<s>[INST] Finding a row-reduced echelon matrix. $A =\begin{pmatrix} 1& 2& 1& 0\\ -1 &0 &3 &5\\  1& -2& 1& 1\end{pmatrix}$.
I should find a row-reduced echelon matrix $R$ which is row equivalent to $A$ and an invertible $3 \times 3$ matrix $P$ such that $R = PA$.
I know that if a matrix is row equivalent to another that means that we can obtain such a matrix by using elementary row operations. But how should i apply this to such question? 
 [/INST] This another approach, however; Don's way explains it to you in a brief solid form.
$A =\begin{pmatrix} 1& 2& 1& 0\\ -1 &0 &3 &5\\  1& -2& 1& 1\end{pmatrix}\xrightarrow{R_1+R_2\mapsto R_2}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\  1& -2& 1& 1\end{pmatrix}
\xrightarrow{-R_1+R_3\mapsto R_3}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\  0& -4& 0& 1\end{pmatrix}\\\xrightarrow{-R_2+R_1\mapsto R_1}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\  0& -4& 0& 1\end{pmatrix}\xrightarrow{2R_2+R_3\mapsto R_3}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\  0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{-1}{2}R_3+R_2\mapsto R_2}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &0 &-0.5\\  0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{3}{8}R_3+R_1\mapsto R_1}\begin{pmatrix} 1& 0& 0& \frac{-13}{8}\\ 0 &2 &0 &-0.5\\  0& 0& 8& 11\end{pmatrix}$ 
and the last matrix is the row-reduced echelon form of $A$.
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