Answer
$\begin{bmatrix}
1 &\frac{1}{4}\\
0 & 1\\
0 & 0
\end{bmatrix}$
Work Step by Step
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
-4 & -1\\
0& 3\\
-3 & 7
\end{bmatrix} \approx^1 \begin{bmatrix}
1 & \frac{1}{4}\\
0& 3\\
-3 & 7
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & \frac{1}{4}\\
0& 3\\
0 & \frac{31}{4}
\end{bmatrix} \approx^3 \begin{bmatrix}
1 & \frac{1}{4}\\
0& 1\\
0 & \frac{31}{4}
\end{bmatrix} \approx^4 \begin{bmatrix}
1 & \frac{1}{4}\\
0& 1\\
0 & 0
\end{bmatrix}$
$1. M_1(- \frac{1}{4})$
$2.A_{13}(3)$
$3.M_2( \frac{1}{3})$
$4. A_{23}(\frac{-31}{4})$
So, elementary matrices are $A_{13}(3),M_2( \frac{1}{3}),A_{13}(3),A_{23}(\frac{-31}{4})=\begin{bmatrix}
1 &\frac{1}{4}\\
0 & 1\\
0 & 0
\end{bmatrix}$
