Question 10.1.8: Solving a System of Equations by Gauss–Jordan Elimination So...
Solving a System of Equations by Gauss–Jordan Elimination
Solve the system given in Example 4 by Gauss–Jordan elimination.
\left\{\begin{array}{rr}x-y-z= & 1 \\2 x-3 y+z= & 10 \\x+y-2 z= & 0\end{array}\right. The given system
\left[\begin{array}{rrr|r}1 & -1 & -1 & 1 \\0 & 1 & -3 & -8 \\0 & 0 & 5 & 15\end{array}\right] The final augmented matrix of the system in Example 4
\underrightarrow{\frac{1}{5} R_{3}} \left[\begin{array}{rrr|r}1 & -1 & -1 & 1 \\0 & 1 & -3 & -8 \\0 & 0 & 1 & 3\end{array}\right] The matrix is now in row-echelon form.
\underrightarrow{R_{2}+R_{1} \rightarrow R_{1}} \left[\begin{array}{rrr|r} 1 & 0 & -4 & -7 \\ 0 & 1 & -3 & -8 \\ 0 & 0 & 1 & 3 \end{array}\right] Produces only one nonzero value in column 2.
\begin {aligned} \underrightarrow{4 R_{3}+R_{1} \rightarrow R_{1}} \\ \underrightarrow{3 R_{3}+R_{2} \rightarrow R_{2}} \end{aligned} \left[\begin{array}{lll|l} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 3 \end{array}\right] Produces only one nonzero value in column 3.
We now have an equivalent matrix in reduced row-echelon form. The corresponding system of equations for the last augmented matrix is
\left\{\begin{array}{l}x=5 \\y=1 \\z=3\end{array}\right.
Therefore, the solution set is {(5, 1, 3)} as in Example 4.
