Answer
See answers below
Work Step by Step
a) $\begin{bmatrix}
4 & 7\\
-2 & 5
\end{bmatrix} \approx^1 \begin{bmatrix}
1 & \frac{7}{4}\\
-2 & 5
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & \frac{7}{4}\\
0 & \frac{17}{2}
\end{bmatrix} \approx^3 \begin{bmatrix}
1 & \frac{7}{4}\\
0 & 1
\end{bmatrix} \approx^4 \begin{bmatrix}
1 &0\\
0 & 1
\end{bmatrix}$
$1.M_1(\frac{1}{4})$
$2.A_{12}(2)$
$3.M_2(\frac{2}{17})$
$4.A_{21}(-\frac{7}{4})$
Since $E_4E_3E_2E_1A=I_2$
we have $A=E_4^{-1}E_3^{-1}E_2^{-1}E_1^{-1}=\begin{bmatrix}
4 & 0\\
0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
-2 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0 & \frac{2}{17}
\end{bmatrix}\begin{bmatrix}
1 & -\frac{7}{4}\\
0 & 1
\end{bmatrix}$
b) $\begin{bmatrix}
4 & 7\\
-2 & 5
\end{bmatrix} \approx^1\begin{bmatrix}
4 & 7\\
0 & \frac{17}{2}
\end{bmatrix} $
$1.A_{12}(\frac{1}{2}) \rightarrow m_{12}=-\frac{1}{2}$
$LU=\begin{bmatrix}
1 &0\\
-\frac{1}{2}& 1
\end{bmatrix}\begin{bmatrix}
4 & 7\\
0 & \frac{17}{2}
\end{bmatrix} $
