Answer:
See below
Step-by-step explanation:
[tex]\left[\begin{array}{ccc}2&-8&22\\1&7&0\end{array}\right] \longrightarrow \left[\begin{array}{ccc}1&-4&11\\1&7&0\end{array}\right][/tex]
Above we have [tex]\dfrac{1}{2}\cdot R_1 \rightarrow R_1[/tex]
Here we multiplied the first row by [tex]\dfrac{1}{2}[/tex]
Then, [tex]R_2-1\cdot R_1 \rightarrow R_2[/tex]
[tex]\left[\begin{array}{ccc}1&-4&11\\1&7&11\end{array}\right] \longrightarrow \left[\begin{array}{ccc}1&-4&11\\0&11&-11\end{array}\right][/tex]
Later, [tex]\dfrac{1}{11}\cdot R_2 \rightarrow R_2[/tex]
[tex]\left[\begin{array}{ccc}1&-4&11\\0&11&-11\end{array}\right] \longrightarrow \left[\begin{array}{ccc}1&-4&11\\0&1&-1\end{array}\right][/tex]
Finally, [tex]R_1+4\cdot R_2 \rightarrow R_1[/tex]
[tex]\left[\begin{array}{ccc}1&-4&11\\0&1&-1\end{array}\right]\longrightarrow \left[\begin{array}{ccc}1&0&7\\0&1&-1\end{array}\right][/tex]
Therefore, the complete process is
[tex]\left[\begin{array}{ccc}2&-8&22\\1&7&0\end{array}\right] \longrightarrow \left[\begin{array}{ccc}1&-4&11\\1&7&0\end{array}\right] \longrightarrow \left[\begin{array}{ccc}1&-4&11\\0&11&-11\end{array}\right][/tex]
[tex]\longrightarrow \left[\begin{array}{ccc}1&-4&11\\0&1&-1\end{array}\right] \longrightarrow \left[\begin{array}{ccc}1&0&7\\0&1&-1\end{array}\right][/tex]
