Answer
$\sim \left[\begin{array}{ r r r r }4&\ \ -5&-1&\ \ \ \ \ 8\\0&-\frac{13}{4}&-13/4&\ \ -4\\0&\ \ \ \ 0&0&-82/13\\0&\ \ \ \ 0&0&\ \ \ \ \ \ 0\end{array}\right]$
With pivots only in the first three rows, the original matrix has columns that do not span $R^4$, by Theorem 4.
Work Step by Step
$\sim \left[\begin{array}{ r r r r }4&-5&-1&8\\3&-7&-4&2\\5&-6&-1&4\\9&\ \ 1&10&7\end{array}\right]\sim \left[\begin{array}{ r r r r }4&\ \ \ -5&-1&\ \ \ 8\\0&-13/4&-13/4&-4\\0&\ \ \ \ 1/4&1/4&-6\\0&\ \ 49/4&49/4&-11\end{array}\right]\sim \left[\begin{array}{ r r r r }4&\ \ -5&-1&\ \ \ \ \ 8\\0&-\frac{13}{4}&-13/4&\ \ -4\\0&\ \ \ \ 0&0&-82/13\\0&\ \ \ \ 0&0&\ \ \ \ \ \ 0\end{array}\right]$
With pivots only in the first three rows, the original matrix has columns that do not span $R^4$, by Theorem 4.
