Answer
$\sim \left[\begin{array}{ r r r r r }10&\ -7&1&\ \ \ \ 4&6\\0&\ -8/5&\ \ \ -26/5&-34/5&9/5\\0&\ \ \ \ \ 0&-193/8&-193/8&49/16\\0&\ \ \ \ \ 0&0&\ \ \ \ \ 0&4715/386\end{array}\right]$
The original matrix has a pivot in every row, so its columns span $R^4$, by Theorem 4.
Work Step by Step
$\left[\begin{array}{ r r r r r }10&-7&1&4&6\\-8&4&-6&-10&-3\\-7&11&-5&-1&-8\\3&-1&10&12&12\end{array}\right]\sim \left[\begin{array}{ r r r r r }10&-7&1&4&6\\0&-8/5&-26/5&-34/5&9/5\\0&61/10&-43/10&9/5&-19/5\\0&11/10&97/10&54/5&51/5\end{array}\right]$
$\sim \left[\begin{array}{ r r r r r }10&-7&1&4&6\\0&-8/5&-26/5&-34/5&9/5\\0&0&-193/8&-193/8&49/16\\0&0&49/8&49/8&183/16\end{array}\right]\sim \left[\begin{array}{ r r r r r }10&\ -7&1&\ \ \ \ 4&6\\0&\ -8/5&\ \ \ -26/5&-34/5&9/5\\0&\ \ \ \ \ 0&-193/8&-193/8&49/16\\0&\ \ \ \ \ 0&0&\ \ \ \ \ 0&4715/386\end{array}\right]$
The original matrix has a pivot in every row, so its columns span $R^4$, by Theorem 4.
