K kayesal New member Joined Mar 24, 2006 Messages 12 Mar 24, 2006 #1 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
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
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Mar 24, 2006 #2 Re: Use Gaussian elimination Hello, kayesal! Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists. \(\displaystyle \begin{array}{ccc}x\,+\,y\,+\,2\;=\;9\\ 2x\,-\,3y\,+\,4z\;=\;7 \\x\,-\,4y\,+\,3x\;=\;-2\end{array}\) Click to expand... 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? Last edited by a moderator: Nov 25, 2017
Re: Use Gaussian elimination Hello, kayesal! Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists. \(\displaystyle \begin{array}{ccc}x\,+\,y\,+\,2\;=\;9\\ 2x\,-\,3y\,+\,4z\;=\;7 \\x\,-\,4y\,+\,3x\;=\;-2\end{array}\) Click to expand... 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?
