Answer
$A^{-1}=\left[\begin{array}{rrr}
-7/5 & 2\\
4/5 & -1
\end{array}\right]$
Work Step by Step
ALGORITHM FOR FINDING $A^{-1}$
Row reduce the augmented matrix $[A\ I]$.
If $A$ is row equivalent to $I$, then
$[A\ I]$ is row equivalent to $[I\ A^{-1}]$.
Otherwise, $A$ does not have an inverse.
---
$\left[\begin{array}{llll}
5 & 10 & 1 & 0\\
4 & 7 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
-R_{2}.\\
.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 3 & 1 & -1\\
4 & 7 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
-4R_{1}
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 3 & 1 & -1\\
0 & -5 & -4 & 5
\end{array}\right]\left\{\begin{array}{l}
.\\
\div(-5)
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 3 & 1 & -1\\
0 & 1 & 4/5 & -1
\end{array}\right]\left\{\begin{array}{l}
-3R_{2}.\\
.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & -7/5 & 2\\
0 & 1 & 4/5 & -1
\end{array}\right]$= $[I\ A^{-1}]$.
$A^{-1}=\left[\begin{array}{rrr}
-7/5 & 2\\
4/5 & -1
\end{array}\right]$
