From this 2006 thread:

Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.

x+y+2=9
2x-3y+4z=7
x-4y+3x=-2

The first equation can be simplifed to get x + y = 7.

So then we have: \(\displaystyle \,\begin{vmatrix}1 & 1 & 0 & | & 7 \\ 2 & -3 & 4 & | & 7\\ 1 & -4 & 3 & | & -2 \end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ R_2-2\cdot R_1 \longrightarrow \\ R_3-R_1 \longrightarrow \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 4 & | & -7\\ 0 & -5 & 3 & | & -9\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ \\ R_3-R_2 \longrightarrow \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 4 & | & -7 \\ 0 & 0 & -1 & | & -2\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ \\ -R_3 \longrightarrow \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 4 & | & -7 \\ 0 & 0 & 1 & | & 2\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ R_2-4\cdot R_3 \longrightarrow \\ \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 0 & | & -15 \\ 0 & 0 & 1 & | & 2\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ -5\cdot R_2 \longrightarrow \\ \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & | & 2\end{vmatrix}\)

Where does this lead?

I know a little about this, but it still stumps me in some parts.

 

[stapel]

From this 2006 thread:

Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.

x+y+2=9
2x-3y+4z=7
x-4y+3x=-2

The first equation can be simplifed to get x + y = 7.

So then we have: \(\displaystyle \,\begin{vmatrix}1 & 1 & 0 & | & 7 \\ 2 & -3 & 4 & | & 7\\ 1 & -4 & 3 & | & -2 \end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ R_2-2\cdot R_1 \longrightarrow \\ R_3-R_1 \longrightarrow \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 4 & | & -7\\ 0 & -5 & 3 & | & -9\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ \\ R_3-R_2 \longrightarrow \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 4 & | & -7 \\ 0 & 0 & -1 & | & -2\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ \\ -R_3 \longrightarrow \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 4 & | & -7 \\ 0 & 0 & 1 & | & 2\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ R_2-4\cdot R_3 \longrightarrow \\ \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & -5 & 0 & | & -15 \\ 0 & 0 & 1 & | & 2\end{vmatrix}\)

\(\displaystyle \begin{array}{rrrr} \\ -5\cdot R_2 \longrightarrow \\ \\ \end{array}\,\begin{vmatrix}1 & 1 & 0 & | & 7\\ 0 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & | & 2\end{vmatrix}\)

Where does this lead?

I know a little about this, but it still stumps me in some parts.

Okay... Did you have a specific question, or are you requesting lesson instruction? Please be specific. Thank you!