Reducing a matrix to reduced row echelon form without introducing fractions

Question

I've been trying to figure out how to reduce this matrix without introducing fractions in the intermediate stages but can't figure out how to do it. $$\begin{bmatrix}2&1&3\\ 0&-2&-29\\ 3&4&5\end{bmatrix}$$

Answer 1

Subtract the first from the last line. This gives a $1$. Use this $1$ to eliminate evrything else on the first column. For the second column, the same trick, subtract first to get a $1$.

Answer 2

Here's one way.

$\begin{bmatrix}2&1&3\\ 0&-2&-29\\ 3&4&5\end{bmatrix} \xrightarrow {R3-R1} \begin{bmatrix}2&1&3\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {R1-2R3} \begin{bmatrix}0&-5&-1\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {-R1} \begin{bmatrix}0&5&1\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {R1+2R2} \begin{bmatrix}0&1&-57\\ 0&-2&-29\\ 1&3&2 \end{bmatrix} \xrightarrow {R2+2R1} \begin{bmatrix}0&1&-57\\ 0&0&-143\\ 1&3&2 \end{bmatrix} \xrightarrow{-R2/143} \begin{bmatrix}0&1&-57\\ 0&0&1\\ 1&3&2 \end{bmatrix} \xrightarrow{R1+57R2} \begin{bmatrix}0&1&0\\ 0&0&1\\ 1&3&2 \end{bmatrix} \xrightarrow[R3-2R2]{R3-3R1} \begin{bmatrix}0&1&0\\ 0&0&1\\ 1&0&0 \end{bmatrix} \xrightarrow[R2\leftrightarrow R3]{R2\leftrightarrow R3} \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} $