Answer
$\left[\begin{array}{ r r r r }7&2&-5&8\\0&-1.57&.429&-3.29\\0&0&4.55&-17.2\\0&0&0&0\end{array}\right],$to three significant figures. The original matrix does not have a pivot in every row, so its columns do not span R4 by Theorem 4.
Work Step by Step
$\left[\begin{array}{ r r r r }7&2&-5&8\\-5&-3&4&-9\\6&10&-2&7\\-7&9&2&15\end{array}\right]\sim \left[\begin{array}{ r r r r }7&2&-5&8\\0&-11/7&3/7&-23/7\\0&58/7&16/7&1/7\\0&11&-3&23\end{array}\right]\sim \left[\begin{array}{ r r r r }7&2&-5&8\\0&-11/7&3/7&-23/7\\0&0&50/11&-189/11\\0&0&0&0\end{array}\right]$
or, approximately$\left[\begin{array}{ r r r r }7&2&-5&8\\0&-1.57&.429&-3.29\\0&0&4.55&-17.2\\0&0&0&0\end{array}\right]$, to three significant figures. The original matrix does not have a pivot in every row, so its columns do not span $R^4$, by Theorem 4
