Answer
See answers below
Work Step by Step
a) $\begin{bmatrix}
4&7\\
-2& 5
\end{bmatrix} \approx^1\begin{bmatrix}
1&\frac{7}{4}\\
-2& 5
\end{bmatrix} \approx^2\begin{bmatrix}
1&\frac{7}{4}\\
0& \frac{17}{2}
\end{bmatrix} \approx^3 \begin{bmatrix}
1&\frac{7}{4}\\
0& 1
\end{bmatrix}$
$1.M_{1}(\frac{1}{4})$
$2.A_{12}(2)$
$3.M_2(\frac{2}{17})$
b) $rank(A)=2$
c) Use the GaussJordan Technique to determine the inverse of A:
$\begin{bmatrix}
4&7 | 1 & 0\\
-2& 5 | 0 & 1
\end{bmatrix} \approx^1\begin{bmatrix}
1&\frac{7}{4} | \frac{1}{4} & 0\\
-2& 5 | 0 & 1
\end{bmatrix} \approx^2\begin{bmatrix}
1&\frac{7}{4} | \frac{1}{4} & 0\\
0& \frac{17}{2} | \frac{1}{2} & 1
\end{bmatrix} \approx^3 \begin{bmatrix}
1&\frac{7}{4} | \frac{1}{4} & 0\\
0& 1 | \frac{1}{17} & \frac{2}{17}
\end{bmatrix} \approx^4 \begin{bmatrix}
1&0 | \frac{5}{34} & -\frac{7}{34}\\
0& 1 | \frac{1}{17} & \frac{2}{17}
\end{bmatrix}$
$1.M_{1}(\frac{1}{4})$
$2.A_{12}(2)$
$3.M_2(\frac{2}{17})$
$4.A_{21}(-\frac{7}{4})$
Hence here, $A^{-1}=\begin{bmatrix}
\frac{5}{4}&-\frac{7}{34}\\
\frac{1}{17}& \frac{2}{17}
\end{bmatrix}$
