Answer
$E_1=\begin{bmatrix}
0&1\\
1 &0
\end{bmatrix}, E_2=\begin{bmatrix}
1&0\\
-2 &1
\end{bmatrix}, E_3=\begin{bmatrix}
1&0\\
0 &-\frac{1}{7}
\end{bmatrix}, E_4=\begin{bmatrix}
1&-3\\
0 &1
\end{bmatrix}$
Work Step by Step
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
2 & -1\\
1&3
\end{bmatrix} \approx^1\begin{bmatrix}
1&3\\
2 & -1
\end{bmatrix} \approx^2\begin{bmatrix}
1&3\\
0 & 7
\end{bmatrix} \approx^3\begin{bmatrix}
1&3\\
0& 1
\end{bmatrix} \approx^4\begin{bmatrix}
1&0\\
0& 1
\end{bmatrix}$
$1.P_{12}(-3)$
$3.A_{12}(-2)$
$4.M_{2}(-\frac{1}{7})$
$5.A_{21}(-3)$
So, elementary matrices are $P_{12}(-3),A_{12}(-2),M_{2}(-\frac{1}{7}),A_{21}(-3)$
or $E_1=\begin{bmatrix}
0&1\\
1 &0
\end{bmatrix}, E_2=\begin{bmatrix}
1&0\\
-2 &1
\end{bmatrix}, E_3=\begin{bmatrix}
1&0\\
0 &-\frac{1}{7}
\end{bmatrix}, E_4=\begin{bmatrix}
1&-3\\
0 &1
\end{bmatrix}$
To verify that $E_1E_2E_3E_4A=I_2$
$\rightarrow A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}=\begin{bmatrix}
1 & -3\\
0 &1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0 &-\frac{1}{7}
\end{bmatrix} \begin{bmatrix}
1 &0\\
-2 &1
\end{bmatrix} \begin{bmatrix}
0& 1\\
1 &0
\end{bmatrix}\begin{bmatrix}
2 & -1\\
1 &3
\end{bmatrix} $
$=\begin{bmatrix}
1 & 0\\
0 &1
\end{bmatrix} $
