# 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?

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Pambos thank you for editing ;)) – Yigit Can Nov 23 '12 at 13:00
The usual name for the process you need is Gaussian elimination. Chances are good that your textbook will have a longish explanation of it. – Henning Makholm Nov 23 '12 at 13:18
I know how I can use Gaussian elimination but how can I associate Gaussian elimination for finding P? – Yigit Can Nov 23 '12 at 13:23

This another approach, however; Don's way explains it to you in a brief solid form. $A =\begin{pmatrix} 1& 2& 1& 0\\ -1 &0 &3 &5\\ 1& -2& 1& 1\end{pmatrix}\xrightarrow{R_1+R_2\mapsto R_2}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\ 1& -2& 1& 1\end{pmatrix} \xrightarrow{-R_1+R_3\mapsto R_3}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& -4& 0& 1\end{pmatrix}\\\xrightarrow{-R_2+R_1\mapsto R_2}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\ 0& -4& 0& 1\end{pmatrix}\xrightarrow{2R_2+R_3\mapsto R_3}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\ 0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{-1}{2}R_3+R_2\mapsto R_2}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &0 &-0.5\\ 0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{3}{8}R_3+R_1\mapsto R_1}\begin{pmatrix} 1& 0& 0& \frac{-13}{8}\\ 0 &2 &0 &-0.5\\ 0& 0& 8& 11\end{pmatrix}$ and the last matrix is the row-reduced echelon matrix of $A$.

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ah thank you, as we talked above I'm going to apply these elementary row operations to the I matrix and get P1 from them – Yigit Can Nov 23 '12 at 14:01

$$A =\begin{pmatrix} \;1& 2& 1& 0\\ \!\!-1 &0 &3 &5\\ \;1& \!\!-2& 1& 1\end{pmatrix}\stackrel{R_2+R_1\,,\,R_3-R1}\longrightarrow\begin{pmatrix} \;1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& \!\!-4& 0& 1\end{pmatrix}\stackrel{R_3+2R_2}\longrightarrow \begin{pmatrix} \;1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& 0& 8& 11\end{pmatrix}$$

Now, for example: to add 1st. row to 2nd one and to substract 1st. one from 3rd. one is to multiply $\,A\,$ from the left by matrix

$$P_1:=\begin{pmatrix}1&0&0\\1&1&0\\\!\!\!-1&0&1\end{pmatrix}$$

Take it from here, and show that each step you get an $invertible$ $\,3\times 3\,$ matrix, so at the end $\,P\,$ is a product of these matrices and, thus, an invertible one.

$$P_1A=\begin{pmatrix} \;1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& \!\!-4& 0& 1\end{pmatrix}$$
Now you try to generalize the above to ge the matrix $\,P_2\,$ s.t. $\,P_2P_1A\,$ gives us the rightmost matrix above (the already reduced form)