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